Question

a particle starting from rest undergoes accl. given by a is equal to (t-2) meter per second. where t is time in second. find velocity of particle after 4 sec. ​

Answer 1

Answer:

a = dv/dt

So dv = adt

Integrating both sides,

V= ✓tdt - ✓2dt

= t^2/2 - 2t

Putting t = 4

V = 0

Answer 2

$\huge \fcolorbox{red}{pink}{Solution :)}$

Given ,

Acceleration = (t - 2) m/s²

Time = 4 sec

We know that , the acceleration is given by

$\large \sf \fbox{Acceleration = \frac{dv}{dt} }$

Thus ,

➡(t - 2) = dv/dt

➡dv = (t - 2)dt

At t = 4 , we get

➡v = (4 - 2)2

➡v = (2)2

➡v = 4 m/s

Hence , the velocity of particle is 4 m/s

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