Question

38. A particle moves along a straight line with an initial
velocity(v0), and acceleration('a/v') where 'a' is constant
and 'v' is its instantaneous velocity. The velocity of
particle at time t is related as
(1) v² = v0^2+ a/v*t. (2) v^2/2=v0^2/2+ at

(3) v^2/2=v0^2-at
(4) v^2/2 = v0^2+ at​

Answer 1

Answer:

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acc = dv/dt....

a/V = dv/dt....

So,

a(dt) = V(dv)....

So,

at(integrating from 0 to t) =V^2/2(integrating from Vo to V)

So,

V^2 - Vo^2 = 2 at

V^2 = Vo^2 + 2 at.....

Option (b) is correct....

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