Question

A car is travelling with a speed of 36 km/h. The driver applied the brakes and retards the car uniformly. The car is stopped in 5 sec. Find (i) The acceleration of car and (ii) Distance before it stops after applying breaks.

Answer 1

Answer:

Explanation:

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Answer 2

Given:

Initial velocity of car, u= 36 km/h

Final velocity of car, v= 0 m/s

(Since, driver stops the car)

Time taken by car, t= 5 s

To Find:

Acceleration of car and Distance covered by car before it stops after applying brakes

Solution:

We know that,

• According to first equation of motion for constant acceleration,

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

• According to third equation of motion for constant acceleration,

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

where,

v is final velocity

u is initial velocity

a is acceleration

t is time taken

s is displacement

$\rule{190}{1}$

It is given that,

$\longrightarrow\rm{u=36\:km/h=36\times\dfrac{5}{18}\:m/s=10\:m/s}$

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

Let the acceleration of car be 'a'

On using first equation of motion on car, we get

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

$\longrightarrow\rm{0=10+a(5)}$

$\longrightarrow\rm{-10=5a}$

$\longrightarrow\rm{5a=-10}$

$\longrightarrow\rm{a=\dfrac{-10}{5}}$

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

$\rule{190}{1}$

Let the distance travelled by car after applying brakes be d

So, on applying third equation of motion on lorry, we get

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

$\longrightarrow\rm{(0)^{2}-(10)^{2}=2(-2)d}$

$\longrightarrow\rm{-100=-4d}$

$\longrightarrow\rm{-4d=-100}$

$\longrightarrow\rm{d=\dfrac{-100}{-4}}$

$\longrightarrow\rm\green{d=25\:m}$

Hence, acceleration of car is -2 m/s² and  distance travelled by car after applying brakes is 25 m.