Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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
Kaushik Ram 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)

```
6 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^2..so the minimum time required is  about 14 sec.
```
6 years ago
510 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
```
6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Mechanics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details