Question

A body starts from the origin and moves along the x-axis such that velocity at any instant is given by $(4t^{3} - 2t)$, where t is in second and velocity is in m/s. What is the acceleration of the particle, when it is 2m from the origin?

• A. $28m/s^{2}$
• B. $22m/s^{2}$
• C. $12m/s^{2}$
• D. $10m/s^{2}$

Answer 1

Answer: (B)

$22m/s^{2}$

Answer 2

Given that $\upsilon = 4t^{3} - 2t$

$x = \int_{}^{}{\upsilon dt}$, $x = t^{4} - t^{2} + C$, at $t = 0,x = 0 \Rightarrow C = 0$

When particle is 2m away from the origin

$2 = t^{4} - t^{2}$ ⇒ $t^{4} - t^{2} - 2 = 0$ ⇒ $(t^{2} - 2)(t^{2} + 1) = 0$

⇒ $t = \sqrt{2}$sec

$a = \frac{d\upsilon}{dt} = \frac{d}{dt}\left( 4t^{3} - 2t \right) = 12t^{2} - 2$ ⇒ $a = 12t^{2} - 2$

for $t = \sqrt{2}$sec ⇒ $a = 12 \times \left( \sqrt{2} \right)^{2} - 2$⇒ $a = 22m/s^{2}$