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
`         									A particle starts from the origin with a velocity of 10 m/s and moves with a constant acceleration till the velocity increases to 50 m/s. At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returns to the starting point?												(1)			Zero							(2)			10 m/s							(3)			50 m/s							(4)						70 m/s			 			At the instant, the acceleration is reversed, v= 50 so to come back to the starting point it will first have to attain 0 velocity, isnt it?						`
one year ago

763 Points
```							10 m/s is the correct answer .we can calculate it by using equation of kinematics .let the acceleration be : a m/s2by using 3rd equation of motion, 2as=v^2 – u^2 .     . calculate the distance covered in attaining the velocity of 50 m/s.  2as=v^2-u^2 , putting the value of v and u we get , s=1200/a m.now, when the acceleration is reversed then the body will slow and then reverse the direction of motion.calculating the velocity of body after covering 1200/a m will give the velocity of object at starting point .using 3rd  equation of motion , 2as=v^2-u^2 , putting the value os a , s and u we get ,-2400=v^2-2500 ,   v^2 = 100 , hence v = 10m/s.
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Physical Chemistry

View all Questions »

### Course Features

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

### Course Features

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