Solve the following :


A particle starting from rest moves with constant acceleration. If it takes $5.0$ s to reach the speed $18.0$ $\mathrm{km} / \mathrm{h}$


(a) The average velocity during this period

(b) The distance travelled by the particle during this period.


$u=0 ; t=5 s e c ; v=18 \times 5 / 18=5 \mathrm{~m} / \mathrm{s}$



$a=1 \mathrm{~m} / \mathrm{s}^{2}$

$S=u t+\frac{1}{2} a t^{2}$


$\mathrm{V}_{\mathrm{avg}}=$ distance/time $=\frac{12.5}{5}$

$\mathrm{V}_{\mathrm{avg}}=2.5 \mathrm{~m} / \mathrm{s}$


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