Question

A car slows from 25.21 m/s to rest in 4.36 s. How far did it travel in this time?

Answer 1

In order to determine the distance traveled by the car, use the followin formula from kinematics:

`x=v_ot+(1)/(2)at^2`

where,

vo: initial speed = 25.21m/s

t: time = 4.36s

a: acceleration = ?

x: distance = ?

To calculate the distance x it is necessary to calculate the deceleration of the car. Use the following formula:

`a=(v-v_o)/(t)=(0(m)/(s)-25.21(m)/(s))/(4.36s)\approx-5.78(m)/(s^2)`

Then, replace the values of t, vo and a into the expression for x and simplify:

`x=(25.21(m)/(s))(4.36s)+(1)/(2)(-5.78(m)/(s^2))(4.36s)^2\approx54.98m`

Hence, the distance traveled by the car before it stops is approximately 54.98m