First slide

Average Velocity

Question

A particle moves with constant speed v along a regular hexagonal A B C D E F in same order (i.e., A to B, B to C, C to D, D to E, E to F and F to A). The magnitude of average velocity for its motion from A to C is

Moderate

A

v
B

v/2
C

$\sqrt{3} v / 2$
D

3v /2

Solution

See figure. The average velocity is given by
$Average velocity = \frac{total displacement}{total time}$
$â´ v_{av} = \frac{AC}{t} or |v_{av}| = \frac{AC}{t}$
The time taken by the particle to reach from A to C is
$t = \frac{AB + BC}{V} = \frac{2 a}{V} . . . . (1)$
From $â³ ABC$ we have
$\cos â¡ 120^{â} = \frac{a^{2} + a^{2} â (AC)^{2}}{2 a^{2}}$
or - $\frac{1}{2} = \frac{2 a^{2}}{2 a^{2}} â \frac{(AC)^{2}}{2 a^{2}}$
Solving we get, $AC = \sqrt{3} a . . . (2)$
Now $|v_{av}| = \frac{\sqrt{3} a}{2 a / v} = \frac{\sqrt{3} v}{2}$

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