Question

A particle is moving along positive $x-$ axis with velocity $v=A-Bt$, where $A$ and $B$ are positive constants and '$t$' is in seconds. If the particle is at the origin initially, then find the coordinates of the position of the particle at $t=\frac{3A}{B}$.

• A. $(0,\frac{3A^2}{2B})$
• B. $(\frac{3A^2}{2B},0)$
• C. $(0,\frac{-3A^2}{2B})$
• D. $(\frac{-3A^2}{2B},0)$

Answer 1

The correct option is D $(\frac{-3A^2}{2B},0)$



Given that ,

$v=A-Bt$

When $v_x=0\text{m/s},t=\frac{A}{B}\text{sec}$

Comparing this with $v=u+at$, we get
$u_x=A\text{m/s},a_x=-B{\text{m/s}}^2$

Using kinematic relation,
$s_1=ut+\frac{1}{2}a{t}^2$
$x_1=u_{x}t+\frac{1}{2}{a}_x{t}^2,y_1=0$
[because particle is moving along x-axis]

At $t=\frac{A}{B}\text{sec}$, we get

$x_1=A\left(\frac{A}{B}\right)-\frac{B}{2}{\left(\frac{A}{B}\right)}^2$

$\Rightarrow x_1=\frac{A^2}{B}-\frac{A^2}{2B}=\frac{A^2}{2B}......\left(1\right)$

After which the particle will retrace its path.
Now, to find the position of particle after $t^{\prime }=\frac{3A}{B}$, we again use

$s_2=ut+\frac{1}{2}a{t}^2$
$\Rightarrow x_2=v_{x}t+\frac{1}{2}{a}_x{t}^2,y_2=0$

$x_2=0+\frac{1}{2}\times (-B)\times {\left(\frac{2A}{B}\right)}^2=\frac{-2A^2}{B}.....\left(2\right)$

Total displacement, $s=s_1+s_2$
$\Rightarrow x=x_1+x_2,y=y_1+y_2$

Using $\left(1\right)$ and $\left(2\right)$, we get

$x=\frac{A^2}{2B}-\frac{2A^2}{B}=\frac{-3A^2}{2B}$ and $y=0$

$\therefore$ at $t=\frac{3A}{B}\text{sec},$
$(x,y)=(\frac{-3A^2}{2B},0)$