Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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/s 2 )


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



 


Grade:10

2 Answers

Aniket Patra
48 Points
10 years ago

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^2..so the minimum time required is  about 14 sec.

vikas askiitian expert
509 Points
10 years ago

 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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free