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 :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{-(15\ m/s)}{2.5\ s}[/tex]
[tex]a=-6\ m/s^2[/tex]
Using third equation of motion as :
[tex]v^2-u^2=2as[/tex], s is the distance
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{-(15)^2}{2\times (-6)}[/tex]
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.
We have that the distance in meters before the stop sign must she apply her brakes in order to stop at the stop sign
s=18.8m
From the question we are told that
It slows uniformly from 15.0 m/s to 0.0
in 2.50s
Generally the Newtons equation for acceleration the is mathematically given as
[tex]a=\frac{u-v}{t}\\\\a=\frac{15-0}{2.50}\\\\a=6[/tex]
Generally the Newtons equation for distance the is mathematically given as
For more information on this visit
https://brainly.com/question/12319416?referrer=searchResults
