Question

A car is moving with a speed of 20m/s can be stopped by applying brakes , after travelling a distance of 2 m . The retardation of the car is 

a) 50 m/s²
b) 100 m/s² 
c) 150 m/s² 
d) 200 m/s²

Answer 1

$\large{\red{\underline{\underline{\bold{Given:-}}}}}$

• Initial velocity = 20m/s

• Final velocity = 0m/s

• Distance covered = 2m

$\large{\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• The retardation of the car

$\large{\green{\underline{\underline{\bold{Solution:-}}}}}$

According to the third equation of motion

$\large\boxed{ \rm \pink{ \implies{v}^{2} - {u}^{2} = 2as}}$

Here:-

• v = final velocity

Here v = 0m/s because the car comes to rest after applying the brakes

• u = initial velocity

• a = acceleration ( retardation)

• s = distance covered

Substituting the values:-

${ \rm { \implies{0}^{2} - {20}^{2} = 2a(2)}}$

${ \rm { \implies - 400 = 4a}}$

${ \rm { \implies \dfrac{ - 400}{4} = a}}$

${ \rm { \implies - 100 = a}}$

Here negative sign indicates retardation

Hence;

$\large \boxed{ \rm \purple{ \therefore Retardation \: = 100m/{s}^{2}} }$

Answer 2

Given:

• Speed of car = 20 m/s
• It is stopped by applying brakes
• And travelled 2m

To find:

• Retardation of car

Solution:

Using the third equation of motion

v² - u² = 2as

Where

• v → final velocity = 0 m/s
• u → initial velocity = 20 m/s
• a → acceleration or retardation = ?
• s → distance travelled = 2m

Here , final velocity ( v ) = 0 m/s because the car is stopped by applying brakes

Substituting the values we have

→ 0² - 20² = 2(a)(2)

→ - 400 = 4a

→ a = -400/4

→ a = -100 m/s²

Here , -ve symbol indicates retardation

Hence , retardation of the car is 100m/s²