Question

Three points are located at the vertices of an equilateral triangle having each side as $\alpha$. All the points move simultaneously with speed $u$ such that first point continually heads for second, the second for the third and the third for the first. Time taken by the points to meet at the centre is

• A. $\frac{\alpha }{3u}$
• B. $\frac{2\alpha }{3u}$
• C. $\frac{{\alpha }^2}{u^3}$
• D. $\frac{3\alpha }{2u}$

Answer 1

The correct option is B $\frac{2\alpha }{3u}$

All three points were initially at same distance from the centroid of the original equilateral triangle of the side $\alpha$

At any time $t$, the three points still form an equilateral triangle, moving with the same speed

The distance between two points is decreasing at the same rate as at the beginning.

$\therefore$ The total amount of time taken for the points to meet is $\alpha$ divided by the inital rate at which two points approach each other

Taking any two points it shows first point moves in the direction of the second at speed $\mu$

While the second moves in direction of first (taking component along one of the triangle edges) at $\mu \cos 60°=\frac{\mu }{2}$

$\therefore$ The rate at which the distance between two points is decreasing is $\frac{3\mu }{2}$

The time taken by the three points to meet is thus $2\alpha$ $3\mu$