Question

Acceleration of the particle in a rectilinear motion is given as $a=4t+3$ in $\left(m/s^2\right)$. The velocity and displacement as a function of time $t$ respectively are
(Given that at $t=0,u=3m/s,x=0m$)

• A. $(t^2+3),(\frac{t^3}{3}+\frac{3t^2}{2}+3t)$
• B. $(t^2+3t),(\frac{t^3}{3}+\frac{3t^2}{2})$
• C. $(2t^2+3t+3),(\frac{2t^3}{3}+\frac{3t^2}{2}+3t)$
• D. $(2t^2+3t),(\frac{2t^3}{3}+\frac{3t^2}{2})$

Answer 1

The correct option is C $(2t^2+3t+3),(\frac{2t^3}{3}+\frac{3t^2}{2}+3t)$

Given, $a\left(t\right)=4t+3$
As we know that the velocity as a fuction of time is given by
${\int }_u^{v}dv={\int }_0^t(4t+3)dt$
$\Rightarrow {\left[v\right]}_u^v=[2t^2+3t{]}_0^t$
$\Rightarrow v-u=2t^2+3t$
Or, $\Rightarrow v-3=2t^2+3t\Rightarrow v\left(t\right)=2t^2+3t+3$
Now, displacement is given by
$x\left(t\right)={\int }_0^{t}v\left(t\right)dt={\int }_0^t(2t^2+3t+3)dt$
$\Rightarrow x\left(t\right)={[\frac{2t^3}{3}+\frac{3t^2}{2}+3t]}_0^t$
$\Rightarrow x\left(t\right)=[\frac{2t^3}{3}+\frac{3t^2}{2}+3t]$