particles colliding, prove with equations, help!?

Question

A particle P, moving along a straight line with constant acceleration 0.3m/s^2 passes a point A on the line with a velocity of 20m/s. At the instant when P passes point A, a second particle Q is 20m behind A and moving with velocity of 30m/s. prove that unless the motion of P or Q changes they will collide

Answer 1

Let point A be the zero point for the vertical motion.
the equation of motion for Q will be

s= -20m + 30[m/s]*t

The equation of motion for P will be

s= 0 + 20[m/s]*t + ½*[3/10]*t^2[m/s^2]

Subtract the last equation from the first and get:
0= -20[m]+10[m/s]*t - (3/20)t^2[m/s^2]

solving for t will give:

t=2,06[s]; t=64,60[s]

interpretation of the result.

Particle Q wil hit P after 2,06[s]
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Answer 2

Set up an expression for the distance between the points as a function of time. Solve for d = 0 and if there are any real solutions, the proof is completed.

In this case the discriminate of the quadratic in t is positive and there are in fact 2 roots, meaning the particles will collide twice. The first time at .6689 sec (Q overtaking P) and later at 199.331 sec when P overtakes Q.

Answer 3

Integrate the acceleration of p to get dp. In order to collide q must travel 20 meters farther:

dq = 20 + vq*t, va is known

dp = at^2/2 + vp*t, where a and vp are given.

careful, don't set them equal ! what's equal is

dq-20 = dp , where t=0 at the the only logical place, when p passes point a.

Satisfy yourself that these make sense.

Answer 4

nicely you may write the equation for distance from A (call it D), with time (t), of particle Q some thing like Dq(t) = 30t - 20 then Dp(t)=20t+0.15t^2 and discover out if there is any t the position Dq is larger than Dp 30t-20-20t-0.15t^2 > 0 10t - (.15t^2 +20) >0 feels like it (e.g. t =3 )