Kinematics Football Problem -- 2 football players colliding....

  • Thread starter Balsam
  • Start date
  • Tags
    Kinematics
In summary, two football players, separated by 42m, are running towards each other. Player 1 starts from rest and accelerates at 2.4m/s^2, while player 2 moves uniformly at 5.4m/s. When attempting to solve for the time it takes for the players to collide, the equations used were displacement=vi(t)+1/2a(t)^2 for player 1 and displacement= (vf+vi/2)t for player 2. However, the equations were not set up with respect to the same coordinate system, resulting in an incorrect answer.
  • #1
Balsam
226
8

Homework Statement


Two football players separated by 42m run directly toward each other. Football player 1 starts from rest and accelerates at 2.4m/s^2, and football player 2 moves uniformly at 5.4m/s.
a) How long does it take for the two players to collide?
b) how far does each player move?
c) How fast is football player 1 moving when the players collide?

For part a- player 1: displacement=42m, a=2.4m/s^2, vi=0m/s.
Player 2: vi=vf=5.4m/s, displacement=42m
.

Homework Equations


I'm only at part a: For player 1, I used the eqn. displacement=vi(t)+1/2a(t)^2
For player 2, I used displacement= (vf+vi/2)t.

The Attempt at a Solution


I plugged in all known values for both players. I tried this twice- the first time, I plugged in the displacement for each player. The second time, I did not. But, for both attempts, I set the 2 equations equal to each other, moved all variables over to one side of the equals sign, setting the equation equal to 0. Then, I used the quadratic formula and solved for time. I got the same answer both times, 4.5seconds. But, the correct answer is 4.1s. I have the solution for this problem, but it's very confusing. What did I do wrong?

I would type up all of my work, but it's over a page long and would look very confusing/messy if I tried to type itHere's what I did:

player 1: displacement=o(t)+1/2(2.4m/s^2)t^2

displacement=1.2m/s t^2Player2: d=5.4m/s(t)

Then, I set the 2 equations equal to each other:

5.4m/s^2
(t)=1.2m/s(t)^2

I moved all terms over to one side of the equals sign: 1.2m/s
(t)^2-5.4m/s^2(t)=0
Then, I used the quadratic formula to solve for t.

x=5.4+√29.16/2.4

x1=4.5 seconds. x2=0seconds

Therefore, time=4.5seconds. But, the correct answer is 4.1seconds.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
We can't critique what you don't post. So you'll have to show your work in order to get help with it.

One suggestion: Make sure that your equations of motion for each player are with respect to the same coordinate system. They should give both give positions versus time that are a true reflection of their locations in that system (Do both players start at x = 0? Are both player's velocities positive?).

Out of curiosity, what were you trying to accomplish with the odd formatting?
 
  • Like
Likes gracy
  • #3
gneill said:
We can't critique what you don't post. So you'll have to show your work in order to get help with it.

One suggestion: Make sure that your equations of motion for each player are with respect to the same coordinate system. They should give both give positions versus time that are a true reflection of their locations in that system (Do both players start at x = 0? Are both player's velocities positive?).

Out of curiosity, what were you trying to accomplish with the odd formatting?

If they were with respect to the same coordinate system, one player would start at 0m and one would start at 42 m, right?
I didn't format it like that, it just turned out that way and I have no idea why.
 
  • #4
Balsam said:
If they were with respect to the same coordinate system, one player would start at 0m and one would start at 42 m, right?
Right.
I didn't format it like that, it just turned out that way and I have no idea why.
I've removed the BBcode tags from the OP that were causing the odd formatting.
 
  • #5
Here's hwta
gneill said:
Right.

I've removed the BBcode tags from the OP that were causing the odd formatting.
I added my solution in the original post, but the formatting is weird again
 
  • #6
Balsam said:
I added my solution in the original post, but the formatting is weird again
You must be cutting and pasting formatted text from some document that includes formatting tags. Try pasting as "unformatted text" into a text document in a simple editor (like Notepad) first.
 
  • Like
Likes gracy
  • #7
gneill said:
You must be cutting and pasting formatted text from some document that includes formatting tags. Try pasting as "unformatted text" into a text document in a simple editor (like Notepad) first.
It's fixed now. Can you check my solution?
 
  • #8
gneill said:
Right.

I've removed the BBcode tags from the OP that were causing the odd formatting.
So anytime you have a question with more than one frame of motion, like this, you write all of your known values with reference to the same coordinate system?
 
