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. v
2
– u
2
= 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),
a
1
=
\(\frac{0^2-20^2}{2(50)}\)
= -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.