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.
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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.
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