Question

A 2250 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. How long does it take the car to decelerate if the force on the car is 8450 N to the east? How far does the car travel during the deceleration?

Answer 1

Given data

*The given mass of the car is m = 2250 kg

*The initial speed of the car is u = 5.00 m/s

*The given final speed of the car is v = 20.0 m/s

*The force on the car is F = 8450 N

The formula for the time taken by the car to decelerate is given as

$\begin{gathered} F=ma \\ =m\times(\frac{v-u}{t}) \end{gathered}$

*Here a is the acceleration of the car

Substitute the known values in the above expression as

$\begin{gathered} 8450=2250\times(\frac{20.0-5.00}{t}) \\ t=4.0\text{ s} \end{gathered}$

Hence, the time taken by the car to decelerate is t = 4.0 s

The formula for the distance travel by the car during the deceleration is given by the kinematic equation of motion as

$\begin{gathered} d=ut+\frac{1}{2}at^2 \\ =ut+\frac{1}{2}(\frac{v-u}{t})t^2 \end{gathered}$

Substitute the known values in the above expression as

$\begin{gathered} d=(5.0)(4.0)+\frac{1}{2}(\frac{20.0-5.0}{4.0})(4.0)^2 \\ =50.0\text{ m} \end{gathered}$

Hence, the distance travel by the car during the deceleration is d = 50.0 m