  • #9
Balsam said:
It's fixed now. Can you check my solution?
It does not incorporate my suggestion in post #2; You have the second player starting from an initial displacement of zero, and the two players are not moving towards each other.
 
  • Like
Likes gracy
  • #10
Balsam said:
So anytime you have a question with more than one frame of motion, like this, you write all of your known values with reference to the same coordinate system?
Yes. Otherwise how can you say where one is with respect to the other?

In this case there is only one frame of reference, that of the football field. It's up to you to lay down a coordinate system in the frame of reference and express both player's motions in that coordinate system.
 
  • Like
Likes gracy
  • #11
gneill said:
Yes. Otherwise how can you say where one is with respect to the other?

In this case there is only one frame of reference, that of the football field. It's up to you to lay down a coordinate system in the frame of reference and express both player's motions in that coordinate system.
makes sense, I'll try the problem again
 
  • #12
Balsam said:

Homework Statement


Two football players separated by 42m run directly toward each other. Football player 1 starts from rest and accelerates at 2.4m/s^2, and football player 2 moves uniformly at 5.4m/s.
a) How long does it take for the two players to collide?
b) how far does each player move?
c) How fast is football player 1 moving when the players collide?

For part a- player 1: displacement=42m, a=2.4m/s^2, vi=0m/s.
Player 2: vi=vf=5.4m/s, displacement=42m
.

Homework Equations


I'm only at part a: For player 1, I used the eqn. displacement=vi(t)+1/2a(t)^2
For player 2, I used displacement= (vf+vi/2)t.

The Attempt at a Solution


I plugged in all known values for both players. I tried this twice- the first time, I plugged in the displacement for each player. The second time, I did not. But, for both attempts, I set the 2 equations equal to each other, moved all variables over to one side of the equals sign, setting the equation equal to 0. Then, I used the quadratic formula and solved for time. I got the same answer both times, 4.5seconds. But, the correct answer is 4.1s. I have the solution for this problem, but it's very confusing. What did I do wrong?

I would type up all of my work, but it's over a page long and would look very confusing/messy if I tried to type itHere's what I did:

player 1: displacement=o(t)+1/2(2.4m/s^2)t^2

displacement=1.2m/s t^2Player2: d=5.4m/s(t)

Then, I set the 2 equations equal to each other:

5.4m/s^2
(t)=1.2m/s(t)^2

Since Player 2 moves at a constant velocity of 5.4 m/s throughout his run, then the distance he travels is simply d = v * t.

Writing 5.4 m/s2 t suggests he is accelerating, which is not the case here, and the units are not correct in any event.
(What units are used to measure distance?)
 
  • #13
Balsam said:
makes sense, I'll try the problem again
But when you set 2 equations equal to each other, don't you have to have 2 variables in each equation that you don't plug anything in for? That's what I remember from math class. So, shouldn't we not plug anything in for displacement?
 
  • #14
Balsam said:
But when you set 2 equations equal to each other, don't you have to have 2 variables in each equation that you don't plug anything in for? That's what I remember from math class. So, shouldn't we not plug anything in for displacement?
It's not clear what you mean here.

For this problem, the two players are running toward one another after being separated initially by a distance of 42 meters. Each second the players are running, there is a different distance being shaved off this initial separation. The amount being shaved off is related by the time spent running, which elapses the same for each player. After a certain amount of time, the players are going to collide with one another, as the distance between them will reach zero.

It always helps to make a sketch of a problem to clarify relationships and what is changing in time.
 
  • #15
SteamKing said:
It's not clear what you mean here.

For this problem, the two players are running toward one another after being separated initially by a distance of 42 meters. Each second the players are running, there is a different distance being shaved off this initial separation. The amount being shaved off is related by the time spent running, which elapses the same for each player. After a certain amount of time, the players are going to collide with one another, as the distance between them will reach zero.

It always helps to make a sketch of a problem to clarify relationships and what is changing in time.

