Question

# A bus accelerates uniformly from $54km/hr$ to $72km/hr$ in $10s$. Calculate the distance covered by the bus in metres during interval and also find the acceleration in metre per second square.

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Solution

## Step 1: GivenInitial velocity, $u=54km/hr=15m/s$Final velocity, $v=72km/hr=20m/s$Time, $t=10s$Step 2: Calculate the accelerationUsing first equation of motion,$a=\frac{v-u}{t}\phantom{\rule{0ex}{0ex}}=\frac{20-15}{10}\phantom{\rule{0ex}{0ex}}=0.5m/{s}^{2}$Step 3: Calculate the distance traveled by busUsing the third equation of motion, the distance $s$ can be calculated ${v}^{2}={u}^{2}+2as\phantom{\rule{0ex}{0ex}}{20}^{2}={15}^{2}+2\left(0.5\right)s\phantom{\rule{0ex}{0ex}}s=\frac{{20}^{2}-{15}^{2}}{2\left(0.5\right)}\phantom{\rule{0ex}{0ex}}s=175m$Therefore, the distance covered is $175m$ and the acceleration is $0.5m/{s}^{2}$

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