Question

A particle starts from position x=${-x}_0$ at time t=0 in positive x direction according tot he give velocity-time (v-t) graph. The position (x) of particle at any time t can be represented as :

• A. $x=x_0+v_{0}t+\frac{1}{2}\frac{v_0}{t_0}{t}^2$
• B. $x=-x_0+\frac{1}{2}\frac{v_0}{t_0}{t}^2$
• C. $x=x_0+\frac{1}{2}\frac{v_0}{t_0}{t}^2$
• D. $x=-x_0-\frac{1}{2}\frac{v_0}{t_0}{t}^2$

Answer 1

The correct option is B $x=-x_0+\frac{1}{2}\frac{v_0}{t_0}{t}^2$

Given: $x=-x_0;t=t_0$

Use equation of motion: $v=u+at$

$v_0=\left(0\right)+a{t}_0$

$v_0=a{t}_0$

$a=\frac{v_0}{t_0}......Eq:01$

Now, use equation of motion $s=ut+\frac{1}{2}a{t}^2$ for distance/position (x) at any time (t);

$x=\left(0\right)t+\frac{1}{2}a{t}^2$

$x=\frac{1}{2}a{t}^2$

Put the value of 'a' from Eq:01;

$x=\frac{1}{2}\left(\frac{v_0}{t_0}\right)t^2$

Required position of particle at any time (t): $x=-x_0+\frac{1}{2}\left(\frac{v_0}{t_0}\right)t^2$

Option (B) is correct.