Particles P and Q move along the same straight line; positions are measured in meters from a common origin, with the positive direction to the right. Their motions begin simultaneously at t=0 seconds and satisfy the data below.
• Particle P: acceleration aP(t)=6t−4, velocity vP(0)=2 m/s, position xP(0)=1 m.
• Particle Q: acceleration aQ(t)=2t+1, velocity vQ(0)=−3 m/s (leftward), position xQ(0)=10 m.
(1) Derive explicit expressions for vP(t) and xP(t), then for vQ(t) and xQ(t). Include constants of integration determined from the initial conditions.
(2) Show that the two particles meet exactly once. Determine the time t (to three decimal places) and position x at which this meeting occurs.
(3) Compute the velocity of each particle at the instant found in part (b). State the direction of motion for each particle and classify the encounter as a head-on collision or an overtaking event.
(4) Let D(t)=xQ(t)−xP(t) denote the separation between the particles. Determine all intervals of time on which Q is ahead of P (that is, D(t)>0). Justify your answer using properties of D and D′.