Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 10

A car starts from rest. It has to cover a distance of 500 m. the coefficient of friction between the road and

the tyre is 0.5. The minimum time in which the car can cover this distance is (g=10m/s2


9 years ago

Answers : (2)

Aniket Patra
48 Points

As soon as the car sets into motion the friction force acting on its tyre would be equal to the limiting value of the static friction which is equal to 5m (where m is the mass of the car).So this force provides the acc which is equal to 5m/s^ the minimum time required is  about 14 sec.

9 years ago
vikas askiitian expert
509 Points

 friction = u(mg)

     ma = umg

        a = ug =5m/s2             (retardation)

now we have

                 V2 = U2 - 2aS      

             S=500 , V=0 , U=? ,a =5 

 after substituting these we get

   U = (5000)1/2

now we have ,

                       V = U -at

             V =0 , U = (5000)1/2 , a=5

    substituting these we get

 t = (200)1/2 = 10sqrt2 sec

9 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details