Math CentralQuandaries & 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,


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.

Stephen La Rocque.

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