Question

If velocity of particle varies with time is given as v=3tsquare_2t+5,then find acceleration of particle at t=2sec.

Answer 1

acceleration=dv/dt

we are given ...

v=3t^2-2t+5

so differentiation with respect to time..

dv/dt =6t-2
a=6t-2

so acceleration at t=2sec

a=6×2-2
=10m/s^2

Answer 2

HEYA!!
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Given that ,

V = 3t^2 - 2t + 5

Now, Differentiate eq with respect to time ,


DV /DT. = D ( 3t^2 - 2t + 5 ) /DT

We have ,

6t - 2

[ as differentiation of a constant is 0 and for x^n we have nx^n-1 ]

Now , Acceleration at t = 2

Acceleration = dv/dt

Put t = 2 in the differentiated equation ..

6 (2) - 2

12 - 2

= 10 ms-2


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