Question

A toy friction car is propelled forward from rest by a girl's hand which causes it to reach a forward velocity of 3.5m/s in 0.4 seconds. it is then rreleased and later stops in 5.6 seconds.
a) what is the acceleration with which the car speeded up?

b)what is the acceleration with which the car slowed down?

c) how far did the toy car travel?

Answer 1

Answer:

a)  Acceleration = $8.75m/s^2$

b) Slow down acceleration = 0.625$m/s^2$

c) Car travels 9.8 m

Explanation:

  Acceleration = Change in velocity/Time

a) Change in velocity = 3.5-0 = 3.5 m/s

   Time = 0.4 seconds

    Acceleration = $3.5/0.4=8.75m/s^2$

b) Change in velocity = 0 - 3.5 = -3.5 m/s

   Time = 5.6 seconds

   Acceleration = $-3.5/5.6=-0.625m/s^2$

   Slow down acceleration = 0.625$m/s^2$

c) We have equation of motion, $v^2=u^2+2as$, where u is the initial velocity, u is the final velocity, s is the displacement and a is the acceleration.

 Here v = 0 m/s, u = 3.5 m/s, a = -0.625$m/s^2$

 Substituting

    $0^2=3.5^2-2*0.625*s\\ \\ s=9.8m$

  So, car travels 9.8 m