Question

A particle starts from rest and has acceleration $a=2t-4$ in $\text{m}/s^2$. The displacement of the particle as a function of time is
[Take $x=0$ at $t=0$]

• A. $x\left(t\right)=\frac{2t^3}{3}-2t^2$
• B. $x\left(t\right)=\frac{t^3}{3}+t^2$
• C. $x\left(t\right)=\frac{t^3}{3}-2t^2$
• D. $x\left(t\right)=t^3-2t^2$

Answer 1

The correct option is C $x\left(t\right)=\frac{t^3}{3}-2t^2$

Given, $a\left(t\right)=2t-4$
As we know that $a=\frac{dv}{dt}\Rightarrow {\int }_0^{v}dv={\int }_0^{t}a\left(t\right)dt$
and also $v=\frac{dx}{dt}\Rightarrow {\int }_0^{x}dx={\int }_0^{t}v\left(t\right)dt$
Thus, $v={\int }_0^{t}a\left(t\right)dt={\int }_0^t(2t-4)dt$
$\Rightarrow v\left(t\right)=t^2-4t$
also, $x={\int }_0^{t}v\left(t\right)dt={\int }_0^t(t^2-4t)dt$
$\Rightarrow x\left(t\right)=\frac{t^3}{3}-2t^2$
Hence, displacement of the particle as a function of time is $x\left(t\right)=\frac{t^3}{3}-2t^2$