Question

# A car accelerates uniformly from $18km/hr$ to $36km/hr$ in $5s$. Calculate the acceleration and the distance covered by the car in that time.

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Solution

## Step 1: GivenInitial velocity, $u=18km/hr=5m/s$Final velocity, $v=36km/hr=10m/s$Time, $t=5s$Step 2: Formula usedThe first equation of motion is given as,$v=u+at$The second equation of motion is given as, $s=ut+\frac{1}{2}a{t}^{2}$Step 3: Determine the accelerationUsing the first equation of motion,$v=u+at\phantom{\rule{0ex}{0ex}}10=5+a\left(5\right)\phantom{\rule{0ex}{0ex}}a=1m/{s}^{2}$Step 4: Determine the distance coveredUsing the second equation of motion, determine the distance s.$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=5\left(5\right)+\frac{1}{2}\left(1\right){\left(5\right)}^{2}=37.5m$Therefore, the acceleration and the distance covered by the car in that time are $1m/{s}^{2}$ and $37.5m$ respectively.

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