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 = ¡2m=s2. 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
amax, 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.

Answer 1

Case 1

motion equation

Romeo $x=5t$

Juliette $x=6-t^2$

They will cross on $t=1; x=5$

Case 2

motion equation

Romeo $x=5t$

Juliette $x=6+\frac{at^2}{2}$

Find the a - acceleration for Juliette which Romeo will be once cross with Juliette.

We have equation

$$5=at$$

Then

$$5t=6+\frac{5t}{2}$$

Time, when romeo cross Juliette

$$t=\frac{12}{5}$$

from it the acceleration

$$a_{max}=\frac{25}{12}$$

motion equation

Romeo $x=5t$

Juliette $x=6+\frac{25}{24}t^2$

​

They will cross on $t=2.4; x=12$

Analyzing this equation

$$\frac{25}{24}t^2-5t+6=0$$ $$\frac{a}{2}t^2-vt+c=0$$ $$D=\sqrt{v^2-4\cdot\frac{a}{2}\cdot c}$$ $$D=\sqrt{5^2-4\cdot\frac{25}{24}\cdot6}=0$$

solution $t_{1,2}=\frac{5}{2\cdot\frac{25}{24}}=\frac{12}{5}=2.4$

If acceleration $a<\frac{25}{12}$ then $D \neq0$ and we have two roots