Question

When a car driver traveling at speed of 10meter per second applies breaks and brings the car to be rest in 20second than retardation will be?
Fast pls!!!?​

Answer 1

Answer:

• Retardation = 0.5 m/s²

Given:

• Initial Speed = 10 m/s
• Time taken for the car to stop = 20 seconds

To Find:

• Retardation = ?

Solution:

Formula to calculate Retardation is:

$\boxed{\text{Retardation} = \dfrac{v-u}{t}}$

where, 'v' is the final velocity, 'u' is the initial velocity and 't' is the time taken.

According to the question, the car is coming to rest in 20 seconds. Hence the final velocity 'v' of the car is 0 m/s.

Hence Substituting the values in the formula we get:

$\implies a = \dfrac{0-10}{20}\\\\\\\implies a = \dfrac{-10}{20}\\\\\\\implies \boxed{ a = -0.5\:m/s^2}$

Hence the acceleration of the car would be -0.5 m/s², where '-' sign indicates retardation.

Answer 2

$\Large\bf{\color{cyan}GiVeN,}$

• A car driver travelling at speed of 10 m/s.

$\longmapsto\:\:\bf\blue{Initial\:velocity\:(u)\:=\:10\:m/s}$

• Driver applies breaks and brings the car to be rest in 20 second.

☆ Driver applies breaks, means car is being stopped after 20 s, i.e.

$\longmapsto\:\:\bf\purple{Final\:velocity\:(v)\:=\:0\:m/s}$

$\longmapsto\:\:\bf\pink{Time\:(t)\:=\:20\:s}$

$\red\bigstar\:\:{\color{olive}{\boxed {\boxed {\bf{\color{peru}v\:=\:u\:+\:a\:t\:}}}}}$

$\bf\orange{Where,}$

• a be the acceleration, which acts as retardation in this case. Because breaks are applied, that's why the tires of the car exert an force in backward direction. So acceleration results a negative value, which is known as retardation.

$:\implies\:\:\bf{0\:=\:10\:+\:a\times{20}\:}$

$:\implies\:\:\bf{20a\:=\:-10\:}$

$:\implies\:\:\bf{a\:=\:\dfrac{-10}{20}\:}$

$:\implies\:\:\bf{a\:=\:-\dfrac{1}{2}\:}$

$:\implies\:\:\bf\green{a\:=\:-0.5\:m/s^2\:}$

$\Large\bold\therefore$ The retardation of car is -0.5 m/s².