Question

# A body is travelling with uniform acceleration and travels $84m$ in the first $6seconds$ and $180m$ in the next $5seconds$. Find the initial velocity and acceleration of the body.

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Solution

## Step1: Given dataA body is travelling with uniform acceleration and travels 84 m in the first 6 seconds and 180 m in the next 5 seconds.Step 2: Formula usedThe motion of an object in one dimension is defined by three equations of motion. They're utilized to calculate displacement, velocity, and acceleration components. The second law of motion establishes the relationship between a body's displacement, starting velocity, time taken, and acceleration.$s=ut+\frac{1}{2}a{t}^{2}\left[wheres=displacement,u=initialvelocity,t=time,a=acceleration\right]$Step 3: Calculating the initial velocity and acceleration Let initial velocity be $u$ and a is acceleration.According to the given data, the first distance covered by the body is 84 m in 6 seconds. ${s}_{1}=84m,{t}_{1}=6s$Then the body covers the next 180 m in 5 seconds.${s}_{2}=180m,{t}_{2}=5s$Therefore the net displacement and total time will be,${s}_{net}={s}_{1}+{s}_{2}=84+180=264m\phantom{\rule{0ex}{0ex}}{t}_{totat}={t}_{1}+{t}_{2}=6+5=11s$Displacement covered in 6 seconds will be,$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}84=u×6+\frac{1}{2}a×{\left(6\right)}^{2}...............\left(i\right)$Displacement covered in 11seconds will be,$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}264=u×11+\frac{1}{2}a×{\left(11\right)}^{2}...............\left(ii\right)$On solving equations (i) and (ii)we will get, $u=2m/sanda=4m/{s}^{2}$Hence, the initial velocity will be $2m/s$and the acceleration will be $4m/{s}^{2}$.

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