Question:
An object having a velocity $4.0 \mathrm{~m} / \mathrm{s}$ is accelerated at the rate of $1.2 \mathrm{~m} / \mathrm{s}^{2}$ for $5.0 \mathrm{~s}$. Find the distance travelled during the period of acceleration.
Solution:
$\mathrm{u}=4 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=1.2 \mathrm{~m} / \mathrm{s}^{2} ; \mathrm{t}=5 \mathrm{sec}$.
Distance travelled
$s=u t+\frac{1}{2} a t^{2}$
$s=(4)(5)+\frac{\frac{1}{2}}{2}(1.2)(5)^{2}$
$=35 \mathrm{~m}$
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