Kinematics of Particles

1. www.ekeeda.com Contact : 9029006464 Kinematics of Particles Kinematics of Particles Email : care@ekeeda.com P INTRODUCTION In this chapter we will do motion analysis of moving particles without taking into account the forces responsible for the motion. We shall study motion of a particle and get to know terms like position, displacement, velocity, acceleration and time. These are the parameters by which we measure motion. We shall study in detail the rectilinear motion and curvilinear motion of a particle. Special cases of motion viz., motion under gravity and projectile motion form an interesting part of this chapter. Study of motion of a particle with respect to a moving frame of reference would be studied, and finally an alternate solution to rectilinear motion problems, using graphical approach will be dealt with. RECTILINEAR MOTION Motion of a particle in a straight line is known as a rectilinear motion. A car moving on a straight highway, lift traveling in a vertical well, stone falling from the top of a building, is examples of rectilinear motion. Position, Displacement and Distance For a moving particle, the information of its position, at various instants, is a very important data in motion analysis. Position means the location of a particlewith respect to a fixed reference point. Such -fixed point is usually referred to as the origin ‘o’. The point ‘o’ can be marked anywhere on the particle’s straight path. In one direction of the origin, the position is taken as positive and therefore the other side of origin implies negative position. Fig. (a) Shows a particle occupying position x = 5 m while in Fig. (b) The particle occupies a position x = - 3 m. 1

2. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com Displacement is defined as a change in position of the particle. It is a vector quantity. If a particle occupies position x1 at some time t1 and a new position x2 at a time t2, then the displacement during the time (t2– t1) is given by: AX = X1-X2 …………[9.1] Displacement is therefore a straight line vector connecting the initial position to the final position and has no relation with the actual distance travelled by the particle. Velocity ‘How fast’ a particle moves is the velocity of the particle in motion. Consider a particle occupying position x at time t and after a small time interval of At occupies a new position x + At. ‘How fast’ the particle has moved during the time interval At is known as the average velocity of the particle.  Displacement Averagevelocity= Time x v t     x x t x       ............ 9.2 Av  is made smaller and smaller, the average velocity If the time interval t will become instantaneous velocity, i.e. velocity at a particular instant. Instantaneous velocity is usually referred to as the velocity of the particle and denoted as v.   X t  lim t   V 0   X t Now by definition, is the derivative of x with respect to t. lim t   0 dx dt   …… [9.2] V 2

3. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com S.I Unit of velocity v is metres/sec (m/s). The magnitude of velocity is known as the speed of the particle. A positive value of v indicates that the particle is moving in the positive direction i.e. the position x increases with time. A negative value of velocity indicates that the particle is moving in the negative direction i.e. the position x decreases with time. For example, if vertically upward direction is taken as positive and a stone is thrown vertically up, it will have +ve velocity during its upward motion till the peak, while it will have -ve velocity during its return downwards. Acceleration A moving particle has velocity at every instant of its motion. If its velocity changes, the rate of change of velocity with time is the acceleration of the particle. If a particle has a velocity ‘v’ at a certain instant and its velocity changes to  during a time interval of t  , the average acceleration of the particle v v during this time interval is   v+Δv -v Δt changeinvelocity time Averageacceleration= = Δv a  ……[9.6] =Δt av  is made smaller and smaller, the average If the time interval t acceleration will become instantaneous acceleration. Instantaneous acceleration is usually referred to as the acceleration of the particle is and denoted as 'a'.   v t  lim t   a 0   v t By definition,  is the derivative of v with respect to t lim t   a 0 dv dt   …… [9.5] a SI unit of acceleration 'a' is m/s2. A positive value of acceleration is indicative of increase in the magnitude of velocity with time i.e. the body is moving faster in the positive direction. A negative value of acceleration indicates that the particle is moving more slowly in the +ve direction or moves more faster in the –ve direction. 3

4. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com DIFFERENT RECTILINEAR MOTIONS Different types of rectilinear motions possible are 1)Motion with Uniform Velocity 2)Motion with Uniform Acceleration 3)Motion with Variable Acceleration. These are further explained in detail. Uniform Velocity Motion For a particle whose velocity remains the same throughout the motion is said to undergo a uniform velocity motion. For example, motion of sound, a train traveling during a certain interval at a constant speed, packages moving on a conveyor belt etc. perform uniform velocity motion. dx Weknowvelocity v=dt dx=vdt or x  t  2 2 Integrating dx= vdt x t 1 x -x 1     =v t -t 2 1 2 1 Here   distance traveled s (since velocity is constant) x t x t  x 2 1   time interval t and 2 1  s = vt s s …….[9.6] or v= uniformvelocityEquation Uniform Acceleration Motion A particle is said to perform uniform acceleration motion if its velocity changes at a uniform rate. If u is the initial velocity of a particle, v is the final velocity and t is the time interval, then the acceleration a of the particle is, v-u t a= v=u+atuniformAccelarationEquation1 …..[9.7 (a)] or 4

5. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com If s is the distance travelled during this time interval then s =Averagevelocity×t u vt  s = 2     u u at  s t 2 1 2 s ut   2 …..[9.7(b)] or uniformAccelerationEquation2 at From Equation 9.7 (a) , we have  v u  t a Substituting in Equation 9.7 (b), we get, 2             v-u a 1 2 v-u a s=u + a 2 2 v -u 2a s = 2 orv =u +2 asuniformAccelerationEquation3 Motion Under Gravity (M.U.G) Any object projected vertically up in the air or projected vertically down towards the earth performs a rectilinear motion with uniform acceleration. The acceleration is constant and its magnitude is g = 9.81 m/s2. M. U. G therefore is a special case of uniform acceleration motion. Consider a ball thrown vertically up with an initial velocity v0 from the top of a tower at A and of height h. The ball would travel vertically up performance rectilinear motion. The velocity keeps on reducing at a uniform rate of 9.81 m/s every sec. i.e. 9.81 m/s2 till it becomes zero at B. This is the maximum height ymax reached by the ball. The downward motion of the ball now begins and the velocity goes on increasing from zero at a uniform rate of again 9.81 m/s2 5

6. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com While passing point A it will have the same speed v0 but pointing downwards. The ball would finally land on the ground at c with a velocity vc after time t sec. Knowing v0 and h we can find the maximum height ymax, time of flight t and final velocity of landing vc using equations of uniform acceleration. For solving problems on M.U.G follow the guidelines listed below. 1)Take the starting point as the origin and take all directions either  +ve or  +ve. 2)By directions, we mean the direction of displacement, velocity and acceleration. 3)With proper sign convention use the three equations of uniform accelerations viz. 1 v=u+at, s=ut+ at , 2 Here, u is the velocity at the start of M.U.G 2 2 2 v =u +2as v is the linear velocity s is the displacement of the particle t is the time interval a is the uniform acceleration due to gravity whose magnitude is 9.81 m/s2 6

7. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 1-A 1.The motion of a car during its 90 minutes journey on a straight road was recorded as below; Travels at 50kmph for the first 20 minutes. Accelerates uniformly to a maximum speed of 90kmph during the next 10 minutes. Travels at this maximum speed for the next 4O minutes. Accelerates uniformly to a stop in the remaining 20 minutes. 1)Acceleration 2)Deceleration 3)total distance traveled 4)average speed during entire motion. 2.Two trains A and B leave the same station on parallel lines. Train A starts with a uniform acceleration of 0.3 m/s2 and attains a speed of 36kmph after which it maintains the same velocity. Train B leaves 2 minutes later with a uniform accelerator of 0.45 m/s2 to attain a maximum speed of 90kmph. When and where will B overtake A. 3.Two elevators in adjoining wells are vertically 60 m apart start from res at the same instant and approach each other. The up moving elevator travels up with a uniform acceleration of 0.3 m/s2. The down moving elevator travels down with an acceleration of 0.15m/s2 1)When do the elevators cross each other? What is the distance traveled by each of them at this instant. 2)Solve the problem if up moving lift starts 2 sec after start of down moving elevator. 4.Cars A and B travel on a straight highway. At t = 0, car A travelling at 54kmph decelerates at 0.5 mls2, while car B, 50 m behind car A, travelling at 36kmph accelerates at 0.8 m/s2. Find the distances and time taken by the two cars before E overtakes A. 5.A particle starts from rest and accelerates at a constant rate of 0.5 m/s2 for some time. Thereafter it decelerates at a constant rate of 0.3 m/s2 and comes to rest. If the particle was in motion for 2 minutes, find the maximum velocity acquired by it. 7

8. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 6.A block is released from rest from the top of an inclined plane and it takes 5 sec to reach the bottom of the plane. What is the time it takes to travel one fifth the distance from the top. 7.An athlete running a 100 m run starts from rest and reaches his maximum speed in a distance of 15 m. He runs the remaining distance with that velocity and reaches the finish in 9'6 sec' Determine the initial acceleration and the maximum velocity acquired by the athlete. 8.Track repairs are going on a 4 km length of railway track. The maximum the train is speed of 108kmph. The speed over the repair track is 3okmph. If the train approaching the repair track decelerates uniformly from a full speed of 108kmph to 3Okmph in a distance of 1200 m and after covering the repair track accelerates uniformly to full speed of 108kmph in a distance of 1500 m, find the time lost due to reduction of the speed in the repair track. 9.An automobile starts from rest and travels on a straight path al2 mf szfor some time. After which it decelerates at 1 m/s2, till it comes to a halt. If the distance covered is 300 m, find the maximum velocit5r of the automobile and the total time of travel. EXERCISE 1-B 1.A stone is thrown vertically up from the top of the tower 20 m high with a speed of 15m/s. find 1)Max. height reached by the stone above ground. 2)Velocity with which the stone hits the ground. 3)Total time of flight. 2.A ball is thrown vertically upwards with a velocity 9 m/s from the edge of a cliff 15 m above the sea level. What is the highest point above sea level reached? How long does it take to hit the water? with what velocity does it hit the water? 3.A stone is thrown vertically up from the top of the tower 40 m high with a velocity of 20 m/s. Three seconds later another stone is thrown vertically up from the round with a velocity of 30 m/s. Calculate when and where the two stones will meet from the foot of the tower. 8

9. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 4.A stone is dropped from the top of a tower. When it has fallen a distance of 10 m, another stone is dropped from a point 38 m below the top of the tower. If both the stones reach the ground at the same time calculate. 1)the height of the tower 2)the velocity of the stones when they reach the ground. 5.Water drips from a tap at a rate of 5 drops per second. The tap is 900 mm above the basin. When one drop strikes the basin, how far is the next drop above the basin. 6.Water leaks from a ceiling 16 m high, at the rate of 5 drops per seconds between first and second drop when the first drop has just touched the ground. 7.Drops of water leak from a tap 2 m above the basin at regular intervals. As the firs: drop strikes the basin the fourth drop starts its fall. What is the spacing between the drops 1 and 2, drop 2 and 3, drops 3 and 4 at this instant. 8.A stone released from rest falls freely under gravity. The distance covered by it in the last second of its motion equals the distance covered in the first four seconds of its motion. Find the time the stone was in motion. 9.A ball thrown vertically up was found to travel a distance of 5 m during 3'd second of its travel. What is the initial velocity of the ball? 10.A halogen gas filled balloon released from the ground, travels vertically up with a constant acceleration of 2 m/s2. Three seconds later, a stone is projected vertically u; from the ground to hit the ascending balloon. The stone just manages to touch th: balloon at the peak of its path. Find the velocity with which the stone was projected and the height at which the stone touches the balloon. 9

10. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com VARIABLE ACCELERATION MOTION Variable acceleration implies that rate of change of velocity is nc: uniform. Rectilinear motions are not always uniformly accelerated, but more often undergo variable acceleration. For example suspend a block from a vertical spring, stretch it and then release. The block would oscillate up and down undergoing variable acceleration motion. The acceleration here is variable acceleration is proportional to the deformation of the spring. Variable acceleration motion is usually defined by acceleration written as a function of time or velocity or position. For the solution of variable acceleration motion, we make use of the basic three differential relations of velocity and acceleration given below. Two of them have been derived earlier as equations 9.3 and 9.5. dx  ……. [9.4] v=dt dv ……. [9.5] a=dt dv from 1and2 a=vdx .…… [9.6] 10

11. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 2 1.The rectilinear motion of a particle has its position defined by the relation x= t3-7t2+20t-10m.Determine, 1)Position, velocity and acceleration at t = 2sec. 2)Minimum velocity and the corresponding time. 2.The motion of a particle moving in straight line is given by a relation. 3 2 s=t -3t +2t+5 where's' is the displacement in metres and 't' is time in seconds. Determine 1)Velocity and acceleration after 4 sec. 2)Maximum or minimum velocity and corresponding displacement and time at which velocity is zero. 3.The position of a particle moving a-long a straight line is defined by the 3 2 X=t -9t +15t+18where x is expressed in meters and t in seconds. relation Determine the time, position and acceleration of the particle when its velocity becomes zero. 4.The car starts from the rest and moves in a straight line such that for a 2 v=(9t +2t)m/s. Where t is in seconds. short time its velocity is defined by Determine its position and acceleration when t = 3 sec. 5.Acceleration of a particle is directly proportional to the square of time t When t = 0, particle is at x= 24 m. Knowing that at t = 6 sec, x = 96 m and v = 18 mi: find expressions of x and v in terms of t. Also find velocity and position at t = 2 sec. 6.The acceleration of a particle is given by a = - kx-3 m/s2. Knowing at x = 2 m, v= O and at x= 0.5 m, v= 3 m/s. Determine a) the value of k b) the particle's velocity at x=1m. 7.The acceleration of an oscillating particle is defined by the relation a=-kx m/s2.Determine 1)The value of k such that v = 12m/s at x=2m and v=0 at x=6m 2)Velocity at x=4 3)Maximum velocity. 11

12. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 8.A particle performing rectilinear motion starts from rest from origin and has, acceleration defined by a = 25 – v2 m/s2. Determine the time and the particle’s displacement, when v = 4 m/s. 9.A particle moving in the + ve x direction has an acceleration a=100-4v2 m/s2 Determine the time interval and displacement of a particle when speed changes from 1 m/s to 3 m/s. 10.The motion of a particle is defined by the relation a = 0.8 t m/s2. It is found that at x= 5 m, v = 12 m/s when t = 2 sec. Find position and velocity at t = 6 sec. 11.The velocity relation of a rectilinear moving particle is defined as = 4 t2- 3t- 1 m/s. At t = 0, X = - 4m. Determine a)The time at which the particle reverses its sense of motion b)At t = 3 sec. 1)Acceleration 2)Position 12.A particle starting from rest moves in a straight line, whose acceleration is 3) displacement 4) Distance traveled. given by a = 8 - 0.003s2, where a is in m/s2 and s is in meters. Determine 1)Velocity of the particle when it has traveled 40m and 2)Distance traveled by the particle when its comes to rest 13.A point moves along a straight line such that its displacement is s = 8t2 + 3t where s is in m and t is in seconds. Find the displacement and velocity at t = 4 sec. Also find the distance traveled in the 10th second and the change in velocity from t = 4 sec till t= 10 sec. 14.The acceleration of the particle is defined by the relation a= 25 – 3x2 mm/s2. The article starts with no initial velocity at the position x = O. a)Determine the velocity when x = 2mm b)the position when velocity is again zero c)Position where the maximum velocity is maximum and the corresponding velocity. 15.A particle performing rectilinear motion has its acceleration given by 1 = (2 - 5 t) m/s2. At t = 4 sec its velocity is 15 m/s and at t = 8 sec its position is 60 m. find at t = O, the particles position, velocity and acceleration. Also find the time when its velocity becomes zero. 12

13. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 16.A stone is projected from the ground vertically up with an initial velocity of 20 m/s 1)Knowing that the air resistance causes an additional deceleration of 1.01v2 m/s2, determine the maximum height reached by stone above the ground. Hint: a = - 9.81 - 0.01v2 m/s2) 2)Solve for maximum height if air resistance is neglected. 17.A particle starting from the origin with an initial velocity u and travelling in a straight path has an accelerating of (5t + 6) m/s2. The position of the particle at t = 2 sec is 1O m. Calculate the particles initial velocity u and its position at t = 4 sec. 18.The acceleration of a vehicle at any instant is given by the expression 10 4 v    displacement is 125 m. Also find its displacement when its velocity is 10 m/s.     m/s2 vehicle starts from rest, find its velocity w89hen its a 2 -8 s 19.A particle moves in along a straight line with acceleration where a is a= 2 in m/s2 and s in meters. When t = 1sec, s = 4m and v = 2 m/ s. Determine acceleration where t = 3 sec. 13

14. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com CURVILINEAR MOTION A particle which travels on a curved path is said to be performing curvilinear motion. Figure shows a particle P moving on a curved path. Let us understand the terms viz. position, velocity and acceleration of a particle performing curvilinear motion. Position It is represented by a position vector r which extends from the origin of the fixed reference axis to the particle. It describes the location of the particlew.r.t the origin. as the particle travels along the curved path, the value of Fkeeps on changing. Velocity Let a particle occupying position P move to P' as the particle performs rvilinear motion. Simultaneously ' the position vector r also moves to r .  . Let the time interval be t The vector joining P and P' is the change in position  . The average velocity is  during the time interval vector r t  r t  V AV If the time interval is made smaller, the point P' will come closer to P and  would be   , the vector r therefore for the limit 0 t tangent to the path at p. The average velocity now becomes instantaneous velocity at P Δr limΔt  Δt 0 dr  V=dt ……[9.9(a)] Thus as shown in figure (b), velocity of a particle performing curvilinear motion is always tangent to the curved path at every moment. 14

15. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com Acceleration Since the direction of velocity is continuously changing in curvilinear motion, it results in acceleration being present at every instant. Let the velocity of the particle at position P be v and at position P' be v' then, the average acceleration is given by   V t a  AV Making At smaller and smaller results in instantaneous acceleration. ΔV Δt a = lim  Δt 0 dv dt or a = The acceleration vector can be at any angle as shown. Curvilinear Motion - Rectangular System Curvilinear motion can be split into motion along x direction, y direction and z direction, which can be independently worked as three rectilinear motions along the x, y and z direction. For a curvilinear motion we therefore write position, velocity and acceleration in the vector form as 2 2 2 …… [9.10 (a)] r=Xi +Yj+Zk andmagnituder= x +Y +z dr dt 2 x 2 y 2 z …… [9.10 (b)] V= =V i+V j+V k andmagnitudeV= V +V +v x y z dv dt 2 x 2 y 2 z …… [9.10 (c)] a= =a i+a j+a k andmagnitudea= a +a +a x y z For a particle moving in the xy plane, its rectangular components are shown. 15

16. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com Curvilinear Motion - N - T System In a curvilinear motion the acceleration vector can also be split along the tangent to the path and normal to the path. The acceleration vector is therefore expressed as 2 [9.11] or a= a +a a 2 n t a a= e + e n t n t An, the normal component of acceleration represents the change in direction c, motion and is always directed towards the centre of curvature. Its magnitude r. given by v = p a Where v is the speed at the instant and p is the radius of curvature For circular curves p is the radius of the circle. For curves which are defined as y = f (x), 3 2 2 dy 1+ dx Ρ= d y dx at, the tangential component of acceleration represents the change in speed of the particle. The direction of at is positive (along velocity vector) or negative (opposite to velocity vector) depending on whether the speed is increasing or decreasing. For Uniform Speed Curvilinear motion at = O For speed changing at uniform rate i.e., Uniform tangential acceleration curvilinear motion, the following relations hold true v=u+at …… [9.14 (a)] 1 s=ut+ at 2 2 2 v =u +2as …… [9.14 (c)] here ‘u' and ‘V' are the initial and final speeds during a time interval of ‘t’. 2 ….. [9.12] n                ….. [9.13] 2 2 2 …… [9.14 (b)] Also ‘s’ is the curved distance traveled by the particle. For speed changing at varying rate i.e varying tangential acceleration motion, 2 d s a …… [9.15] =dt t 16

17. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com Relation between Rectangular and N - T components of Acceleration We may sometimes be required to find the N - T components of acceleration i.e. a, and ay knowing the rectangular components of acceleration i.e a* and ar. This can be done using the relation given below, which is 3 v …… [9.16] Ρ=a v -v a x y x y From the above relation radius of curvature p can be calculated and hence a. and hence an and at can be found out. EXERCISE 3 1.A curvilinear motion is defined by y = 8t3 - 6t m and ax = 4t m/s2.At t = 1 sec x v = 6 m/s. Calculate magnitude of velocity and acceleration at t = 3 sec. 2.A particle starts from rest from the position (- 5, - 2, 4) m. Its acceleration is 2 defined by a=3ti=12t j+5kk m/s2. Find the particle's position, displacement velocity and acceleration at t =2 sec. 3.A particle starting from rest at the position (5, 6, 2) m accelerates 2 a=6ti-24t j+10km/s2. Determine the acceleration, velocity, position and displacement of the particle at the end of 2 seconds. 4.A particle performing curvilinear motion has velocity components given as x v= 32 t- 4 m/s and y v = 4 m/s. At t = 3 sec it occupied the position (5, 12) m. Determine the equation of the path traced by the particle. 5.If x = 1 - t and y = t2 where x and y are in metres and 't' is-in seconds, determine x and y components of velocity and acceleration. Also write equation of the path. 6.A particle moves along a curved path declined by y = l/8 x2. At any instant its x coordinate is given by x=2t2-4t. Determine its velocity and acceleration at x=8m. 7.A particle travels along the path defined by the parabola y = 0.5 x2. If the x x v= 5 t m/s, determine the distance of particle component of velocity is from the origin O and the magnitude of acceleration when t = 1 sec. At t = 0, X = 0 and v = 0. 17

18. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 8.A particle P moves in a circular path of 4 m radius. At an instant the speed is increasing at a rate of 8 m/s2 and its total acceleration is 10 m/s2. Determine the particles speed at this instant. 9.A racing car traveling on a circular curve ABC of 400 m radius increases its speed uniformly from 72kmph at A to 108kmph at C over a distance of 300 m along the curve. What was the total acceleration of the car when it was at B, 250 m from A. Also find the total acceleration of the car when it was at A and at C. 10.A train enters a curve of radius 500 m with a speed of 6Okmph. Determine magnitude of total deceleration at the instant the brakes are applied so that the train stops by covering a distance of 4OO m along the curve. Also determine the time required by the train to come to rest. 11.The motion of a particle is declined by the position vector r = 6t i+ 4t2j where r is in metres and t is in seconds. At the instant when t = 3 sec, find (i) Tangential and normal components of accelerations (ii) Radius of curvature. 12.The position of the charged particle moving in a horizontal plane is measured electronically. This information is fed into a computer which employs a curve fitting techniques to generate analytical expression for its 3 4 r=t i+t j where i is in metres and t is in seconds. For t = l position given by sec, determine (i) the acceleration of the particle in rectangular components (ii) its normal and tangential acceleration and (iii) the radius of curvature of the path. 13.A particle travels along a parabolic shaped track y = 10 + O.4 x2 with a constant speed of 6 m/s. At x = 3 m, find a) components of velocity b) acceleration. x y= 3 the x and y components of its velocity when x = 3 m ? What is the acceleration of the point at this instant? 15.A particle travels along a cubic curve y=o.2x3.At a position y=5.4m its speed was 5 m/s and decreasing at a rate of 0.8 m/s2.Find a) At this instant the particle's total acceleration b) velocity components along x and y directions. c) time when the particle comes to rest. 2 14.A point moves along a path with a constant speed of 8 m/s. What are 18

19. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 16.A car travels along a depression in a road, the equation of depression being 2 x =180y. The speed of the car is constant and equal to 54 km/hr. Find the acceleration when the car is at the deepest point in the depression. What is the radius of curvature of the depression? 17.An airplane travels along a path such that its acceleration is given by, a=10i+6t j m/s2 .If the plane starts from rest from the origin, determine at t=4sec. a) speed of the airplane b) radius of curvature of the path c) position of the airplane.   18.A rocket follows a path such that its acceleration is given by a= 4i+tj . At r=0, it starts from rest. At t = 10 sec. determine a) Speed of the rocket b) Radius of curvature of its path, c) an and at components of acceleration. d) Position of the rocket. a 19.A particle moves in the x-y plane with acceleration component m/s2 x=-5 and av = 2 m/s2. If at t = O, its velocity is 1O m/s directed at 36.870 with the + ve axis, find the radius of curvature at t = 8 sec and the corresponding normal and tangential components of acceleration. 20.A particle is moving in x-y plane and its position is defined by             3 2 2 3 2 3 Find radius or curvature when t = 2 sec. r= t i+ t j 21.a) A. particle moves in x-y plane and its position r = (6t) i + (3t – 4t2) j m. Find the radius of curvature of its path and an and at components of acceleration when it crosses the x axis again. b) Solve for the same conditions given i = (3 t) i + (4 t- 3 t2)j m. 19

20. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com PROJECTILE MOTION A particle freely projected in the air in any direction other than vertical, follows a curved path and this motion is referred to as a projectile motion. The path traced by the projectile is known as its trajectory and is parabolic in nature. Projectile motion is a curvilinear motion and can be worked using rectangular system i.e. splitting the motion along horizontal direction and vertical direction Since gravitational force acts in the vertical direction, it is a uniform accelerated motion in the vertical direction and a uniform velocity motion in the horizontal direction, assuming the air resistance to be negligible. Figure shows a projectile fired with an initial velocity vo at an angle 0 with the horizontal from the top of the tower at A and of height h. The projectile travels along a parabolic trajectory and reaches the peak at B. Fig. At the peak the vertical components of velocity vBy = 0. The downward motion now begins and it finally lands with a velocity vc at C on the ground. Procedure to solve projectile problems [Refer Fig.] step 1:Draw a kinematic diagram showing kinematic parameters (initial Velocity, angle of projection, range, time of flight, velocity of landing, vertical displacement) given or asked in the problem. step 2:Resolve the initial velocity o v into components v and v . Resolve ox oy landing velocity v . into v and v . cx cy cy 20

21. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com step 3:The curvilinear motion is split into horizontal motion (HM) and Vertical motion (VM). Make a table with two columns as shown. The left column (HM) lists the kinematic terms like v, s and t for horizontal motion with uniform velocity. The right column (VM) lists the terms u, v, s, a and t for vertical motion with uniform acceleration of a = 9.81 m/s . Take a sign convention  + ve or  + ve for vertical motion. A sign convention t + ve means all vectors like velocity (u, v), acceleration (a) and displacement (s), acting upwards are + ve. step 4: For HM use kinematic relation v = s/t since horizontal motion is   with uniform velocity. For horizontal motion V v v ox Bx cx For VM use kinematic relations v=u+at 1 s=ut+ at 2 v =u +2as 2 2 2 21

22. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 4 1.A gunman fires a bullet with a velocity of 100 m/s, 500 upwards from the top of hill 30b m high to hit a bird. The bullet misses its target and finally lands on the ground. Calculate (a) the maximum height reached by the bullet above the ground (b) total time of flight (c) horizontal range of the bu1let (d) velocity with which the bullet hits the ground. 2.A ball is projected upwards from the top of a tower 3O m high and strikes the ground after 8 sec at a point 3OO m from the foot of the tower. Determine the velocity of projection and also the maximum height attained by the ball above the ground. 3. A stunt motorcyclist has to clear a ditch. Find the minimum speed it should have at A to clear the ditch. Also find his speed at B and the angle of ramp which should be provided. 4.A ball dropped vertically on an inclined plane at A rebounds with a velocity vo perpendicular to the plane and lands again on the plane at B. Knowing vo = 15 m/s, determine the range R. 5.A ball is projected from the top of a 2O m high building with a velocity of 10 m/s at 3Oo upward with the horizontal. The ball lands on a sloping ground. Find the range R and the time of flight of the bal1. 6.An object is projected so that it just clears two obstacles each 7.5 m high which are situated 50 m from each other. If the time of passing between two obstacles is 2.5 seconds, calculate the complete range of projection and initial velocity of projection. 22

23. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 7.A helicopter moving horizontally at an altitude of 100 m drops a packet. The packet landed on the ground 250 m away from the point on the ground directly below the point of drop. What is the velocity of the helicopter ? 8. An artillery gun can fire a shell with a velocity of 26O m/s. How much maximum it can penetrate into the enemy's territory as it targets the enemy from the border. 9.A ball thrown with speed of 12 m./s at an angle of 60o with a building strikes the ground 11.3 m horizontally from the foot of the building as shown. Determine the height of the building. 10.A ball is projected from the top of a tower 110 m height with the velocity of 1OO m/s at an angle of 250 to the horizontal. Neglecting the air resistance, find (1)The maximum height the ball will rise from the ground. (2) The horizontal distance it will travel before it strikes the ground. (3) The velocity with which it strikes the ground. 11.A ball bearing slides down a 300 plane and leaves the plane with a certain velocity vo. Find for what angle of value of vo, the ball bearing ran pass through a 0.8 m wide slot. 12.A gardener holds a water hose at A causing the water to fall on the plants at B. find a) the discharge velocity vo of water at A. b) velocity of the water as it falls on the plants at B. 23

24. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 13.A fighter plane flying horizontally at 324kmph at an altitude of 35OO m plans to strike an enemy train moving with a constant velocity of 9O km/hr. Find the distance the plane should be behind the train just as it drops the bomb. 14.An aeroplane is flying in horizontal direction with a velocity of 540 km/hr and at height of 22OOm. When it is vertically above the point A on the ground, an object is dropped from it. The object strikes the ground at point B. Calculate the distance AB (ignore air resistance). Also find velocity at B and time taken to reach B. 15.A missile M is fired from a certain position A with a velocity of 100 m/s at an angle of 550 upwards with the horizontal. After some time an antimissile N is fired from the same point A with a velocity of 500 m/s at an angle of 400 upwards to the horizontal to destroy the missile M. Find (a) the time of flight of missile M before it is destroyed. (b) the horizontal and vertical distance from A where the destruction takes place. 16.A box dropped from a helicopter moving horizontally with a constant velocity vH, takes 12 sec to reach the I ground and strike the ground at an angle of 60o. Determine a) the altitude h of the helicopter b) the horizontal distance R traveled by the box c) the velocity vH of the helicopter. 17.Water being discharged from a horizontal pipe fixed at a height of 3 m from the ground, falls at a horizontal distance of 4.5 m from the point of discharge. What is the velocity of water discharge from the pipe. 24

25. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 18.A ball thrown horizontally from the top of a 5Om high building hits the horizontal ground 20m from the base of the building. What is the initial velocity of the ball? 19.Find the initial velocity and the corresponding angle of projection of a projectile such that when projected from the ground it just clears a wall 4.5 m high at a horizontal distance of 6 m and finally lands on the ground at a distance of 35 m beyond the wall. 20.In an Asian games shot put event, a ball of mass 6 kg was thrown from an elevation of 1.85 m at an inclination of 420 with the horizontal. The horizontal distance covered on the ground was measured to be 21.64 m. Find a) the velocity of projection b) The time of flight of the ball c) The kinetic energy of the ball at the instant projection and also at the instant it touches the ground. 21.A projectile is projected from the ground with a velocity of 25m/s. Find a) The maximum horizontal range. b) What is the percentage increase in the maximum horizontal range, if it’s initial velocity is increased by 20%. 22.A gunman standing on the ground fires his gun to hit a bird flying at an altitude of 20m from the ground. The angle of projection being 600 upwards with horizontal. If the bullet hits the bird 2.5 sec after firing, find a)The velocity of the bullet as it left the gun. b) The bird is killed instantly. Find the time it takes to reach the ground. Neglect the height of gunman. 23.A ball is projected on an incline from A with velocity vo at an angle  as shown. The ball lands at C, 80 m away from A. If the maximum height covered above A is 22 m find vo and  . 24.A ball rebounds at A and strikes the incline plane at point B at a distance 76 m as shown in figure. If the ball rises to maximum height h = 19 m above the point of projection, compute the initial velocity and the angle of projection  . 25

26. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 25.Ball bearings leave the horizontal trough with a horizontal velocity v0 to fall through a gap of 60 mm. calculate the permissible range of velocity v0 that will enable the balls to enter the gap. 26.A ball is thrown upward from a high cliff as shown. With a velocity of 6O m/s at The ball strikes the inclined ground at right angles. If inclination of ground is 300 as shown, determine (i) Time after which the ball strikes the ground. (ii) Velocity with which it strikes the ground.(iii) co-ordinates (x, y) of a point of strike w.r.t. point of projection. 27.A jet of water discharging from a nozzle hits a vertical screen placed at a distance of 8m from the nozzle at a height of 3m. When the screen is shifted by 4m further away from nozzle, the jet hits the- screen again at the same point. Find the "angle of projection and velocity of projection of the jet at the nozzle. 28.A boy standing at L2 m in front of a wall attempts to strike a ball at the height h on the wall. Taking vo = 15 m/s and knowing h = 7.5 m find the angle 0 for which the ball strikes the wall at B. 26

27. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com GRAPHICAL ANALYSIS The motion of a particle along a straight path can be represented by motion curves. Common motion curves are position-time (x - t), velocity-time. (v - t), acceleration-time (a - t) and velocity-position (v - x) curves. 1)position - time (x – t) curve This is drawn with position on the ordinate dx and time on abscissa. Since V=dt at any instant of time the slope of x - t curve gives the velocity of the particle at that instant.   V= slopex-tcurve  …… [9.17] 2)velocity - time (v – t) curve This is drawn with velocity on the ordinate dv and time on abscissa Since V=dt, At any instant of time the slope of v – t curve gives the acceleration of the particle at that instant.   a= slopev-tcurve  …… [9.18] dx now, V=dt dx=vdt   dx= vdt  or Let the particle's position be xi at time it and its position change to fx f t at time i t and f t, then the If an elemental strip of width dt is taken between area of this strip is vdt. vdt   and i t  represents the entire area under the v - t curve between f t x  f  dx=areaunderv-tcurve  x x x =areaunderv-tcurve i  f - i   Or …… [9.19] x =x + areaunderv-tcurve f i 27

28. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 3)acceleration - time (a – t) curve This is drawn with acceleration on the ordinate and time on the abscissa. dv Weknow, a=dt dv=adt    or dv= adt Let the particle's velocity be iv at the time it and its velocity be fv at time ft . If an elemental strip of width dt is taken between it and ft , then the area of this strip is adt   adtrerepresents the entire area under the a - t curve between t and t i f v  f dv=area under a - t curve  v v -v =area under a - t curve i  f i     or v =v + areaundera-tcurve ....... 9.20 f i From an a - t curve the Particle' position instant X carl also be known at any f ft , knowing the particles position xi and velocity vi at an prior instant it, using the area moment method formula given below.    i G x =x +v t+ areaundera-tcurve t-t t t tis the  t= - and …….[9.21]here f i f i G abscissa of the centroid G of the area under a-t curve. 28

29. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 4)velocity - Position (v - x) curve This is drawn with velocity on the ordinate and position on abscissa. vdv dx  a= We know,   ……. [9.22] a=v× slopev-xcurve From v - x curve we can find the particle's velocity v and the corresponding slope at a given position x and hence using equation 9.22 we can find the particles acceleration. Let us tabulate the uses of various motion curves Standard Motion Curves No. 1. Motion curve v-t Use Formula Slope of x - t curve gives V =(slope x-t curve) velocity 2. a-t a) Slope of v t curve gives a=(slope v-t curve) acceleration i x+ (area under v - t f x = b) Area under v – t change curve) in position and hence the new position.   3. a)area under a - t curve v =v + areaundera-tcurve f i gives change in velocity and hence the new velocity b) Area under a - t curve also helps in finding the Particle's x =x +v t+  areaundera-tcurve t-t f i i Position   G   4. v-x Slope of v - x curve helps a= slopev-tcurve in finding the particle's acceleration 29

30. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com We know that rectilinear moving particles either move with uniform velocity or uniform acceleration or may move with variable acceleration. The general or standard motion curves viz. (a - t, v - t, X - t) for these different rectilinear motions are given here. The particle's motion may be easily identified if we relate the given motion curve problem with these standard curves. For example, if the particles a - t curve is parallel to acceleration axis, the particle is in uniform acceleration motion. Similarly if the x - t curve has a cubic equation it indicates variable acceleration motion. a)Uniform Velocity Motion curves b)uniform Acceleration Motion curves c)variable Acceleration (Linear Variation) Motion curves 30

31. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 5 1.From graph (1) find position and velocity at t= 4, 1O and 15 sec. 2.Refer graph (2). Find acceleration at t =10, 30, 45 and 50 sec. 3.From graph (2) find position at t = 20,30, 40, 5O and 60 sec knowing xo = 0. 4.From graph (3) find position and velocity at t = 6, 10 and 19 sec knowing xo = 10m,v o = 5m/s. 5.Figure shows a - t curve for a article performing rectilinear motion knowing that at t = 0, v = 4 m/s and x = 20 m, find graphically the velocity and position of the particle at t = 9 sec. 6.The race car starts from rest and travels along a straight road until it reaches a speed of 42 m/s in 50seconds s shown by v-t graph. Determine the distance travelled by race car in 50 seconds. Draw x-t and a-t graph. 7.For a particle performing rectilinear motion, the v - t diagram is shown. Draw a- t and x - t diagrams for the motion if at t = 2 sec, x = 2O m. What is the v - x graph of a rectilinear moving particle is shown. Find acceleration of the particle at 2O m, 80m and 200 m displacement during 6 - 10 sec ? Also find the total distance traveled. 31

32. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 8.The v - x graph of a rectilinear moving particle is shown. find acceleration of the particle at 2O m, 80m and 200 m. 9.Figure shows an a - t graph for a particle moving along the x axis. Draw v- t and x - t graphs and find the speed and position of the particle at t = 9O sec. Also find the maximum speed attained by the particle. 10.Figure shows an a - t graph for are rectilinear moving particle. Construct v- t and x - t graphs for the motion. At t= 0 the particle is at x = 0 and it velocity is v=3m/s. 11.A car moves along a straight road such that its velocity is described by the graph shown in figure. For the first 10 seconds the velocity variation is parabolic and between 1O seconds to 3O seconds the variation is linear.   Construct the s - t and a – t graphs for the time period 0 t 30 (Hint: area 1 3 under parabolic curve of base a and height h= ah 32

33. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 12.The car starts from rest and travels along a straight track such that it accelerates at a constant rate for 10 sec. and then decelerates at a constant rate. Draw the v-t and s-t graphs and determine the time t' needed to stop the car. How far has the car travelled? 13.Velocity time graph for a particle moving along a straight line is shown. Draw displacement-time and acceleration-time graphs. Also find the maximum displacement of the particle. 33

34. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com RELATIVE MOTION So far the motion analysis was done from a fixed frame of reference, fixed to the reference, However if the motion analysis is undertaken from a moving frame of reference i.e. observations by a person also in motion, then such analysis comes under relative motion. Few examples of situations where relative motion analysis is done are 1)A person .in a moving train observes another moving train on a parallel track. 2)A captain of a moving ship observes the motion of other ships close to it. 3)The pilot of a fighter plane observing a moving target before striking. Figure shows two particles A and B moving independent of each other. Let their positions defined from a fixed frame of reference xoy, fixed at 0. If now A observes B, then A will find B to be occupying the position B A measured from a moving reference located at A. From the vector triangle so formed, we may write r r r   Ar and Br be r . This position is B A B A ….. [9.23] or r =r -r Relativepositionrelation B A B A Differentiating the above relation w.r.t time, we have or v =v -v Relativevelocityrelation ….. [9.24] B A B A Further differentiating w.r.t time, we get or a =a -a Relativeaccelerationrelation….. [9.24] B A r B A Here are the relative position, velocity and v and a B A B A B A acceleration of B w.r.t A whereas v anda are the absolute position, velocity and r B B B acceleration of particle B. In general the absolute motion of any moving particle say B is the sum of absolute motion of another moving particle say A and the relative motion of B w.r.t A. 34

35. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 6 1. Wedge B travels to the left with acceleration of 2 m/s2, while block A travels up the slope of the wedge with an acceleration of 1 m/s2 relative to the wedge. Determine the true velocity of the block at t = 5 sec after starting from rest. 2. a) A train running at 60 kmph to the right is struck by a stone thrown at right angles to the train with a speed of 10 m/s. Find the velocity and direction with which the stone appears to strike the train, to a person sitting at the window in the train. b) Solve if velocity of train is 36kmph and velocity of stone =18kmph. 3. Trains A and B are traveling on parallel tracks in opposite directions. Velocity of A is thrice the velocity of B. The trains take 20 sec to cross each other. Determine the velocities of the trains knowing that train A is 260 m long and train B is 300 m long. 4. At t = 0 the location of cars A and B are shown. The cars travel towards the intersection. Car A has a speed of 10 m/s and a deceleration of 1.5 m/s2. Car B starts from the rest and accelerates at 2 m/s2.Determine at t= 3 sec the relative position, velocity and acceleration of car B w.r.t car A. 5. Two ships A and B leave a port at the same time. A travels at 36kmph in the North- 'Vest direction while B travels at 50" south of west at 27kmph. a) Determine the relative velocity of B w.r.t A. c) After how much time will the two ships be 120 km apart. 6. a) Two ships move from a port at the same time. Ship ‘A' has velocity of 3O km/hr and is moving North 3OoWest, while ship ‘B' is moving in South-West direction with a velocity of 40 km/hr. Determine relative velocity of A'w.r.t. ‘B'. Also find the gap between the ships 2 hrs. Later. 35

36. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 7. Ship A has a velocity of 45kmph due east relative to ship B, which in turn has a velocity of 60kmph at 300 South of West relative to ship C. Find the velocity of A relative to C. 8. A man walking on a straight path at 15kmph finds that the rain is falling at an angle of 45" to the vertical. As the man increases his speed to 20kmph, he finds the rain now falling at 30o with the vertical. What is the true velocity of the rain. Refer Figure. 36

37. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com DEPENDENT MOTION Let us first understand independent motion. Vehicles moving on a highway, lifts travelling in parallel vertical shafts, ships moving in the sea etc. are examples of independent motion of particles. Motion of independent moving particles are related using relative motion equations. In a dependent motion system, the motion of a particle is dependent on motion of another or several other particles in the system. The particles forming a dependent motion system are usually connected to each other by one or more ropes/strings. In Fig., motion of counter weight W and the lift C form a dependent motion system. In Fig., blocks A, B and C form a dependent motion system. To find a relation between the position, velocities and acceleration of the dependent particles, we have a method known as constant string length method(C S L M). C S L M This method is used to relate the position, velocity and acceleration of two or more particles connected by a common string. This method is based on the principle "The total length of the connecting string in terms of variable position: of the various particles connected to it is a constant whatever be the positions”. The following steps are used to establish kinematics relations using CSLM for dependent motion. Let’s take a example of three moving blocks A, B and C whose position, velocities and acceleration relations are required to be found out. Take a fixed reference axis perpendicular to the direction of motion of the moving particles. If the particles move in the same direction only one reference axis will do. If they move in different direction, for every direction a for every direction a reference axis is required. 37

38. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com In the given example blocks A and B move in the vertical direction hence for them we take a horizontal fixed reference (1). Block C moves horizontally for which-we have taken a vertical fixed reference (2).From the t t t of A x and horizontal reference (1) mark the variable position t= - and f i G blocks A and B respectively. From the vertical reference (2) mark the variable position xc of block C. x x x Here Measure the length L of the string in terms of variables , and A B C x x x .........(1) Note that some constants which are L= + 2 + ±constant A B C added are string portions wrapped over the pulley, while some constants are subtracted like constant a and b which are lengths from the centre of pulley to the reference axis or moving particles. correction to equation (1) In this step a negative sign is attached to the variable which decreases with time during the motion. In the example taken up, say if B was moving down, then A would travel B x increases up and C would travel to the left. In the process the variable A xand xC decrease with time. We therefore with time, while variables correct equation (1) and get equation (2). Differentiate the corrected relation (2) w.r.t. time ……. (3) 0 =-V +2V -V The above equation (3) is the relation between the velocities of particles A B C A,B and C. Differentiate equation (3) again w.r.t time ……. (4) 0 =-a +2a -a The above equation (4) is the relation between the acceleration of particles A B C A, B and C. Note that the position, velocity and acceleration relations developed through equations (2), (3) and (4) are scalar relations (relating only the magnitude), since the direction of motion of the particles have already been accounted by correction to equation (1). 38

39. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 7 A a = 2.5 m/s2 down for the 1. Determine the acceleration of block knowing arrangements shown. 2. Blocks A and B are connected by an inextensible string as shown. If block B moves to the right, starting from rest with constant acceleration of 3 m/s2. a) Find the acceleration of block A. b) The velocity of block A at t = 8 sec. 3. Knowing velocity of A is 3 m/s to the right and that of B is 4 m/s to the left, determine the velocity of block C. 4. a) Knowing aB = 3 mf s2 t, determine acceleration of block A. b) Knowing acceleration of block B w.r.t. A is 9 m/s2  determine the accelerations of blocks A and B. Knowing block A has an acceleration of 0.8 m/s2 , 5. determine the acceleration of block B. 39

40. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com EXERCISE 8 Theory Questions Q.1.State the different types of Particle Motion. Q.2.What are the different types of Rectilinear Motions and list the corresponding equations applicable. Q.3.What is normal acceleration and tangential acceleration. Q.4.Derive expression for maximum height and maximum range for a projectile or horizontal surface. Q.5.Write a short note on Motion Curves. Q.6.Draw motion curves for constant velocity case Q.7.Prove that area under a-t curve gives velocity and area under v-t curve gives displacement. UNIVERSITY QUESTIONS 1.During a test, the car, moves in a straight line such that its velocity is  defined by v - 0-3 2 ) t m/s, where T is in seconds. Determine the 2 (9 t position and acceleration when t - 3 sec. Take at t = 0, x = 0. (5 Marks) 2.A man in a balloon is rising with a constant velocity of 5 m/s propels a bail upwards with a velocity of 2 m/s relative to the balloon. After what time interval will the ball return to the balloon. (5 Marks) 3.A point moves along a curved path y   m its speed is 6 m/s 2 0.4 x Atx , 2 increasing at m s At this instant find — (5 Marks) 2 3 / (i) Velocity components along x and y directions, (ii) Its acceleration. 40

41. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 2s for 6 sec 4.A train leaves station ‘A’ and attains speed at the rate of 4 m/ and then 6 m/ 2s till it reaches a velocity of 48 m/s. Further the velocity remains constant, then brakes are applied giving the train a constant deacceleration stopping it in 6 sec. If the total running time between two stations is 40 sec. plot a-t, v-t and x-t curve. Determine distance between two stations. (10 Marks) 5.A motorist is travelling at 90 kmph, when he observes a traffic signal 250 m ahead of him turns red. The traffic signal is timed to stay red for 12 sec. If the motorist whishes to pass the signal without stopping just as it turns green. Determine i) The required uniform deceleration of the motor. Ii) The speed of motor as it passes the signal. (4 Marks) 6.A particle of mass 1 kg is acted upon by a force F which varies as shown in fig. If initial velocity of the particle is 10 m/s determine i) what is the maximum velocity attained by the particle. Ii) Time when particle will be at the point of reversal. (6 Marks) 7.In Asian games, for 100 m event an athlete accelerates uniformly from the start to his maximum velocity in a distance of 4 m and runs the remaining distance with that velocity. If the athlete finishes the race in 10.4 sec, determine i) his initial acceleration, ii) his maximum velocity. (6 Marks) 8.A ship A travels in the north making an angle of 450 to the West with a velocity of 18 km/ hr and ship B travels in the East with a velocity of 9 km/hr. Find the relative velocity of B w.r.t. ship A. 9.A curviline motion of a particle is defind by At t=0, x=0. Find out position, velocity and acceleration at t = 4 sec. (4 Marks) y=48-3t m 2 V =25-8tm/sand x (4 Marks) 41

42. www.ekeeda.com Contact : 9029006464 Email : care@ekeeda.com 10.A stone is thrown vertioally upwards and returns to the starting point at the ground in 6 sec. Find out max. height and initial velocity of stone . (6 Marks) 2 2 11.For a particle in rectilinear motion α=-0.05V m/s , at v=20m/s,x=0.Find x n at v = 15 m/s and (4 Marks) acc atx =50m. 12.Acceleration of a particle moving along a straight line is represented by the relation a = 30-4.5 x m/s . The starts with zero initial velocity at x = 0. 2 2 Determine (a) the velocity when x = 3 m (b) the position when the velocity is again zero (c) the position when the velocity is maximum. (4 Marks) 13.A particle is projected from the top of a tower of height 50m with a velocity of 20 m/sec at an angle 30 degrees to the horizontal. Determine: (6 Marks) 1)Horizontal distance AB it travel from the foot of the tower. 2)The velocity with which it strikes the ground at B 3)Total time taken to reach point B. 14.A sprinter in a 100m race accelerates uniformly for the 35m and then runs with constant velocity. If the sprinter’s time for the first 35m is 5.4 seconds, determine his time for the race. (4 Marks) 0 15.A gunman fires a bullet with a velocity of 100m/s, 50 upwards from the top of a hill 300m high to hit a bird. The bullet misses its target and finally lands on the ground. Calculate (a) the maximum height reached by the bullet above the ground (b) total time of flight (c) velocity with which the bullet hits the ground. (6 Marks) 2 16.A point moves along the path y=x /3witha constant speed of 8m/s. What are the x and y components of the Velocities when x=3. What is the acceleration of the point when x=3 (4 Marks) 42