Question

When Maggie applies the brakes to her car, it slows uniformly from 15.0 m/s to 0.0 m/s in 2.50s. How many meters before the stop sign must she apply her brakes in order to stop at the stop sign?

Answer 1

From the given information, we can find that the (average) acceleration of the car while applying the brakes is

$a=\dfrac{0-15.0\,\frac{\mathrm m}{\mathrm s}}{2.50\,\mathrm s}=-6.00\,\dfrac{\mathrm m}{\mathrm s^2}$

Then the distance traveled $\Delta x$ is given by

$0^2-\left(15.0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=18.8\,\mathrm m$

Answer 2

Answer:

Distance, s = 18.75 meters

Explanation:

It is given that,

Initial speed of the car, u = 15 m/s

Final speed of the car, v = 0

Time taken, t = 2.5 s

We need to find the distance covered by the car before the stop sign must she apply her brakes in order to stop at the stop sign. Firstly, we will find the acceleration of the car as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{-(15\ m/s)}{2.5\ s}$  

$a=-6\ m/s^2$

Using third equation of motion as :

$v^2-u^2=2as$, s is the distance

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(15)^2}{2\times (-6)}$

s = 18.75 m

So, the distance covered by the car before it comes to rest is 18.75 meters. Hence, this is the required solution.