\[(a)\;1.5 m/s \quad (b)\;2 m/s \quad (c)\;1.25m/s \quad (d)\;1 m/s\]

$V_{av}=\large\frac{displacement}{time}$

$\qquad=\large\frac{PQ}{t}$

$\qquad=\large\frac{\sqrt 2 R}{t}$

$\qquad=\large\frac{2}{t}$ (since $ R=\sqrt 2 m)$

now $\theta=\large\frac{1}{2} \times a t^2$

$\qquad \large \frac{\pi}{2}=\large \frac{1}{2}$$ \times \large \frac{\pi}{4} $$t^2$

$\qquad t^2=4$

$\qquad t=2$

$v_{av}=\large \frac{2}{t}=\frac{2}{2}$$=1 m/s$

Hence d is the correct answer.

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