Question

At $t=0$, a particle starts from $(4,0)$ and moves towards positive $x-$axis with speed of $v=8t+3\text{m/s}$. The final position of the particle as a function of time is

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

Answer 1

The correct option is C $4t^2+3t+4\text{m}$

Given, particle starts from $(4,0)$, so the initial position of the particle is $x_i=4$
As we know that,
$\int dx={\int }_0^{t}v\left(t\right)dt$
$\Rightarrow {\int }_{x_i}^{x_f}dx={\int }_0^t(8t+3)dt$
$\Rightarrow (x_f-x_i)=[4t^2+3t{]}_0^t$
$\Rightarrow x_f-4=4t^2+3t\Rightarrow x_f=4t^2+3t+4\text{m}$