Answer :
deceleration of the car is [tex]a=8.03\ m/s^2[/tex]
Explanation:
It is given that,
Initial speed of the car, u = 15 m/s
Final speed of the car, v = 0 (it comes to rest)
Distance covered by the car, d = 14 m
Let a is the acceleration of the car. We know that the rate of change of velocity of an object is called its acceleration. Mathematically, it can be calculated as :
[tex]a=\dfrac{v^2-u^2}{2d}[/tex]
[tex]a=\dfrac{-(15)^2}{2\times 14}[/tex]
[tex]a=-8.03\ m/s^2[/tex]
So, the deceleration of the car is [tex]a=8.03\ m/s^2[/tex]. Hence, this is the required solution.
