Romeo is at x = 0 at t = 0 when he sees Juliet at x = 6m. (a) He begins to run t

Question

Romeo is at x = 0 at t = 0 when he sees Juliet at x = 6m. (a) He begins to run towards her at v = 5m/s. She in turn begins to accelerate towards him at a = -2 m/s^2. When and where will they cross? Sketch their motions measuring time horizontally and position vertically. (b) Suppose instead she moved away from him with positive acceleration a. Find a_max, the maximum a for which he will ever catch up with her. For this case ¯nd the time t of their contact. Show that for smaller values of a these star crossed lovers will cross twice. Draw a sketch for this case. Explain in words why they cross twice.

I am really only looking for help on part b, as I can get part a correct. I'd especially appreciate an explanation of how you find it and not just the answer.

Note:

Explanation / Answer

at the time of crossing the position of the romeo and juliet will be same

s = ut + 0.5 * at^2

so,

distance travelled by romeo = 6 + 5t

distance travelled by juliet = 0.5 * 2 * t^2

6 + 5t = 0.5 * 2 * t^2

on solving we'll get t = 1 sec

they will meet at t = 1 s

The signed distance here is 6 - 5t + (a/2) t^2. At amax=25/12, this becomes 6 - 5t + (25/24)t^2 = 6 (1- (5/6)t)^2, which only has one solution. for a < amax, it has two solutions; the first where R catches up with J, and the second where J passes R, accelerating

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