Question

A particle moves over the sides of an equilateral triangle of side l with constant speed v as shown in figure. The magnitude of average acceleration as it moves from A to C is (A)

Answer 1

If the particle moves on AC then avg acceleration = 0.
But if it moves on AB and then on BC :
Avg acceleration = sqrt3 /2 * v/l.

See pic.

Answer 2

The magnitude of average acceleration from A to C is (a) zero if it directly goes from A to C and (b) $a=\frac{v^{2} \sqrt{3}}{2 l}$ if it goes via point B.

Solution:

An equilateral triangle has same length for all the three sides say, AB = BC = AC. The angle in equilateral triangle is 60° each. If the particle directly moves from A to C with constant speed v then the average acceleration will be zero as there is no change in the velocity in this path.

But if the particle reaches C via B point, then the average acceleration can be determined using the following method.

$a=\frac{|v-u|}{t}\quad \rightarrow(1)$

Here, a is the ‘average acceleration’, v is the ‘final velocity’, u is the ‘initial velocity’ and t is the ‘time taken’ to cover the distance.

So the time taken can be found as $t=\frac{d i s t a n c e}{v e l o c i t y}$

$\Rightarrow t=\frac{\text {distance from } A \text { to } B}{\text {velocity}}+\frac{\text {distance from } B \text { to } C}{\text {velocity}}$

As the sides are of equal length, distance from A to B = distance from B to C = l and velocity is constant as v

So,

$t=\frac{l}{v}+\frac{l}{v}=\frac{2 l}{v} \quad \rightarrow(2)$

Also, $v=-v \sin 60=\frac{-v \sqrt{3}}{2} \text { and } u=v \sin 60=\frac{v \sqrt{3}}{2}$

$|v-u|=\left|\frac{-v \sqrt{3}}{2}-\frac{v \sqrt{3}}{2}\right|=v \sqrt{3} \quad \rightarrow(3)$

So substitute eqn (2) and (3) in eqn (1), we get $\bold{a=\frac{v \sqrt{3}}{2 l / v}}$

$a=\frac{v^{2} \sqrt{3}}{2 l}$