Question:
A particle starting from rest moves with constant acceleration. If it takes $5.0$ s to reach the speed $18.0$ $\mathrm{km} / \mathrm{h}$
Find
(a) The average velocity during this period
(b) The distance travelled by the particle during this period.
Solution:
$\mathrm{u}=0 ; \mathrm{t}=5 \mathrm{sec} ; \mathrm{v}=18 \times 5 / 18=5 \mathrm{~m} / \mathrm{s}$
$\mathrm{v}=\mathrm{u}+\mathrm{at}$
$5=0+a(5)$
$a=1 \mathrm{~m} / \mathrm{s}^{2}$
$S=u t^{\frac{1}{2}} a t^{2}$
$\mathrm{S}=0+\frac{\frac{1}{2}}{2}(\mathrm{a})(5)^{2}$
$\mathrm{S}=12.5 \mathrm{~m}$
$\mathrm{V}_{\mathrm{avg}}=$ distance/time $=\frac{12.5}{5}$
$V_{a v g}=2.5 \mathrm{~m} / \mathrm{s}$