Question

A BODY STARTS WITH AN INITIAL VELOCITY OF 10M/S AND MOVES ALONG A STRAIGHT LINE WITH A CONSTANT ACCELERATION. WHEN THE VELOCITY OF THE PARTICLE IS 50M/S THE ACCELERATION IS REVERSED IN THE DIRECTION. FIND THE VELOCITY OF THE PARTICLE WHEN IT REACHES THE STARTING POINT.

Answer 1

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Answer 2

Answer:

The velocity of the particle is 70 m/s.

Explanation:

Given that,

Initial velocity u = 10 m/s

Final velocity v = 50 m/s

When the body accelerated

Using equation of motion

$v^2-u^2=2as$

$(50)^2-10^2=2as$

$2as=2500-100$

$2as=2400$....(I)

Now,The acceleration is reversed

Final velocity = 0

Initial velocity = 50 m/s

Using equation of motion again

$v^2-u^2=-2as'$

$-2as'=0-2500$

$2as'=2500$.....(II)

The body starts moving in opposite direction from that point. after coming to the initial point.

The total distance s'' = s+s'

Let the velocity of the body is v'

Using equation of motion again,

$v^2-u^2=2as''$

$v'^2-0=2a(s+s')$

$v'^2=2as+2as'$

$v'^2=2400+2500$

$v'=\sqrt{2400+2500}$

$v'=70\ m/s$

Hence, The velocity of the particle is 70 m/s.