Help Needed with Physics Problem (2 runners on circular track)?

Question

Two runners start in opposite directions at the northernmost point of a circular track 20.0 m in radius. If the speed of each runner is constant and is 5.31 km/h for Runner A and 3.44 km/h for Runner B, where will they meet? Where will they meet the second time? Assume that they go around the track on essentially the same path, with Runner A initially headed west and Runner B initially headed east.

Can anyone help? Thank you!!!

Answer 1

You posted this twice ... there's a question about the radius of the track ... a 20 m radius is pretty small (125 m circumference), but a 20 km radius is very large (125 km -- about 78 miles).

I think we'll go with what you have, and change the speeds from km/hr to m/s.

Okay, here we go.

Circumference of track = 2 pi r = 40 pi = 125.664 m

Runner A speed = 5.31 km/hr x 1000 m/km / 3600 sec/hr = 1.475 m/s

Runner B speed = 3.44 (1000/3600) = 0.956 m/s

They first meet at time t where

1.475t + 0.956t = 125.664
2.431t = 125.664
t = 51.70 sec.

They'll meet the first time after 51.7 seconds. But you need to know they'll meet. We'll come back to that.

First, let's see they'll meet the second time. That will be 51.70 seconds after the first meeting.

To get the two locations, let's just concentrate on Runner B. In 51.7 seconds, he runs 51.7 * 0.956 = 49.40 m (first meeting), and in 103.4 seconds, he runs 98.80 m (second meeting).

Since the track circumference is 125.66 m, the first meeting occurs at

(49.40 / 125.66) * 360 degrees = 141.5 degrees

and the second meeting is at

(98.8 / 125.66) * 360 degrees = 283.0 degrees

along the track, where zero degrees is north and 90 degrees is east.

Those are your two answers.

Answer 2

They will meet around 76 meters from northermost point in direction of runner A (runner B will have travelled around 49 meters)
And again, same thing for the second time, treat that point as starting point, and they will meet another 76 meters in direction of A's travel (by this time, A will have made a full circle but B would not have yet)
To understand the problem, lets simplify it by saying that the speed of A is 2 KM/hr and B is 1 KM/hr. For any given time constant, A will travel twice the distance than B i.e. speed of A divided by speed of B. So when they meet, the total distance travelled by A will be double than that of A. So, if the total distance travelled is 3 KM, then 2 KM was covered by A and 1 KM was covered by B.
In this question, using the same logic, when they meet for the first time, they both have travelled for the same 'time'. We also know that they have together made a full circle, so the total distance travelled is the circumference of the circle i.e. 2*pi*radius = 2*22/7*20 = approx 125 meters.
And now to get the distance travelled by A, simply get the ratio of A's speed with is (5.31/ (5.31+3.44)) = approx 0.6068
Multiply this number by the total distance of 125 meters and you get the distance travelled by A when it meets B, which is about 76 meters.

Though it would have been easier to expain the answer diagramattically, but important point to note here is the ratio of speed of A and B and that they will be travelling for the same time when they meet....
You can try this question with putting simple whole numbers for speed ..
(This question would have been little difficult if both A and B were travelling in the same direction... but even then the same logic could be used with some variation)

Answer 3

The track is 125.66 m. 2xPix20m

The runners will need to travel a total of 125.66m before they meet.

Use this equation to work out how far A will travel before they meet:

Distance A=Speed A x total distance/(speed A + speed B)

Distance A = 1.475 m/s x 125.66 m/ (1.475 + 0.956) m/s
= 76.24 m from the starting position in the direction A is running (anticlockwise direction).

To find out where they meet again just add 76.24m to the previous total i.e. 152.49 m from starting point in direction of A (anticlockwise direction) which is really just 26.83 m from the starting position (1 lap plus 152.49m - 125.66m).

Answer 4

they are going to fulfill around seventy six meters from northermost factor in process runner A (runner B could have travelled around 40 9 meters) And returned, comparable undertaking for the 2nd time, handle that factor as beginning factor, and that they are going to fulfill yet another seventy six meters in process A's holiday (by making use of this time, A could have made an entire circle yet B does not have yet) to comprehend the priority, shall we simplify it by making use of asserting that the fee of A is two KM/hr and B is a million KM/hr. For any given time consistent, A will holiday two times the gap than B i.e. velocity of A divided by making use of velocity of B. So whilst they meet, the whole distance travelled by making use of A would be double than that of A. So, if the whole distance travelled is 3 KM, then 2 KM became into coated by making use of A and a million KM became into coated by making use of B. in this question, making use of an identical good judgment, whilst they meet for the 1st time, they the two have travelled for an identical 'time'. We additionally be attentive to that they have got jointly made an entire circle, so the whole distance travelled is the circumference of the circle i.e. 2*pi*radius = 2*22/7*20 = approx a hundred twenty five meters. And now to get the gap travelled by making use of A, basically get the ratio of A's velocity it is (5.31/ (5.31+3.40 4)) = approx 0.6068 Multiply this huge sort by making use of the whole distance of a hundred twenty five meters and you get the gap travelled by making use of A whilst it meets B, it is approximately seventy six meters. even though it could have been much less stressful to expain the respond diagramattically, yet significant factor to be conscious right it is the ratio of velocity of A and B and that they are going to be vacationing for an identical time whilst they meet.... you could attempt this question with putting straightforward entire numbers for velocity .. (this question could have been little complicated if the two A and B have been vacationing in an identical course... yet even then an identical good judgment must be used with some version)