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2.Motion in Straight Line
hard
$x-t$ graph for a uniformly accelerated particle is as shown in the figure. Then find the average velocity between points $(i)$ and $(ii)$ ......... $ms^{-1}$
A$3$
B$2$
C$4$
D$1.5$
Solution
$\mathrm{v}_{\text {avenge }}=\frac{\mathrm{x}_{2}-\mathrm{x}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}=\frac{\mathrm{ut}_{2}+\frac{1}{2} \mathrm{at}_{2}^{2}-\left(\mathrm{ut}_{1}+\frac{1}{2} \mathrm{at}_{1}^{2}\right)}{\mathrm{t}_{2}-\mathrm{t}_{1}}$
$=\frac{u\left(t_{2}-t_{1}\right)+\frac{1}{2} a\left(t_{2}-t_{1}\right)\left(t_{2}+t_{1}\right)}{t_{2}-t_{1}}$
$=\frac{2 u+a\left(t_{2}+t_{1}\right)}{2}=\frac{\left(u+a t_{1}\right)+\left(u+a t_{2}\right)}{2}=\frac{v_{1}+v_{2}}{2}$
$=\frac{\tan \theta_{1}+\tan \theta_{2}}{2}=\frac{1+3}{2}=2 \mathrm{ms}^{-1}$
$O R$
For uniformly acceleration motion
$\mathrm{v}_{\text {avenge }}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}=\frac{1+3}{2}=2 \mathrm{ms}^{-1}$
$=\frac{u\left(t_{2}-t_{1}\right)+\frac{1}{2} a\left(t_{2}-t_{1}\right)\left(t_{2}+t_{1}\right)}{t_{2}-t_{1}}$
$=\frac{2 u+a\left(t_{2}+t_{1}\right)}{2}=\frac{\left(u+a t_{1}\right)+\left(u+a t_{2}\right)}{2}=\frac{v_{1}+v_{2}}{2}$
$=\frac{\tan \theta_{1}+\tan \theta_{2}}{2}=\frac{1+3}{2}=2 \mathrm{ms}^{-1}$
$O R$
For uniformly acceleration motion
$\mathrm{v}_{\text {avenge }}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}=\frac{1+3}{2}=2 \mathrm{ms}^{-1}$
Standard 11
Physics
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