Calculating Impact Time in Two Rugby Players Collision

[TheShapeOfTime]

I don't have any work yet to show because each question has some element that I don't know how to resolve.

#1. Two rugby players are running towards each other. They are 37 m apart. If one is accelerating from rest at 0.5 m/s^2 and the other was already moving at 3.1 m/s and maintains her speed, how long before they crunch together?


* I did a questioin like this but both people where going at a constant speed. How do I account for the person accelerating?

[tex] \frac{x}{?} = \frac{37 - x}{3.1}[/tex]


#2. Superwoman is hovering above the ground when a person free-falling goes by her at terminal velocity of 140 km/h. Unfortunately, the parachute does not open. If it takes her 1.9 s to realize the person is in distress, what must her acceleration be if she is to catch the person just before she hits the ground 1000m below?

* I guess I could find the distance she traveled in the 1.9 s and substract it from 1000. Not sure where to go from there.

 

[Parth Dave]

1. Try thinking about the problem as if one of the two people were not moving at all.

2. I guess I could find the distance she traveled in the 1.9 s and substract it from 1000. Thats a good start. If superwoman (wonder woman, superman?) is to catch the person right before the person hits the ground than they both must reach the ground at the same time. Which means the time it takes for both of them must be the same. Try finding out the time and than see what you can say about superwomans acceleration.

 

[poolwin2001]

1.Let the players crunch(meet) after time t.As they meet,they are at the same position.So the distance traveled by the 1st person=distance traveled by the 2nd + initial distance b/w them(here,37m)
Now i think you can do it.Just use expression for distance(x=u*t +.5at^2)

 

[TheShapeOfTime]

1.Let the players crunch(meet) after time t.As they meet,they are at the same position.So the distance traveled by the 1st person=distance traveled by the 2nd + initial distance b/w them(here,37m)
Now i think you can do it.Just use expression for distance(x=u*t +.5at^2)

Are you sure their distance traveled (before colliding) will be the same if they're moving at different speeds?

 

[TheShapeOfTime]

1. Try thinking about the problem as if one of the two people were not moving at all.

2. I guess I could find the distance she traveled in the 1.9 s and substract it from 1000. Thats a good start. If superwoman (wonder woman, superman?) is to catch the person right before the person hits the ground than they both must reach the ground at the same time. Which means the time it takes for both of them must be the same. Try finding out the time and than see what you can say about superwomans acceleration.

1. Still not sure what to do here, I'll keep thinking.

2. So if I find the time before the person hits, I can just use that and the distance to find superwoman's acceleration? Doesn't this exclude the 1.9 s entirely? Does it even matter in the first place?