Question

The velocity of the particle is given by $v=3t^2+2t$ in $\text{m/s}$. The acceleration and displacement of the particle as a function of time respectively are
[Take $x=0$ at $t=0$]

• A. $(t+2)m/s^2,(3t^3+t^2)m$
• B. $(6t+2)m/s^2,(t^3+t^2)m$
• C. $(6t^2+2)m/s^2,(t^3+t^2)m$
• D. $(t+2)m/s^2,(t^3+t)m$

Answer 1

The correct option is B $(6t+2)m/s^2,(t^3+t^2)m$

Given, $v\left(t\right)=3t^2+2t$
As we know that acceleration of the particle is given by
$a\left(t\right)=\frac{dv}{dt}=\frac{d(3t^2+2t)}{dt}=(6t+2)m/s^2$
Also, $v=\frac{dx}{dt}\Rightarrow x\left(t\right)={\int }_0^{t}v\left(t\right)dt$
$\Rightarrow x\left(t\right)={\int }_0^t(3t^2+2t)dt$
$\Rightarrow x\left(t\right)=(t^3+t^2)m$