Question

A car travelling at 15m/s comes to a rest in a distance of 14m when the brakes are applied. Calculate the deceleration of the car
Marking brainliest :)

Answer 1

Answer:

force of the breaks is 6650 N, direction opposite to direction of movement

Explanation:

Answer 2

Answer :

deceleration of the car is $a=8.03\ m/s^2$

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 :

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

$a=\dfrac{-(15)^2}{2\times 14}$

$a=-8.03\ m/s^2$

So, the deceleration of the car is $a=8.03\ m/s^2$. Hence, this is the required solution.