For example, if you set the equation of a parabola and a line equal to each other, you can find the point of intersection. You would get an equation like y=3x^2 and y=4x+2, where you have 2 unknowns in each quation, y and x, and you plug in one equation for y in the other equation. so, shouldn't our 2 equations have both d and t as the 2 variables without any values plugged in? And then, we can sub one equation in for d in the other equation, if that makes sense
 
  • #16
Balsam said:
For example, if you set the equation of a parabola and a line equal to each other, you can find the point of intersection. You would get an equation like y=3x^2 and y=4x+2, where you have 2 unknowns in each quation, y and x, and you plug in one equation for y in the other equation. so, shouldn't our 2 equations have both d and t as the 2 variables without any values plugged in? And then, we can sub one equation in for d in the other equation, if that makes sense
That's exactly what you're doing here. You have two equations for position versus time. You take the expression for one position and substitute it for the position variable in the other equation.
 
  • Like
Likes gracy
  • #17
gneill said:
That's exactly what you're doing here. You have two equations for position versus time. You take the expression for one position and substitute it for the position variable in the other equation.
But, we plugged something in for d, which is like the variable y in those equations
 
  • #18
Balsam said:
But, we plugged something in for d, which is like the variable y in those equations
I don't understand your confusion. You have two variables, d and t, in each equation. How you go about solving the two equations in two unknowns is up to you. You can do the easy thing and equate the two positions (d's) and solve for t first, or substitute t from one into the other and solve for d first. Either way its a standard algebra problem at that point.
 
  • Like
Likes gracy
  • #19
gneill said:
I don't understand your confusion. You have two variables, d and t, in each equation. How you go about solving the two equations in two unknowns is up to you. You can do the easy thing and equate the two positions (d's) and solve for t first, or substitute t from one into the other and solve for d first. Either way its a standard algebra problem at that point.
I tried this problem again, plugging in +42m for d for player 1 and -42m for d for player 2 and I got the same answer as before
 
  • #20
Balsam said:
I tried this problem again, plugging in +42m for d for player 1 and -42m for d for player 2 and I got the same answer as before
Why did you plug in values for d? It's an unknown. The only positions you know are the initial positions, neither of which are "d".
 
  • Like
Likes gracy
  • #21
gneill said:
Why did you plug in values for d? It's an unknown. The only positions you know are the initial positions, neither of which are "d".
It doesn't matter anyways, because they end up cancelling out. If you didn't plug anything in for d, you'd get the same answer
 
  • #22
You should show your attempts in detail. What are the two equations for the positions of the players that you are starting with?
 
  • Like
Likes gracy
  • #23
gneill said:
You should show your attempts in detail. What are the two equations for the positions of the players that you are starting with?
I stated those in the original post and I stated what the equations look like after I plug in the values
 
  • #24
Balsam said:
I stated those in the original post and I stated what the equations look like after I plug in the values
But you were told that the equation for player 2 there was incorrect as it didn't include his initial position and his velocity direction was wrong. If you are still getting incorrect answers it must be because something is still wrong. We can't tell what that is if we can't see exactly what you're working with.
 
  • Like
Likes gracy
  • #25
gneill said:
But you were told that the equation for player 2 there was incorrect as it didn't include his initial position and his velocity direction was wrong. If you are still getting incorrect answers it must be because something is still wrong. We can't tell what that is if we can't see exactly what you're working with.
So you change displacement to df-di?
 
  • #26
Balsam said:
So you change displacement to df-di?
Could be. It depends upon what you do with it. Write out the equation you have in mind.
 

Related to Kinematics Football Problem -- 2 football players colliding....

1. What is the Kinematics Football Problem?

The Kinematics Football Problem is a physics problem that involves two football players colliding while running at each other. It is used to demonstrate the principles of kinematics, which is the study of motion without considering the forces causing the motion.

2. How do you solve the Kinematics Football Problem?

To solve the Kinematics Football Problem, you need to use the principles of kinematics, including displacement, velocity, and acceleration. You will also need to use the equations of motion, such as the distance formula, to calculate the position and speed of each player before and after the collision.

3. What information do you need to solve the Kinematics Football Problem?

To solve the Kinematics Football Problem, you will need to know the starting positions and velocities of both players, as well as the mass and speed of the ball. You will also need to know the time it takes for the collision to occur.

4. What are some real-life applications of the Kinematics Football Problem?

The Kinematics Football Problem has many real-life applications, such as in sports science and accident reconstruction. It can also be used to study the motion of objects in collisions, such as cars or other moving objects.

5. What are some other examples of kinematics problems?

Some other examples of kinematics problems include calculating the trajectory of a projectile, determining the motion of a pendulum, and analyzing the speed and position of a moving car. Kinematics is a fundamental concept in physics and has many applications in various fields of science and engineering.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
5K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
912
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
821
Back
Top