What time did they both set off?

AI Thread Summary
Willy and Milly leave their starting points at the same time, heading towards each other, and meet at noon. Milly, who is faster, arrives at point A by 1:00 pm, while Willy reaches point B by 2:15 pm. The problem involves calculating their departure times using their speeds and distances, leading to two equations with three unknowns. Substituting one equation into the other helps to solve for the unknowns. The calculations confirm that both arrived at their destinations on time.
GoghAway

Homework Statement


Willy leaves point A and Milly leaves point B, both of them are heading toward each other's starting point. They both have constant speeds, but Milly is faster, and they pass each other at noon without stopping. Milly arrives at point A at 1:00 pm, and Willy arrives at point B at 2:15 pm. At what time did they both set off?

Homework Equations


v = d / t
v = (d2 - d1) / (t2 - t1)

The Attempt at a Solution


I'm putting the data here because it's what I think the problem is telling me, but I know I could be wrong.

Willy
d = D
v = Vw
t = (14.25 hours - t1)

Vw = D / (14.25 h - t1)

Milly
d = D
v = Vm
t = (13.0 hours - t1)

Vm = D / (13.0 h - t1)

After noon
Willy
d = (D - d1)
v = Vw
t = 2.25 hours

Vw = (D - d1) / 2.25 h

Milly
d = d1
v = Vm
t = 1 hour

Vm = d1 / (1 h)

Since Milly is going faster noon, or when they meet would be after the midpoint between point A and B for her, and before the midpoint for Willy.

Since Willy's velocity is always the same:
D/(14.25 h - t1) = (D - d1)/2.25 h
D (2.25 h) = (D - d1)(14.25 h - t1)
D = ((D - d1)(14.25 h - t1)) / (2.25 h)

Since Milly's velocity is always the same:
D/(13.0 h - t1) = d1/(1 h)
D (1 h) = d1 (13.0 h - t1)
D = (d1 (13.0 h - t1)) / (1 h)
d1= (D (1 h)) / (13.0 h - t1)


I'm not sure whether I can sub what D = or what d1 = from what I have since Willy's velocity is always the same with what I have since Milly's velocity is always the same or vice versa. Either way, I don't think I know enough of the unknowns to come up with a numerical value as an answer, but I don't know how to get the information that I need.
 
Physics news on Phys.org
Welcome to PF.

Your method will work. You have gotten to the two equations

D = ((D - d1)(14.25 h - t1)) / (2.25 h)

d1= (D (1 h)) / (13.0 h - t1)

It looks like you have two equations with three unknowns. But see what happens if you substitute the expression for d1 of the second equation into the first equation.
 
  • Like
Likes GoghAway
I believe we are given that they both set off at the same time.
 
Thank you TSny! I did as you suggested, and checked the times to see if both Milly and Willy had the correct amount of time after they meet at noon so they would arrive at the correct times, which they did!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top