Question

a car travelling at a speed of 10m/s brought to rest in 20s by applying brakes.calculate the acceleration and distance travelled during this time.​

Answer 1

Given :

• Initial velocity,u = 10 m/s
• Final velocity,v = 0 m/s
• Time taken,t = 20 s

To Find :

• The acceleration
• And distance travelled during this time

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2as$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

${\underline{\sf{Answer}}}$

Case 1 : We have to Find the acceleration of thd car .

From the first equation of motion,we have

v = u + at

⇒ 0 = 10 + a × 20

⇒ 0 = 10 + 20a

⇒ -20a = 10

⇒ a = -0.5

Case 2 : We have to Find the distance travelled by the car during 10sec.

From the Second equation of motion,we have

S = ut + 1/2 at²

S= 10 × 20 + 1/2 × -0.5 × ( 20 )²

S= 200 - 5/20 × 400

S= 200 - 5 × 20

S = 200 - 100

⇒S = 100 m

Therefore, The acceleration of car is -0.5 m/s² and the distance travelled by the car in 10 sec is 100 m

Answer 2

Given:

Initial velocity of car,u= 10 m/s

Final velocity of car,v= 0 m/s

(Since, it starts from rest)

Time taken by car,t= 20 s

To Find:

Acceleration of car

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,

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

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

t is time taken

$\rule{190}{1}$

Let the acceleration of car be a and distance travelled by car be s

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

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

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

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

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

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

$\longrightarrow\rm\green{a=-0.5\ m/s^{2}}$

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

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

$\longrightarrow\rm{(0)^{2}-(10)^{2}=2(-0.5)s}$

$\longrightarrow\rm{-100=-s}$

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

Hence, acceleration of car is -0.5 m/s² and distance travelled during this time is 100 m.