#### Thank you for registering.

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

Click to Chat

1800-5470-145

+91 7353221155

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

# A car moving with a speed of 40 km/h can be stopped by applying brakes at least after 2m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance? (a)8m (b)6m (c)4m (d)2m

Jitender Pal
7 years ago

(a)
If F is retarding force and s the stopping distance then
1/2  m2   =Fs
For same retarding force, s a v2
s_2/s_1 = (v_2/v_1 )^2  =((80 km/h)/(40 km/h))^2  =4
s2 = 4s1 = 4 × 2 = 8m

vivek sharma
13 Points
4 years ago
F is retarding force, and x is stopping distance. Then F. x = 1/2 mv² F. 2 =1/2m(40)² F=1/4.m.1600 X’ = mv²/2F X’ = 6400m/1600m = 8 metre
Nazir Hossain
11 Points
3 years ago
First calculate the acceration in first case v ^2= 2aSa=v^2/2s , a= 1600/2*2, a=400m/s^2Now, calculate distance in 2nd caseS= v^2/2a , S= 6400/2*400, S= 8 m
ammalu
13 Points
3 years ago
we known that v2=u2+2as then a=u2/2s =(50* 5/ 18)2 /2*6
s=u2/2a then (100*5/18)       =24
2 * (50*5/ 18)2
12