A car of mass 1.5 ton is running at 72 kmph on a straight horizontal r

1.	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. While running at the same speed, on the same road, the driver observes an accident 50 m in front of him. He immediately applies the brakes and just manages to stop the car at the accident spot. Calculate the braking force.
Answer» 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.

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. While running at the same speed, on the same road, the driver observes an accident 50 m in front of him. He immediately applies the brakes and just manages to stop the car at the accident spot. Calculate the braking force.

Answer»

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.