Question

a motorcyclist along with the machine weighs 160 kg while driving at 72 km per hour he stops his machine over a distance of 8 M find the retarding force of the brakes

Answer 1

Initial velocity, u = 72 km/h = 72× 5/18 = 20 m/s

Final below, v = 0

Distance for stopping, s = 8m

We know that v^2 - u^2 = 2as

0 - 20^2 = 2×a×8

16a = -400

a = -400/16 = -25 m/s^2

So retardation of the motorcycle = 25 m/s^2
Mass = 160 kg
Force = ma = 160×25 = 4000 N

Retarding force is 4000 N

Answer 2

★ Retarding Force = 4000 Newtons ★

Explanation:

Given:

• Initial velocity (u) = 72 km/h
• Final velocity (v) = 0 ( since, machine stops)
• Distance covered (s) = 8 m
• Mass of machine (m) = 160 kg.

To Calculate:

• What is the acceleration (a) = ?

Formula to be used: v² – u² = 2as

• Convert the km/h to m/s

• 72 x 5/18 = 20 m/s

$\implies \: v ^{2} - u ^{2} = 2as \\ \\ \implies \: (0 )^{2} - (20 )^{2} = 2 \times a \times 8 \\ \\ \implies \: - 400 = 16a \\ \\ \implies \: \frac{ - 400}{16} = a \\ \\ \implies \: - 25 \: m/s ^{2} = a$

∴ Retardation

→ – (a) = – ( – 25 ) = 25 m/s²

∴ Retarding force = Mass x Retardation

$\implies \: 160 (kg)\times 25(m/s ^{2} ) \\ \\ \implies \: 4000 \: N \:$