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

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

Step-by-step 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})} \\ &= (\Delta v)/(a) \\ &= \frac{(-24)\; {\rm m\cdot s^(-1)}}{(-5.1)\; {\rm m\cdot s^(-2)}} \\ &\approx 4.7\; {\rm s}\end{aligned}`.