Question

particle moves along a straight line. Its position (x) at any instant is given by x= 32t - 8t3/3, where x is in metre and t in second. Find the acceleration of the particle at the instant particle comes to rest​

Answer 1

The acceleration of the particle at the instant particle comes to rest​ is $-32\ m/s^2$.

Explanation:

A particle moves along a straight line. Its position (x) at any instant is given by :

$x=32t-\dfrac{8t^3}{3}$ ........(1)

Differentiate equation (1) wrt t as :

$\dfrac{dx}{dt}=\dfrac{d(32t-\dfrac{8t^3}{3})}{dt}\\\\since, \dfrac{dx}{dt}=v\\\\v=32-8t^2$ ..........(2)

When the object is at rest, v = 0

$32-8t^2=0\\\\t=2\ s$

Differentiate equation (2) wrt t, then :

$since,\ \dfrac{dx}{dt}=a\\\\a=-16t$

At t = 2 s,

$a=-32\ m/s^2$

So, the acceleration of the particle at the instant particle comes to rest​ is $-32\ m/s^2$.

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Answer 2

be carefull , to understand this question u should know the relation between 'x,v and a'