Question

A car of mass 1200kg going 30m/s applies it's break and comes rest. If friction force between the Tyres and the road is 600n , how far does the car moves before coming to rest ?

Answer 1

Given:-

Mass of car (m) = 1200kg

force (f) = -600N [ due to friction ]

Initial velocity (u)= 30m/s

Final velocity (v)= 0 [break applied]

To find :-

● Distance (s) covered by car before come in rest !

• First find the value of acceleration (a)

• By this formula we can find the value of acceleration

$\boxed{\sf{\dag \ \ Force (f)= Mass (m)\times Acceleration (a)}}$

$\longmapsto\sf f= ma\\ \\ \longmapsto\sf -600= 1200\times a\\ \\ \longmapsto\sf \cancel{\dfrac{-600}{1200}}=a\\ \\ \longmapsto\sf -0.5m/s^2= a\\ \\ \longmapsto {\boxed{\sf {Acceleration (a)= -0.5m/s^2}}}$

◆Now we have to find the distance (s)

$\underline{\bigstar{\sf{\ \ By\ 3rd \ equation \ of \ motion}}}$

$\boxed{\sf{\dag\ \ v^2= u^2+2as}}$

Where ,

✧v= final velocity

✧u=initial Velocity

✧a = acceleration

✧ s= Distance

Find the time !

$\longmapsto\sf v^2=u^2+2as\\ \\ \longmapsto\sf (0)^2= (30)^2= 2\times -0.5\times s\\ \\ \longmapsto\sf -900= -1\times s\\ \\ \longmapsto\sf\cancel{\dfrac{-900}{-1}}=s\\ \\ \longmapsto\sf s= 900m$

$\underline{\sf{\star\ \ The \ car \ moves \ 900m \ before \ come\ in \ rest }}$

Answer 2

Given :

• Initial velocity (u) = 30ms¯¹
• Final velocity (v) = 0ms¯¹
• Mass of car (m) = 1200 kg
• Frictional Force (F) = - 600 N

To Find :

• Distance travelled by the car before coming to rest

Concept :

When breaks are applied, final velocity(v) of car will be 0.

And as the force is frictional force which acts opposite to the direction so its value is taken as negative.

Formula used :

$v = u + at \\ s = ut + \frac{1}{2} a {t}^{2} \\ 2as = {v}^{2} - {u}^{2}$

Solution :

As we know,

F = ma, where F is the force, m is mass and a is acceleration.

Then,

F = ma

-600 N = 1200 ×a

-600/1200 ms¯² = a

-0.5 ms¯² = a

Now, for finding the distance

$2as = {v}^{2} - {u}^{2} \\ 2 \times ( - 0.5) \times s = {0}^{2} - {30}^{2} \\ - 1 \times s = - 900 \\ s = 900 \: m$

Hence, the car moves 900 m before coming to rest.