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When and where do the cars meet? 
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#1
Feb713, 02:23 PM

P: 98

1. The problem statement, all variables and given/known data
Two traffic lights are 100m apart. Light 1 being due west from light 2. Car 1 is moving east at a constant speed of 25m/s. When car 1 passes light 1, car 2 starts from rest, west at a constant acceleration of 2.0 m/s². Where do they pass and how long after the light changes do they pass? 2. Relevant equations I was thinking Vf² = Vi² + 2aΔx 3. The attempt at a solution Well I tried solving for the final velocity of car 2, since Vf of car 1 would be the same as initial. Vf² = Vi² + 2aΔx = 0² + 2(2.0m/s/s)(100) Vf² = 400 Vf = 20 I'm not sure if this was the right approach but can someone point me in the right direction here? 


#2
Feb713, 02:28 PM

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P: 41,316

I assume car 2 starts out from light 2?
See if you can express the position of each car as a function of time. (Hint: Measure the position of each from the same point.) 


#3
Feb713, 02:55 PM

P: 98

Well if the assumption I'm making based on your advice...I'd choose Δx = 1/2(Vf + Vi)t for the formula. So for time I believe it would be t = (2Δx)/(Vf+Vi) I decided to measure from 50m so Car 1 = 2(50)/50 = 2 seconds to reach 50m and assuming I calculated Vf correctly up in the original question (but subbing in 50 for 100) Car 2 = 2(50)/14.14213562 = 7.071067814 seconds to reach 50m So now do I just set these 2 equal to each other? 25 m/s = 7.071067814 m/s divide both sides by 7.071067814 making the time it takes to pass 3.5 s? 


#4
Feb713, 03:07 PM

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P: 41,316

When and where do the cars meet?
You'd be better off using: [tex]x = x_0 + v_0 t + (1/2) a t^2[/tex] That's what I had in mind when I spoke of position as a function of time. Note: The calculation you made for Vf in your first post is not quite relevant. You found the speed of car 2 when it reaches light 1, not when it passes car 1. 


#5
Feb713, 03:24 PM

P: 98

Actually: Δx = Vi*t + (1/2) a t^2...could this also be written as: X  Xo = Vi*t + (1/2) a t^2 X = Vi*t + (1/2) a t^2 + X_o Yeah, that makes sense...okay so lets see Car 1 would be 25m/s * t Car 2 would be (1/2)(2.0m/s/s) t^2 + 100m so (1m/s/s) t^2 + 100m So NOW set them equal to each other? (1m/s/s) t^2 + 100m = 25m/s/s t (1m/s/s) t^2  25m/s/s t + 100m = 0 Ahh a quadratic...hmm well after the calculating I get 28.5 , 3.51 Obviously time can't be negative so I'm assuming 3.51 would be the time they intersect?? 


#6
Feb713, 03:40 PM

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P: 41,316




#7
Feb713, 03:44 PM

P: 98

25 / 7.071067814 comes out to 3.535533905 with 3 sig figs = 3.54 and this other method came to 3.507810594 and here would be 3.51 So with the 3 sig figs it would make a difference...not much at all...but different 


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