Question

If $a=(3t^2+2t+1)m/s^2$ is the expression according to which the acceleration of a particle moving along a straight line varies, then find displacement of the particle 2 seconds after starting.

• A. $26m$
• B. $\frac{26}{3}m$
• C. $\frac{30}{7}m$
• D. $\frac{26}{7}m$

Answer 1

The correct option is B $\frac{26}{3}m$

$a=(3t^2+2t+1)$
${\int }_0^{v}dv={\int }_0^t(3t^2+2t+1dt)(\because a=\frac{dv}{dt})$
$v=t^3+t^2+t$
$v=\frac{ds}{dt}$
$ds=vdt$

Assume it will cover $s$ displacement in 2 seconds.
${\int }_0^{s}ds={\int }_0^2(t^3+t^2+t)dt$
$s={[\frac{t^4}{4}+\frac{t^3}{3}+\frac{t^2}{2}]}_0^2$
$s=4+\frac{8}{3}+2$
$s=\frac{12+8+6}{3}=\frac{26}{3}m$