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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?

 
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?

Grade:12th pass

1 Answers

Adarsh
768 Points
6 years ago
10 m/s is the correct answer .
we can calculate it by using equation of kinematics .
let the acceleration be : a m/s2
by 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.

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