Question

Velocity of a particle moving in a straight line varies with its displacement asv=(4+4S)−−−−−−−√ ms−1 . Find its displacement in the first 2 seconds of its motion
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Answer 1

Answer:

8m

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Answer 2

Explanation:

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Velocity of a particle moving in a straight line varies with its displacement asv=(4+4S) ms−1 . Find its displacement in the first 2 seconds of its motion

$\mathbb{\huge{\underline{\underline{\underline{\underline{AnsWer}}}}}}$

It's a differential equation.

v = (4+4s)${}^{0.5}$

v =ds/dt

ds/dt = (4+ 4s)${}^{0.5}$

ds/(4+4s)${}^{0.5}$ = dt

Integrating both sides we get,

0.5×(4+4s)${}^{0.5}$ + C = t

at t=0, s=0. so we get

C = -1

therefore, at t = 2s we get

s = (t+1)² - 1 = 8m

So,the answer is 8m

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