Question

A body is moving with uniform acceleration covers $200m$ in the first $2s$ and $220m$ in the next $4s$. Find the velocity in $m{s}^{-1}$ after $7s$.

• A. $10$
• B. $15$
• C. $20$
• D. $30$

Answer 1

The correct option is A

$10$

Step 1. Given data

Distance covered in first case $S_1=200m$

Time taken in the first case, $t_1=2s$

Distance covered in second case, $S_2=220m$

Time taken in the second case, $t_2=4s$

Step 2. Finding the acceleration and the initial velocity

By using the formula, $s=ut+\frac{a{t}^2}{2}$ [ where, $s$is the distance covered, $u$ is the initial speed,$t$ is time, $a$ is acceleration in SI units.]

The distance covered in $t_1=2s$,

$S_1=u{t}_1+\frac{a{t}_1^2}{2}$

$200=2u+2a$…$eq\left(1\right)$

The distance covered in $t_2=(4+2)s=6s$,

$S_2=u{t}_2+\frac{a{t_2}^2}{2}$

$220=6u+18a$ ………. $eq\left(2\right)$

Solving both equations, we get,

$a=-15m{s}^{-2}$ ,

$u=115m{s}^{-1}$

Now, Using the first equation of motion, we get

Step 3. Finding the final velocity, $v$ after time, $t=7s$

$v=u+at$

$$\begin{gathered} v=115-15\times 7 \\ =10m{s}^{-1} \end{gathered}$$

Therefore, the velocity is $10m{s}^{-1}$.

Hence, Option $\left(A\right)$ is the correct.