Question

14. Breaks are applied to a car moving at a speed of 72 km/h, it comes to
rest with in 5 sec. Determine retardation produced in the car and
distance covered by it before coming to rest.

Pllzz give me derrivatio. Pic also​

Answer 1

Given:

Initial velocity of the car, u = 72 km/h

= $\rm\frac{72 × 1000 m}{3600 s}$ = 20m/s

Final velocity of the car, v = 0 km/h = 0m/s

time taken, t = 5s

We know that, negative acceleration or retardation = $\rm\frac{v - u}{t}$

= $\rm\frac{0 - 20 ms^{-1}}{5}$

= $\rm\frac{-20 ms^{-1}}{5 s}$ OR -4m/s²