A bus increases its speed from $36\ km/h$ to $72\ km/h$ in $10\ seconds$. Its acceleration is:
(a) $5\ m/s^2$
(b) $2\ m/s^2$
(c) $3.6\ m/s^2$
(d) $1\ m/s^2$

Here initial velocity $u=36\ km/h=36\times\frac{5}{18}=10\ m/s$

Final velocity $v=72\ km/h=72\times\frac{5}{18}\ m/s=20\ m/s$

Time $t=10\ s$

Therefore, acceleration $a=\frac{v-u}{t}$

$=\frac{20\ m/s-10\ m/s}{10\ s}$

$=1\ m/s^2$

So, option (d)  is correct.

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Updated on: 10-Oct-2022

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