Question

A particle moving with a uniform acceleration travels $24m$ and $64m$ in the first two consecutive intervals of $4s$ each. Its initial velocity will be.

• A. $5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• B. $3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• C. $4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• D. $1\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Answer 1

The correct option is D

$1\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step1: Given data and assumptions.

Distance covered by particle $s_1$$=24m$

Distance covered by particle $s_2$$=64m$

Time is taken by distance $s_1$, $t_1=4s$

Time is taken by distance $s_2$, $t_2=4s$

Let the initial velocity be $u$

Step2: Finding the initial velocity.

Formula used:

$s=ut+\frac{1}{2}a{t}^2$

Where $s$ is distance, $u$ is initial velocity, $a$ is acceleration, $t$ is time.

Now,

When $s=s_1=24m$, $t=t_1=4s$ then,

$24=4u+8a$ ….$\left(i\right)$

Again, we have.

Total distance, $s=s_1+s_2=24+64=88m$

Total distance, $t=t_1+t_2=8s$

Since,

$s=ut+\frac{1}{2}a{t}^2$

$88=8u+32a$ …..$\left(ii\right)$

On solving equation $\left(i\right)$ and $\left(ii\right)$ we get.

$u=1\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Thus, its initial velocity will be $u=1\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Hence, option D is correct.