Question

The velocity of a bodyoving along a straight line varies with time as v equal to 2t square e pawer-t the a of the body will be zero at t

Answer 1

Question:

The velocity (v) of a body moving along a straight line varies with time as $\sf v = 2t^2 e^{-t}$ . Find time (t) at which the acceleration (a) of body is zero.

Answer:

$\boxed{\sf Acceleration \ of \ body \ is \ zero \ at \ t = 0 \ and \ 2 \ units }$

Given:

$\sf v = 2 {t}^{2} {e}^{ - t}$

To Find:

Time (t) at which acceleration (a) of body is zero.

Explanation:

As we know;

$\boxed{ \bold{a = \frac{dv}{dt} }}$

$\sf \implies a = \frac{d}{dt} (2 {t}^{2} {e}^{ - t} )$

Factor out constant term:

$\sf \implies a = 2 \frac{d}{dt} ( {t}^{2} {e}^{ - t} )$

Using the chain rule:

$\sf \implies a = 2 ({t}^{2} \frac{d}{dt} ( {e}^{ - t} ) + {e}^{ - t} \frac{d}{dt} ( {t}^{2} ))$

$\sf \implies a = 2 ({t}^{2} {e}^{ - t} \frac{d}{dt} ( - t) + {e}^{ - t} \frac{d}{dt} ( {t}^{2} ))$

$\sf \implies a = 2 ({t}^{2} {e}^{ - t} \times ( - 1)+ {e}^{ - t} \frac{d}{dt} ( {t}^{2} ))$

$\sf \implies a = 2 ( - {t}^{2} {e}^{ - t} + {e}^{ - t} \frac{d}{dt} ( {t}^{2} ))$

$\sf \implies a = 2 ( - {t}^{2} {e}^{ - t} + {e}^{ - t} \times 2t)$

$\sf \implies a = 2 ( - {t}^{2} {e}^{ - t} + 2t{e}^{ - t} )$

At a = 0:

$\sf \implies 2 ( - {t}^{2} {e}^{ - t} + 2t{e}^{ - t} ) = 0$

$\sf \implies - {t}^{2} {e}^{ - t} + 2t{e}^{ - t} = 0$

$\sf \implies t{e}^{ - t} ( 2 - t)= 0$

$\sf \implies t = 0 \ and \ t = 2 \ \: units$

$\therefore$

Time (t) at which acceleration (a) of body is zero = 0 & 2 units

Answer 2

Answer:

By differentiating v and putting a = 0 we get t = 0 and 2