   SEARCH HOME Math Central Quandaries & Queries  Question from laura, a student: Particle A travels at constant speed of 1.0ms-1 in straight line, passing particle B, at which time particle B beings traveling with constant acceleration of 0.2ms-2 until reaching Particle A. What equations represent the displacement of particle A and particle B measure from the point they first pass? How long does it take for particle B to catch up with particle A? What is the speed of particle B when reaching particle A? What distance have both particles traveled from where they first pass to where they meet again? Calculate the displacement of both particle A and B at 2 second intervals between t=0s and t=12s? We have two responses for you

Hi Laura,

I can help get you started. Suppose the point on the line where A passes B is P and the time when A passes B is t = 0 seconds. A is traveling at 1 m/sec so t seconds later A the distance from A to P is sA = 1 × t = t metres. When B starts moving at t = 0 seconds its initial velocity is zero but its acceleration is 0.2 m/sec2. Thus the distance from B to P t seconds later is sB = 1/2 × 0.2 t2 = 0.1 t2 metres. Hence the distance between A and B t seconds after A passed B is the difference between sA and sB.

I hope this helps,
Penny

Hi Laura.

Both particles' displacements are given by the equation

d = v0t + ½ at2

where d is the displacement, v0 is the initial velocity, t is the time and a is the acceleration.

So use this equation twice, substituting for what you know for each particle and simplify.

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.