Question

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

• A. $\frac{t^3}{3}-t^2+2t+4$
• B. $\frac{t^3}{3}-t^2+2t$
• C. $\frac{t^3}{3}-t^2+4$
• D. $\frac{t^3}{3}-t^2$

Answer 1

The correct option is A $\frac{t^3}{3}-t^2+2t+4$

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