Question

7. If a car decelerates at a rate of 5.1 m/s2

, and if its initial speed was 24 m/s. How long will it

take the car to stop?

Answer 1

Answer:

Approximately $4.7\; {\rm s}$.

Explanation:

Acceleration is the rate of change in velocity.

The initial velocity of this vehicle was $v_{0} = 24\; {\rm m\cdot s^{-1}}$. After stopping, the velocity will be $v_{1} = 0\; {\rm m\cdot s^{-1}}$. Hence, the change in the velocity of this vehicle will be:

$\begin{aligned}\Delta v &= v_{1} - v_{0} \\ &= 0\; {\rm m\cdot s^{-1}} - 24\; {\rm m\cdot s^{-1}} \\ &= (-24)\; {\rm m\cdot s^{-1}} \end{aligned}$.

It is given that the vehicle decelerates at a constant rate of $5.1\; {\rm m\cdot s^{-2}}$. Thus, the rate of change in the velocity of this vehicle (acceleration) will be negative (velocity is decreasing) at $a = (-5.1)\; {\rm m\cdot s^{-2}}$.

To find the time required for this change in velocity, divide the change by the rate of change:

$\begin{aligned}(\text{time required}) &= \frac{(\text{change})}{(\text{rate of change})} \\ &= \frac{\Delta v}{a} \\ &= \frac{(-24)\; {\rm m\cdot s^{-1}}}{(-5.1)\; {\rm m\cdot s^{-2}}} \\ &\approx 4.7\; {\rm s}\end{aligned}$.