Question

The position of an object moving along an x axis is given by x=3t-4t^2+t^3 . Find accelaration at t=2 sec and t=4 sec

Answer 1

Answer:

a=4 & a=16

Explanation:

X=3t-4t²+t³

X=t³-4t²+3t

by differentiation→

$average \: velocity \: = \frac{dx}{dt}$

$v = 3 {t}^{2} - 8t + 3$

$acceleration = \frac{dv}{dt}$

$a = 6t - 8$

Now the acceleration will be..

6×2-8=4

6×4-8=16

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