[SOLVED] Kinematic motion problem. (acceleration, displacement, velocity)by ballerina_tee Tags: acceleration, displacement, kinematic, motion, solved, velocity 

#1
Feb1608, 10:16 PM

P: 20

1. The problem statement, all variables and given/known data
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. (brackets are answers from textbook) a) how long before they crunch together? (7.5 seconds) b) how fast was the accelerating players going? (3.8 m/s) c) how far has each player run? (14 m, 23 m) 2. Relevant equations Not sure. 3. The attempt at a solution player 1: a=0.5 m/s^2 initial velocity=0 m/s (started from rest) player 2: speed = 3.1 m/s d= 37 m for b) Would I have to convert the acceleration to speed? Is that even possible? c) is it asking how far each person has to run in order to run the entire 37m? or just when they "crunch" together. I'm not sure where to start. There isn't much information to use the equations. 



#2
Feb1608, 10:38 PM

P: 1,877

So the first thing you will want to do is establish your coordinate system. It would probably be a good idea to make one of the players at the origin. You can make two equations for distance. What would the intersection of the distance equations tell you?




#3
Feb1608, 10:38 PM

P: 235

For a) you need to find when the sum of thier distances travelled is equal to 37.




#4
Feb1608, 10:43 PM

P: 235

[SOLVED] Kinematic motion problem. (acceleration, displacement, velocity)
Mindscrape, Wouldnt that tell you when both football players have run the same distance? But they have not run the same distance when they collide.
Edit: rugby players football players... wateva ;p 



#5
Feb1608, 10:50 PM

P: 1,877

It tells you at what time their positions are equal. Think about it graphically, if a right moving accelerating guy is starts at x=0 then his position will be concave up parabolic looking in time. The other guy will start at x=30 and have a negative linear slope. Where the positions intersect represents the collision and the time when that happens.




#6
Feb1608, 10:52 PM

P: 20

I'm still sort of lost.
The players are running at different speeds... Would this be a start? finding t for the 2nd player. t=d/v =37/3.1 = 11.9 s (that would be the total time it took the player to run 37 m?) EDIT: ^ I just read your post, I understand what you're trying to say now...but how would I find the intersection? 



#7
Feb1608, 10:56 PM

P: 235

Oh yeh, sorry Mindscrape. Thats actually quite a neat way to do it =D I just figured when the sum of their distances added to 37.




#8
Feb1608, 10:58 PM

P: 1,877

How would you find the intersection of two curves otherwise, algebraically? As a reminder, the equation for distance as a function of time looks like
[tex]x(t) = x_i+v_it+\frac{1}{2}at^2[/tex] where xi and vi are the initial speeds and velocities of particles (or football players!). No problem, spoon, it's always good to have learning experiences. 



#9
Feb1608, 11:10 PM

P: 20

Would I have to set up 2 equations ... 1 being quadratic and the other linear? It's been so long I done math, finding intersections.
[tex]x(t) = x_i+v_it+\frac{1}{2}at^2[/tex] since the initial velocity for player 1 is 0... xi and vi*t would cancel? for player 2, initial velocity/speed would be 3.1 so that value would fill the xi and vi of the equation? 



#10
Feb1608, 11:15 PM

P: 1,877

So player one has no initial position or velocity, the way I defined it at least.
x1(t) = .5*.5t^2 Player two has no acceleration, but has a negative velocity and an initial position of 37 in our coordinate system. x2(t) = 373.1*t 



#11
Feb1608, 11:19 PM

P: 20

then, you would set the equations equal to each other and solve for t?
I don't get how you got x(t) = .5*.5t^2 or was it a typo? Wouldn't it be x(t) =.5t^2 .5t^2 = 26.9t^2 ? :S 



#12
Feb1608, 11:26 PM

P: 1,877

His acceleration is .5 and the term has a 1/2 out front anyway, so .5*.5.
You ought to get a quadratic equation that looks like .25t^2+3.1t37=0. Hopefully this all makes sense as I have to take off for the night. 



#13
Feb1608, 11:36 PM

P: 20

now that it's a quadratic formula so i solved for t...(finding roots) it gives me 7.5 :) I understand now! Thanks!
finally to solve for b) I would do 0.5 X 7.5 (aXt = v) from a=v/t = 3.75 (3.8) 



#14
Feb1608, 11:37 PM

P: 20

c) player 2 d=vxt =3.1x7.5 = 23m




#15
Feb1608, 11:42 PM

P: 20

player 1 d=vxt = 3.8x7.5 = 28m ?? if you divide it by 2, it would be 14 but why would you divide by 2?




#16
Feb1708, 12:13 AM

P: 235

player 1:
[tex] x=ut+\frac{1}{2}at^2[/tex] [tex]x=(0)(7.5)+\frac{1}{2}(0.5)(7.5)^2[/tex] [tex]x=14[/tex] You cannot use the average speed equation because the player is accelerating 



#17
Feb1708, 04:51 AM

P: 20

For player 1:
x=0.25t^2 (this player is just accelerating) when you subtitute t by 7.5 you ger 14...as expected... Thank you for the questions...it turns out to be a good preparation for my first physics test 



#18
Feb1708, 08:12 AM

P: 20

Thanks everyone for your contribution! It really helped.
Torresmido, no problem, I hope you ace that test. Do these questions over and over until it's embedded in your mind. I'm doing grade 11 physics at the moment so I'm learning the basic concepts. My textbook/teacher gives really easy examples so when it comes to problems like this, I have no clue where to start! 


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