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

$\huge\underline\mathrm{GivEn:-}$

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

$\huge\underline\mathrm{Need\:To\:Find:-}$

• Find the acceleration and retarding force of the machine = ?

$\huge\underline\mathrm{Formula\:Used:-}$

v² - u² = 2as

Convert km/h into m/s :-

$\implies$ 72 $\times$ $\large{\bf {\frac{5}{18}}}$ = 20 m/s

v² - u² = 2as

$\implies$ (0)² - (20)² = 2 $\times$ a $\times$ 8

$\implies$-400 = 16a

$\implies$ a = $\large{\bf{\frac{-400}{16}}}$ = -25 m/s²

• Retardation

-(a) = -(-25) = 25 m/s²

• Retarding force = Mass $\times$ Retardation

$\implies$ 160 $\times$ 25

$\implies$ 4000 N

$\large{\boxed{\bf{Retarding \:Force = 4000\:N}}}$