Question

The velocity of a body moving along a straight varies with time as $V=t^2-4tm/s$, find the average acceleration of the body during an interval $t=1sec$ to $t=3sec$

• A. $0m/S^2$
• B. $6m/S^2$
• C. $3m/S^2$
• D. $1m/S^2$

Answer 1

The correct option is A $0m/S^2$

Velocity of the body is given by,

$v=t^2-4t$ m/s

Acceleration of the body is the rate of change of velocity w.r.t time.

$\therefore a=\frac{dv}{dt}$

$\therefore a=\frac{d}{dt}[t^2-4t]$

$\therefore a=2t-4$

At $t=1$ sec, $a_1=2\left(1\right)-4$

$\therefore a_1=-2$

At $t=3$ sec, $a_2=2\left(3\right)-4$

$\therefore a_2=6-4=2$

Thus, average acceleration during interval $1$ sec and $3$ sec is,

$a_{avg}=\frac{a_1+a_2}{2}$

$\therefore a_{avg}=\frac{-2+2}{2}$

$\therefore a_{avg}=0$