Question

What is the position at any time, for a body starting
from rest, with an acceleration a = bt2
?​

Answer 1

Answer:

a=dv/dt=at2 0vdv= otat2 dt 0vv = 0tat3/3v= at3/3dx/dt=at3/3x=at4/12

Explanation:

Answer 2

Given that,

Acceleration of the body is :

$a=bt^2$

t is time taken

To find :

Position of the body.

Solution:

Velocity is given by :

$v=\int\limits {adt}\\\\v=\int\limits {(bt^2)dt}\\\\v=\dfrac{bt^3}{3}$

Let x is the position of body. It is given by :

$x=\int\limits{vdt}\\\\x=\dfrac{1}{3}\int\limits{bt^3dt}\\\\x=\dfrac{b}{3}\int\limits{t^3dt}\\\\x=\dfrac{bt^4}{12}$

So, the position of the body is $\dfrac{bt^4}{12}$.

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