Question

A particle start from rest with a velocity of $10\, m / s$ and moves with a constant acceleration till the velocity increases to $100\, m / s$. At an instant the acceleration is simultaneously reversed, what will be the velocity of the particle when it comes back to the starting point?

• A. 10 m / s
• B. 20 m / s
• C. 30 m / s
• D. 40 m / s

Answer 1

Answer: (A)

According to relation $= v ^{2}- u ^{2}=2 as$
$(100)^{2}-(10)^{2} =2 as $
$10000-100 =2 as $
$9900 =2 as$
Now acceleration just reversed $= a =- a$
Particle comes back to original position $v=?$
$v ^{2}- u ^{2}=2$ as
$(v)^{2}-(100)^{2}=2(-a) s $
$v^{2}=10000-2 as $
$v^{2}=10000-2 as$
Put 2as $=9900$
$v ^{2}=100$
$v =\sqrt{100}=10 m / s$