# Thread: Mechanics problem: ratios of velocities, times and distances

1. ## Mechanics problem: ratios of velocities, times and distances

I am working through an old mechanics textbook for UK A-level (1970s/80s) and am having difficulty with following problem, which is from an old A-level paper. I have managed to solve the first part of the question, but am floundering on the second and third parts. Help please!

This is the first part of the question:
A particle moves on a straight line so that its distance s from a fixed point of the line is given by s = b sin kt, where t is the time and b, k are constants. Prove that its velocity v at any point is given by v2=k2(b2 - s2).

and here is my answer:

s = b sin kt so ds/dt = v = bk cos kt
cos kt = v/bk, so sin2 kt = 1 - v2 / b2k2 = (b2k2 - v2) / b2k2

Therefore

s2/b2 = (b2k2 - v2) / b2k2 so k2s2 = b2k2 - v2 i.e. v2 = k2(b2 - s2)

Here are the second and third parts of the question, where I need help:

Particles A and B move along a straight line, each starting from a point O with velocity u. A is acted on by a constant force directed towards O, and B by a force, also towards O, proportional to the distance from O.

(i) If they both cover the same distance L before starting back towards O, show that the times taken tA and tB respectively, are in the ratio 4 : pi, and that their velocities as they pass a point distant s from O are in a ratio given by

vA2 : vB2 = L : (L + s)

(ii) If they both take the same time T between leaving O and returning to O, show that their maximum distances from O are in a ratio given by

LA : LB = pi : 4

I have tried using F=ma and integrating the expressions for acceleration with respect to t and with respect to s, but I can't get to the ratios stated, particularly with the time expressions.

Any help much appreciated.

2. ## Re: Mechanics problem: ratios of velocities, times and distances

OK, I've solved it. Particle A has constant deceleration away from O, so obeys the equations of motion for constant acceleration, and we get

vA2 = u2(L - s)/L

Particle B's motion can be found by integrating with respect to v and s, to get

vB2 = u2(L2 - s2)/L2

The ratio of these two is vA2 : vB2 = L : L + s

For particle A, solving for time gives tA = 2L/u

For particle B, noting the result of the first part of the question yields sB = L sin ut/L and solving for time gives tB = L(pi)/(2u)

The ratio of these two times is tA : tB = 4 : pi