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

Answer 1

a) Acceleration =
`8.75m/s^2`

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

c) Car travels 9.8 m

Step-by-step 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