At t=0, car A starts moving at a constant acceleration of 1.60 m/s^2 north. Afte

Question

At t=0, car A starts moving at a constant acceleration of 1.60 m/s^2 north. After 20.0s, car B leaves the same starting point and starts following at a constant acceleration of 3.6 m/s^2 north. NOTE: both cars start from rest and accelerate with constant acceleration. How far from the starting point has car A travelled before being over taken? What is the final speed of car A? What is the final speed of car B? The two blocks collide in a perfectly elastic collision. Give the corresponding velocities after the collision, if the block on the right originally has a velocity of: 15.0 m/s to the left before the collision. 10.0 m/s to the right before the collision.

Explanation / Answer

Velocity of carA after 20 sec = at = 1.6*20=32 m/s

And distance covered by car A = 0.5at^2 =0.5*1.6*400=320 m

by the relative motion concept

S=ut+0.5at^2

320 = - 32t + 0.5(3.6-1.6)t^2

t=40 sec

Total time taken to carA =20+40=60sec

a.Distance travel at the time of overtake =0.5at^2

=0.5*1.6×3600=2880m

b. Final speed of car A =at =1.6*60=96 m/s

C. Travelling time of car B at the time of overtake=40sec

So final speed of carB =at = 3.6*40=144 m/s

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