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RECTILINEAR KINEMATICS

RECTILINEAR KINEMATICS . Objectives : -find the kinematic quantities of a particle traveling along a straight path. APPLICATIONS. The motion of large objects, such as rockets …, can often be analyzed as if they were particles. Why?. POSITION AND DISPLACEMENT. Define rectilinear motion.

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RECTILINEAR KINEMATICS

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  1. RECTILINEAR KINEMATICS Objectives: -find the kinematic quantities of a particle traveling along a straight path.

  2. APPLICATIONS The motion of large objects, such as rockets …, can often be analyzed as if they were particles. Why?

  3. POSITION AND DISPLACEMENT Define rectilinear motion The position relative to the origin, O, is defined by r, or s, units? The displacement is change in position. Vector : r = r’ - r Scalar :  s = s’ - s The totaldistance, sT, is …

  4. VELOCITY Velocity is... It is a vector quantity, its magnitude is called speed, units m/s The average velocity is vavg = r/t The instantaneous velocity is v = dr/dt Speed is … v = ds/dt Average speed is (vsp)avg = sT/  t

  5. ACCELERATION Acceleration is …, its units are … Vector form: a = dv/dt Scalar form: a = dv/dt = d2s/dt2 We can also express: a ds = v dv

  6. Position: Velocity: v t v s s t ò ò ò ò ò ò = = = dv a dt or v dv a ds ds v dt v o v s s o o o o o Thus • Differentiate position to get velocity and acceleration. v = ds/dt ; a = dv/dt or a = v dv/ds •Integrate acceleration for velocity and position. • so and vo are the initial position and velocity at t = 0.

  7. v t ò ò = = + dv a dt v v a t yields c o c v o o s t ò ò = = + + ds v dt s s v t (1/2)a t yields o o c 2 s o o v s ò ò = = + v dv a ds v (v ) 2a (s - s ) yields 2 2 c o c o v s o o CONSTANTACCELERATION (particular case) Constant acceleration(e.g. gravity ac =g= -9.81 m/s2 ), then

  8. EXAMPLE Given: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6t m/s2. Find: The distance the motorcycle travels before it stops.

  9. 1) Determine the velocity. a = dv / dt => dv = a dt => v – vo = -3t2 => v = -3t2 + vo v t ò ò = - dv ( 6 t ) dt v o o 3) Calculate distance using so = 0: v = ds / dt => ds = v dt => => s – so = -t3 + vot => s – 0 = -(3)3 + (27)(3) => s = 54 m s t ò ò = - + ds ( 3 t v ) dt 2 o s o o EXAMPLE (solution) 2) Find time to stop (v = 0). Use vo = 27 m/s. 0 = -3t2 + 27 => t = 3 s

  10. GROUP PROBLEM SOLVING Given: Ball A is released from rest at a height of 12m at the same time that ball B is thrown upward, 1.5m from the ground. The balls pass one another at a height of 6m Find: The speed at which ball B was thrown upward.

  11. GROUP PROBLEM SOLVING (solution) 1) Time required for ball A to drop to 6m. t = 1.1s 2) Ball B: sBo = 1.5m, a=…, s=… (vB)o = 9.5m/s

  12. End of 12.1-12.2 Let Learning Continue

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