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Chapter 2 Unit Vector

Vector addition by components Vector addition by graphing. Chapter 2 Unit Vector. Displacement ( Vector ). Speed & Velocity. Average and Instantaneous Average Speed = Distance/ Elapsed time ( How fast it moves, but no direction)

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Chapter 2 Unit Vector

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  1. Vector addition by components Vector addition by graphing Chapter 2Unit Vector

  2. Displacement (Vector)

  3. Speed & Velocity • Average and Instantaneous • Average Speed = Distance/ Elapsed time (How fast it moves, but no direction) Units and dimensions

  4. Average Velocity a) m/s b) m/s Time to go to Florida from Atlanta Average Velocity?

  5. Instantaneous Velocity Smaller and smallergives smaller and smaller As 0 it gives the limiting value (In general instantaneous values are given unless noted otherwise.) What if changes ( push the accelerator pedal)

  6. Acceleration Rate of change of velocity is acceleration Average acceleration

  7. Equation of Kinematics Constant acceleration (straight line- motion) Say X0and t0 =0

  8. Constant acceleration Velocity increases What is the average velocity?

  9. If starts at and with constant acceleration.

  10. If you do not know “t”

  11. Summary Table 2.1 Equations of Kinematics for Constant Acceleration

  12. Application of the Equations • Draw the situation • Pick +ve and –ve direction and co-ordinates • Write the known values (with signs) • Look for hidden information • Enough information, what is unknown? • If two objects, are they connected?

  13. No air resistance Constant acceleration due to gravity Free FALL g= 9.80 m/s2 or 32.2 ft/s2 (on earth surface) Gallileo’s test Leaning tower (better on moon) Air filled tube Evacuated tube

  14. Example 10 A Falling Stone After 3.0 sec what is the displacement Known: v0=0a=g (i.e.) t=3.0 y=v0t+(1/2)at2 v=v0+at v2=v02+2ay Which eqn to use? v0=0 m/s t=3.0 sec y=v0t+(1/2)at2 =0+(1/2)(-9.8)*9=(-44.1m)

  15. What is the velocity after 3.0 sec? v=v0+at v2=v02+2ay Can use both----why? Which is easier? v = -9.8m/s2*3sec = -29.4ms -ve sign indicate

  16. Example 12How high the coin reach, if it is tossed up with an initial speed of 5.00 m/s? At the highest point the velocity should be zero. v0=5.0 m/s, a=-9.8 m/s2 v=0.0 m/s, y=? v2=v02+2ay

  17. v=v0+at 0=5-9.8t t=0.5/sec How long will it take to reach the top ? v=0, v0=5, a= -9.8 Which eqn ? How much time it takes to come to the same level as it was thrown ? v0=5.0 a=-9.8 y=0 t=?

  18. t=0 or i.e.t=0 or Why two values? Explain each answer.

  19. symmetry No air resistance—acceleration is the same for the motion in both directions Velocity vector changes continuously Acceleration does not. Time to reach the highest point = time to reach the ground At same height, speed is the same.

  20. Bullet reaches the ground with the same speed irrespective of which direction it was fired first.

  21. v2=v02+2ayv2=(30)2+2(-9.8)*0 v=+/-30 What is the significance of + and – sign? Show the speed of the bullet when it comes down to the same height. What is the velocity if shot up with initial velocity 30 m/s under gravity when it reaches the same level? Known information: v0=30, y=0, Lets say 150 m a=-g Which eqn ?

  22. Harder way How high it reaches before changing direction? v0= 30, a=-g, v=?(hidden information) y=?(needs to find) Which eqn ? v2=v02+2ay 0=(30)2+2(-9.8)*y y=(30*30)/(2*9.8)=45.92 m

  23. v0=0 y=? v=? For the downward motion, v2=v02+2ay =0+2(+9.8)*[(30*30)/(2*9.8)] (why +ve ?) =30*30v=30 m/s (again +/-) Find the time to reach max height and then find the speed when it reaches the same level.

  24. What is the velocity of the bullet when it reaches the ground if we start from the mountain? v=? v0= 30, a=-9.8, y=-150 (why –ve ?) v2=v02+2ayv2=(30)2+2(-9.8)(-150) =(30)2+2.94*103 =3.84*103 v=61.97 m/s (which sign?)

  25. If you drop down with 30 m/s from the mountain, what is the speed? v0= -30, g=-9.8, y=-150 v2=v02+2ay =(30)2+2(-9.8)(-150) =(30)2+2.94*103 =3.84*103 v=61.97 m/s

  26. Example t=t t=0 v0 2v0 v=v0+at 0=v0+at v=2v0+at1 0=2v0+at1

  27. Answer-- 2t, 4y

  28. Graphical analysis x=0 and t=0 Slope =

  29. Example 16 Bicycle Trip

  30. 1st segment 2nd segment 3rd segment

  31. If velocity changes (i.e. acceleration) (assume v0=0 for simplicity) Slope of the tangent Instantaneous velocity v=v0+at

  32. Conceptual questions 2 REASONING AND SOLUTION The buses do not have equal velocities. Velocity is a vector, with both magnitude and direction. In order for two vectors to be equal, they must have the same magnitude and the same direction. The direction of the velocity of each bus points in the direction of motion of the bus. Thus, the directions of the velocities of the buses are different. Therefore, the velocities are not equal, even though the speeds are the same.

  33. Conceptual question 4 REASONING AND SOLUTION Consider the four traffic lights 1, 2, 3 and 4 shown below. Let the distance between lights 1 and 2 be x12, the distance between lights 2 and 3 be x23, and the distance between lights 3 and 4 be x34.

  34. The lights can be timed so that if a car travels with a constant speed v, red lights can be avoided in the following way. Suppose that at time t = 0 s, light 1 turns green while the rest are red. Light 2 must then turn green in a time t12, where t12 = x12/v. Light 3 must turn green in a time t23 after light 2 turns green, where t23 = x23/v. Likewise, light 4 must turn green in a time t34 after light 3 turns green, where t34 = x34/v. Note that the timing of traffic lights is more complicated than indicated here when groups of cars are stopped at light 1. Then the acceleration of the cars, the reaction time of the drivers, and other factors must be considered.

  35. Conceptual question 5 REASONING AND SOLUTION The velocity of the car is a vector quantity with both magnitude and direction The speed of the car is a scalar quantity and has nothing to do with direction. It is possible for a car to drive around a track at constant speed. As the car drives around the track, however, the car must change direction. Therefore, the direction of the velocity changes, and the velocity cannot be constant. The incorrect statement is (a).

  36. Conceptual question 14 REASONING AND SOLUTION The magnitude of the muzzle velocity of the bullet can be found (to a very good approximation) by solving Equation 2.9, With v0 = 0 m/s; that is where a is the acceleration of the bullet and x is the distance traveled by the bullet before it leaves the barrel of the gun (i.e., the length of the barrel).

  37. Since the muzzle velocity of the rifle with the shorter barrel is greater than the muzzle velocity of the rifle with the longer barrel, the product ax must be greater for the bullet in the rifle with the shorter barrel. But x is smaller for the rifle with the shorter barrel, thus the acceleration of the bullet must be larger in the rifle with the shorter barrel.

  38. Problem 4 REASONING The distance traveled by the Space Shuttle is equal to its speed multiplied by the time. The number of football fields is equal to this distance divided by the length L of one football field. SOLUTION The number of football fields is

  39. Problem 8 REASONING AND SOLUTION Let west be the positive direction. The average velocity of the backpacker is .

  40. Combining these equations and solving for xe (suppressing the units) gives The distance traveled is the magnitude of xe , or

  41. Problem 8 solution (2) N N N y y y 6.44 km 6.44 km W W W E E E O A is 6.44 km with an velocity of 2.68 m/s =say t1 x x x S S S O O O Time for 1st segment A A A B is x m with an velocity of .447m/s =say t2 Time for 2nd segment (X in meters) B Time for displacement(O B) with an average velocity of 1.34m/s =say t x

  42. t1+t2=t

  43. Problem 10 ReasoningThe definition of average velocity is given by Equation 2.2 as Average velocity = Displacement/(Elapsed time). The displacement in this expression is the total displacement, which is the sum of the displacements for each part of the trip. Displacement is a vector quantity, and we must be careful to account for the fact that the displacement in the first part of the trip is north, while the displacement in the second part is south.

  44. -17 m/s xs +27 m/s for tn xn Displacement = xn+ xs Solution According to Equation 2.2, the displacement for each part of the trip is the average velocity for that part times the corresponding elapsed time. Designating north as the positive direction, we find for the total displacement that

  45. Where tNorth and tSouth denote, respectively, the times for each part of the trip. Note that the minus sign indicates a direction due south. Noting that the total elapsed time is tNorth + tSouth we can use Equation 2.2 to find the average velocity for the entire trip as follows:

  46. But Therefore, we have that The plus sign indicates that the average velocity for the entire trip points north.

  47. Problem 19 REASONING AND SOLUTION x=v0t+(1/2)at2 Uniform acceleration, starts from rest.

  48. Problem 23 REASONING We know the initial and final velocities of the blood, as well as its displacement. Therefore, Equation 2.9 can be used to find the acceleration of the blood. The time it takes for the blood to reach it final velocity can be found by using Equation 2.7

  49. SOLUTION a.) The acceleration of the blood is b.) The time it takes for the blood, starting from 0 cm/s, to reach a final velocity of +26 cm/s is

  50. Problem 28 REASONING AND SOLUTION The speed of the car at the end of the first (402 m) phase can be obtained as follows:v12 = vo2 + 2a1x1 The speed after the second phase (3.50  102 m) can be obtained in a similar fashion.v22 = v022 + 2a2x2 v2=96.9 m/s

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