Question

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

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

Answer 1

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

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