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

Question

2. Romeo is at x = 0 at t = 0 when he sees Juliet at x = 6m.
(a) He begins to run towards her at constant velocity v = 5m/sec. She in turn
begins to accelerate toward him at a = -2m/sec2 . 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 find
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 have figured out part a but can not figure out how to derive the answer for part b.

Here is what I have for part a:

(a)
Given the information above, we can derive the respective functions for Romeo & Juliet.

Romeo:
VR(t) = 5 *Given
Taking the anti-derivative of this function, we can find the function of Romeo's position.
PR(t) = 5t + C *Anti-derivative with unknown constant.
Using the information provided about Romeo's initial position, we can solve for C.
0 = 5(0) + C ==> Therefore C = 0 and Romeo's position is given by the function PR(t) = 5t.

Juliet:
AJ(t) = -2 *Given
Taking the anti-derivative of this function, we can find the function of Juliet's velocity.
VJ(t) = -2t + C *Anti-derivative with unknown constant.
Using the information provided about Juliet's initial velocity, we can solve for C.
0 = -2(0) + C ==> Therefore C = 0 and Juliet's velocity is given by the function VJ(t) = -2t.
Taking the anti-derivative of this function, we can find the function of Juliet's position.
PJ(t) = -t2 + C *Anti-derivative with unknown constant.
Using the information provided about Juliet's initial position, we can solve for C.
6 = -(0)2 + C ==> Therefore C = 6 and Juliet's position is given by the function PJ(t) = -t2 + 6

Setting both functions equal to each other, we can find the time that they will meet.
5t = -t2 + 6 ==> t2 + 5t - 6 = 0 ==> (t - 1)(t + 6) = 0 ==> so t = 1 second.

Explanation / Answer

1. In t seconds, Romeo travels x=vt=5t. Juliet travels x=.5a(t^2)=.5(2)(t^2)=t^2. Together they travel 6 meters, so 5t+t^2=6. From this you find t=1 or t=5. Use the smaller value because that's when they meet each other, so t=1. 2. In t seconds, Romeo travels x=vt=5t and Juliet travels t^2. We need 5t=t^2+6 for Romeo to catch up with Juliet. Thus t=2 or t=3. Use the smaller value because that's when he first catches up with her, so t=2. When a is smaller, Juliet travels x=.5a(t^2). We need 5t=.5a(t^2)+6. Use the quadratic to find that t = (5 +/- sqrt(25-12a) )/2, so there are two solutions for t. That is because Romeo catches up to Juliet at the smaller t, but she keeps accelerating, so he must continue running to catch up to her another time.

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