x-t graph for a uniformly accelerated particl | vedclass

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Question

$\mathrm{v}_{	ext {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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\mathrm{v}_{	ext {avenge }}=\frac{\mathrm{x}_{2}-\mathrm{x}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}=\frac{\mathrm{ut}_{2}+\frac{1}{2} \mathrm{at}_{2}^{2}-\left(\mathrm{ut}_{1}+\frac{1}{2} \mathrm{at}_{1}^{2}ight)}{\mathrm{t}_{2}-\mathrm{t}_{1}}$

$=\frac{u\left(t_{2}-t_{1}ight)+\frac{1}{2} a\left(t_{2}-t_{1}ight)\left(t_{2}+t_{1}ight)}{t_{2}-t_{1}}$

$=\frac{2 u+a\left(t_{2}+t_{1}ight)}{2}=\frac{\left(u+a t_{1}ight)+\left(u+a t_{2}ight)}{2}=\frac{v_{1}+v_{2}}{2}$

$=\frac{	an 	heta_{1}+	an 	heta_{2}}{2}=\frac{1+3}{2}=2 \mathrm{ms}^{-1}$

$O R$

For uniformly acceleration motion

$\mathrm{v}_{	ext {avenge }}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}=\frac{1+3}{2}=2 \mathrm{ms}^{-1}$

$x-t$ graph for a uniformly accelerated particle is as shown in the figure. Then find the average velocity between points $(i)$ and $(ii)$ ......... $ms^{-1}$

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