
A car having a mass of 1000 Kg is moving at a speed of 30 m/sec. Brakes are applied to stop the car so that it comes to rest. If the frictional force between the tyres of the car and the road surface is 5000 N, the car will stop/comes to rest in
(A) 5 seconds
(B) 10 seconds
(C) 12 seconds
(D) 6 seconds
Answer
161.4k+ views
Hint:Use the first equation of motion and put all the required value from the question in that equation such as initial velocity given, final velocity will be zero as we have to stop the car, force and mass is given we can find the acceleration also and finally after putting all these we get the time required.
Formula used:
First equation of motion: $v = u + at$
Where, v is final velocity.
u is initial velocity.
a is acceleration.
And t is time.
Also, acceleration $a = \dfrac{F}{m}$
Complete answer:
Let start with the given information:
Initial velocity, $u = 30m/s$.
Final velocity, v = 0.
We know that: acceleration $a = \dfrac{{Force}}{{mass}} = \dfrac{F}{m}$
$a = - \dfrac{{5000}}{{1000}} = - 5m/{s^2}$
Now, from first equation of motion;
$v = u + at$
$0 = 30 + ( - 5t)$
By solving;
$t = 6\sec $
Therefore, the time required to stop the car is 6 seconds.
Hence, the correct answer is Option(D).
Note: Here the body given that is car needs to be stopped by applying brakes that’s why the final velocity is zero but it’s not the case all the time so be careful about the same and put the values accordingly from the information provided in the question. Also check the units before putting the values.
Formula used:
First equation of motion: $v = u + at$
Where, v is final velocity.
u is initial velocity.
a is acceleration.
And t is time.
Also, acceleration $a = \dfrac{F}{m}$
Complete answer:
Let start with the given information:
Initial velocity, $u = 30m/s$.
Final velocity, v = 0.
We know that: acceleration $a = \dfrac{{Force}}{{mass}} = \dfrac{F}{m}$
$a = - \dfrac{{5000}}{{1000}} = - 5m/{s^2}$
Now, from first equation of motion;
$v = u + at$
$0 = 30 + ( - 5t)$
By solving;
$t = 6\sec $
Therefore, the time required to stop the car is 6 seconds.
Hence, the correct answer is Option(D).
Note: Here the body given that is car needs to be stopped by applying brakes that’s why the final velocity is zero but it’s not the case all the time so be careful about the same and put the values accordingly from the information provided in the question. Also check the units before putting the values.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Class 11 JEE Main Physics Mock Test 2025

Differentiate between audible and inaudible sounds class 11 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
