Question

A particle moves in a straight line according to the relation
x=t
3
-4t
2+3t
Find the acceleration of the particle at displacement equal to zero.

Answer 1

Answer:

$\bold\red{a=-2}$

$\bold\red{a=10}$

Explanation:

Given,

A particle moving in straight line According yo relation,

$x = {t}^{3} - 4 {t}^{2} + 3t$

To get the acceleration,

we need to differentiate twice the given equation wrt t,

we get,

$= > \frac{dx}{dt} = \frac{d}{dt} ( {t}^{3} - 4 {t}^{2} + 3t) \\ \\ = > \frac{dx}{dt} = 3 {t}^{2} - 8t + 3$

Again differentiating,

we get,

$= > \frac{ {d}^{2}x }{d {t}^{2} } = 6t - 8$

Therefore,

Acceleration as a function of t is,

$= > a(t) = 6t - 8$

Now,

when displacement is zero,

$= > {t}^{3} - 4 {t}^{2} + 3t = 0 \\ \\ = > {t}^{2} - 4t + 3 = 0 \\ \\ = > {t}^{2} - 3t - t + 3 = 0 \\ \\ = > t(t - 3) - 1(t - 3) = 0 \\ \\ = (t - 1)(t - 3) = 0 \\ \\ = > t = 1 \\ \\ and \\ \\ = > t = 3$

Therefore,

$a(t = 1) = (6 \times 1 )- 8 = 6 - 8 = - 2$

and,

$a(t = 3) = (6 \times 3) - 8 = 18 - 8 = 10$

Hence,

the particle acquires zero displacement at two intervals and two acceleration.