Question

A car is travelling with a speed of 72 km/h. The driver applied the brakes and car retards uniformly. The car is stopped in 10 sec. Find (i) The retardation of car and (ii) Distance travel before it stops after appling breaks?

Answer 1

Given:

Initial velocity of car,u= 72 km/h

Final velocity of car,v= 0 km/h

(Since, car stops Finally)

Time taken by car,t= 10 s

To Find:

(i) The retardation of car

(ii) Distance travel before it stops after applying breaks

Solution:

We know that,

• Retardation is equal to acceleration in magnitude but have opposite sign with respect to acceleration

• According to first equation of motion for constant acceleration,

$\purple{\underline{\boxed{\bf{v=u+at}}}}$

• According to third equation of motion for constant acceleration,

$\pink{\underline{\boxed{\bf{v^{2}-u^{2}=2as}}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

t is time taken

$\rule{190}{1}$

It is given that,

$\longrightarrow\rm{u=72\ km/h=72\times\dfrac{5}{18}\ m/s=20\ m/s}$

$\longrightarrow\rm{v=0\ km/h=0\times\dfrac{5}{18}\ m/s=0\ m/s}$

$\rule{190}{1}$

Let acceleration of car be 'a'

So, on applying first equation of motion on car, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{0=20+a(10)}$

$\longrightarrow\rm{-20=10a}$

$\longrightarrow\rm{a=\dfrac{-20}{10}}$

$\longrightarrow\rm{a=-2\ m/s^{2}}$

So,

$\longrightarrow\rm\green{Retardation=2\ m/s^{2}}$

$\rule{190}{1}$

Let the distance covered by car before it stops after applying brakes be 's'

Also, on applying third equation of motion on car, we get

$\longrightarrow\rm{v^{2}-u^{2}=2as}$

$\longrightarrow\rm{(0)^{2}-(20)^{2}=2(-2)s}$

$\longrightarrow\rm{-400=-4s}$

$\longrightarrow\rm{-4s=-400}$

$\longrightarrow\rm{s=\dfrac{-400}{-4}}$

$\longrightarrow\rm\green{s=100\ m}$

Hence, the retardation of car is 2 m/s² and distance covered by car before it stops after applying breaks is 100 m.

Answer 2

First convert the velocity to m/s from km/h. Here, you get it as 20 m/s.

Now, you have final velocity: v = 0 m/s as the body is going to come at rest.

Whole initial velocity: u = 20m/s,

The acceleration is retarding in nature

So, you have a = (-2 m/s²)

Using the Kinematic Equation :  v=u+at  you get t = 2s

And using  v²=u²+2as  

You have  s=v²/2as  = 400/(2)(2),

i.e. s = 100 m

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