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:
$u=0 ; t=5 s e c ; v=18 \times 5 / 18=5 \mathrm{~m} / \mathrm{s}$
$\mathrm{v}=\mathrm{u}+\mathrm{at}$
$5=0+a(5)$
$a=1 \mathrm{~m} / \mathrm{s}^{2}$
$S=u t+\frac{1}{2} a t^{2}$
$\mathrm{S}=0+\frac{1}{2}(\mathrm{a})(5)^{2}$
$\mathrm{V}_{\mathrm{avg}}=$ distance/time $=\frac{12.5}{5}$
$\mathrm{V}_{\mathrm{avg}}=2.5 \mathrm{~m} / \mathrm{s}$