A car of mass 1.5 ton is running at 72 kmph on a straight horizontal road. On turning the engine off, it stops in 20 seconds.

Given: m = 1.5 ton = 1500 kg,

u = 72 kmph = 72 × \(\frac{5}{18}\) m/sm/s = 20 m

s^-1 (on turning engine off),

v = 0, t = 20 s, s = 50 m

To find: Braking force (F)

Formula:

i. v = u + at

ii. v² – u² = 2as

iii. F = ma

Calculation:

On turning the engine off,

From formula (i),

a = \(\frac{0-20}{20}\) = -1 m s^-2

This is frictional retardation (negative acceleration).

After seeing the accident,

From formula (ii),

a₁ = \(\frac{0^2-20^2}{2(50)}\) = -4 m s^-2

This retardation is the combined effect of braking and friction

∴ braking retardation =4 – 1 = 3 m s^-2

From formula (iii), the braking force, F = 1500 × 3 = 4500 N

The braking force is 4500 N.