Question

a bus running at a speed 18 km/hr is stopped in 2.5 seconds by applying breaks calculate the retardation produced.

Answer 1

A bus running at a speed $18\,\mathrm{km/hr}$ is stopped in 2.5 seconds by applying breaks calculate the retardation produced.

Retardation is the synonym of deceleration of bus. Deceleration is the rate at which the velocity of a body decreases with time:

$$a = \frac{\Delta v}{\Delta t}$$

$\Delta v = 18\,\frac{\mathrm{km}}{\mathrm{h}}$ - changing of speed

$\Delta t = 2.5\,\mathrm{s}$ - time of stopping

Therefore:

$$a = \frac{\Delta v}{\Delta t} = \frac{18\,\frac{\mathrm{km}}{\mathrm{h}}}{2.5\,\mathrm{s}} = \frac{\frac{18}{3.6}\,\frac{\mathrm{m}}{\mathrm{s}}}{2.5\,\mathrm{s}} = 2\,\frac{\mathrm{m}}{\mathrm{s}^2}$$

Answer: $2\,\frac{\mathrm{m}}{\mathrm{s}^2}$