Question

A body travelling with uniform acceleration moves 64m in first 4 seconds and 200m in next 7 seconds. Find the acceleration of the body.

• A. $a=\frac{3.5}{80}\frac{m}{s^2}$
• B. $a=\frac{35}{80}\frac{m}{s^2}$
• C. $a=\frac{80}{35}\frac{m}{s^2}$
• D. $a=\frac{0.8}{35}\frac{m}{s^2}$

Answer 1

The correct option is C

$a=\frac{80}{35}\frac{m}{s^2}$

Let, $u$ be the initial velocity of the body, and $a$ be the acceleration of the body.

The distance travelled by the body in the first 4 seconds is 64 m.

Given $S_1=64mand{t}_1=4s$

We know $S=ut+\frac{1}{2}a{t}^2$

$S_1\Rightarrow 64=(u\times 4)+\frac{1}{2}a\times 4^2$

$\Rightarrow u+2a=16\to \left(1\right)$

Given the body travels 200 m in next 7 seconds. So, the total distance travelled by the body in $4s+7s=11s$ is $64m+200m=264m$

$S_2=64+200=264mand{t}_2=4+7=11s$

Also, $S_2\Rightarrow 264=(u\times 11)+\frac{1}{2}a\times {11}^2$

$\Rightarrow u+5.5a=24\to \left(2\right)$

Solving (1) and (2),

equation (2)-(1)

$\Rightarrow 3.5a=8$

$\Rightarrow a=\frac{80}{35}\frac{m}{s^2}$