A car of mass $2000 K g$ moving with a velocity of $72 k m / h$ is brought to rest by the application of brakes within a distance of $20 m$ . Find the average braking force and the time taken for the car to come to rest.

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Solution

Given,

Mass of car, $M = 2000 k g$

Velocity, $V = 72 \times \frac{5}{18} = 20 m s^{- 1}$

Apply kinematic equation of motion

$v^{2} - u^{2} = 2 a s$

$0 - 20^{2} = 2 a \times 20$

$a = - 10 m s^{- 2}$

Breaking force, $F = m a = 2000 \times 10 = 20 k N$

Apply first kinematic equation

$v = u + a t$

$t = \frac{v - u}{a} = \frac{0 - 20}{- 10} = 2 sec$

Breaking force is $20 k N$ and time is $2 sec$ .

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A car of mass 2000 Kg moving with a velocity of 72 km/h is brought to rest by the application of brakes within a distance of 20 m. Find the average braking force and the time taken for the car to come to rest.

Given, Mass of car, M=2000kg Velocity, V=72×518=20ms−1 Apply kinematic equation of motion v2−u2=2as 0−202=2a×20 a=−10ms−2 Breaking force, F=ma=2000×10=20 kN Apply first kinematic equation v=u+at t=v−ua=0−20−10=2 sec Breaking force is 20 kNand time is 2sec.

A car of mass $2000 K g$ moving with a velocity of $72 k m / h$ is brought to rest by the application of brakes within a distance of $20 m$ . Find the average braking force and the time taken for the car to come to rest.