Question

A particle moving along the x-axis has a position given by x = (24t – 2.0t 3 ) m, where t is measured in s. What is the magnitude (m/s2 ) of the acceleration of the particle when the particle is not moving?

Answer 1

The magnitude 24 ($\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving.

Explanation:

Given,

A particle moving along the x-axis has a position given by

$x=(24t-2.0t^3)$ m      ........ (1)

To find, the magnitude ($\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving = ?

Differentiating equation (1) w.r.t, 't', we get

$\dfrac{dx}{dt} =\dfrac{d((24t-2.0t^3))}{dt}$

⇒ $\dfrac{dx}{dt} =24(1)-3(2.0)t^{2} =24-6t^{2}$     ....... (2)

⇒ $24-6t^{2} = 0$

⇒ $t^{2}=2^{2}$

⇒ t = 2 s

Again, differentiating equation (2) w.r.t, 't', we get

$\dfrac{d^2x}{dt^2} =-12t$

Put t = 2, we get

$\dfrac{d^2x}{dt^2} =-12(2)=24$

Thus, the magnitude 24 ($\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving.