Can anyone help me with these two strangely simple yet confusing velocity problems?

  • #1
fphysicsclass
37
0

Homework Statement



I've already done 10 probs, but I can't figure these out. Please help if you can.

1. An automobile traveling v = 90 km/h overtakes a L = 1.00 km long train traveling in the same direction on a track parallel to the road. If the train's speed is 80 km/h, how long does it take the car to pass it, and how far will the car have traveled in this time?
x min
x km
What are the results if the car and train are traveling in opposite directions?
x s
x km

2. An unmarked police car traveling a constant 125 km/h is passed by a speeder traveling 140 km/h. Precisely 1.00 s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 2.50 m/s2, how much time passes after the speeder passes before the police car overtakes the speeder (assumed moving at constant speed)?

x s

Homework Equations



dx=v(0)t+(1/2)at^2
v^2=v(0)^2+2adx

The Attempt at a Solution



1. dx=10t? Tried making it into a rel. vel. problem.

2. (38.9)^2=(34.7)^2+2(2.5)dx ?? I've been working on this one for an hour...
 

Answers and Replies

  • #2
fphysicsclass
37
0


Can anyone help?
 
  • #3
mplayer
152
0


For the first one, I would try thinking about the difference in speed between the two moving objects. Ask yourself how fast the car is moving relative to the train. Then, given that information, how long does it take to travel that 1 km length of the train. You will then have a time to use to figure out the distance traveled on the road.
 
  • #4
azizlwl
1,065
10


1. Both are moving. The train is leading by 1.0km at time t=0. You have to find the time for both the train front fender and car front fender at same position/displacement.
2. The same procedure except one body is accelerating and the other at constant velocity.

Sketch graph of position vs time for question 1 and velocity vs time for question 2 is very helpful.
ISEC (Isaac Newton)- indentify, sketch, execute and finally check
 
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  • #5
fphysicsclass
37
0


I did draw the graphs before, and they are accurate. I'm just bad at using equations. I'll let you know in less than 5 mins. what I tried and got on the first one.
 
  • #6
fphysicsclass
37
0


Also, it wouldn't make sense for the first one to draw a v-t graph because v doesn't tell you anything about position. Should I instead drawn a x-t graph?
 
  • #7
fphysicsclass
37
0


ehh. I can't even figure out the first one. So, I have 90t + 1 = ?

I don't know where to go from there. Oh wait. 90t+1 = delta x?
 
  • #8
fphysicsclass
37
0


I don't know. Can anyone help? this is frustrating, and I almost have to turn it in online. I tried graphing the x-t graph, and can't make sense of it. I understand that one must equate 2 equations - that is how you find the intersection. But my answer is always wrong. I just want some help. Thanks.
 
  • #9
mplayer
152
0


Ok, ask yourself this:

If you were standing on the train, how fast would you perceive the car to be moving?

If you take the perceived speed and use it to figure out how long it takes for the car to travel 1km (the length of the train), what number do you get? It's just as simple as t=d/v.

So then you would have the amount of time that it takes the car to pass the train and can use that and the specified speed to figure out the actual distance the car traveled.
 
  • #10
azizlwl
1,065
10


1. x-t diagram. At t=0, train at position 1km, and car at position 0.
Then the position increases depends on the speed. Both at constant speed, thus a straight lines. One start from x=1km and the other from origin. Where the lines intersect is the place they meet(both front ends).

2. Drawing a parabolic curve is a bit hard especially with my poor free-hand drawing. But if you draw a v-t, a constant velocity is a straight line parallel to x-axis and constant acceleration a slanting one with the gradient equal to the acceleration. The distance move is just v times t. From a constant velocity is just a rectangular area and for constant acceleration it is a triangular area.
 
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  • #11
fphysicsclass
37
0


For 1, I did 10t = 1; t = 6 min. Does anyone (dis)agree, and can help with 1 and/or 2?
 
  • #12
fphysicsclass
37
0


azizlwl - I will try your method, and I think I messed up in drawing the x-t graph for a = c. I drew the v as exponentially increasing. I got it confused with the x-t graph. You know how it is, first day of AP Phys. A little lost. Haven't taken phys in 2 years. Will keep you updated.
 
  • #13
fphysicsclass
37
0


Ok. So for 1, I got 6 min and 9 km. Now I need to do part b.

Also, I will be starting 2 after that.
 
  • #14
mplayer
152
0


Those are the answers I got for 1a. Just use a similar approach for 1b.
 
  • #15
fphysicsclass
37
0


On #1, how should I start out on the opposite directions. I know that the train will be traveling @ -80 kmh, and the car, +90 kmh. I don't know what the relative velocity of the train would be. Maybe -170 kmh?
 
  • #16
fphysicsclass
37
0


I probably should find the rel. vel. of the car. If I did, would I be right to use 170t = 1 to solve for time? I got 0.02 s.
 
  • #17
mplayer
152
0


I probably should find the rel. vel. of the car. If I did, would I be right to use 170t = 1 to solve for time? I got 0.02 s.

Yes.

Well, yes to the equation, but not sure where the 0.02s is coming from.
 
  • #18
fphysicsclass
37
0


170 is in kmh. I converted my answer to seconds because that is what the answer form asked for. I made sure to include it in my original post.
 
  • #19
fphysicsclass
37
0


Which actually should be moved over a couple decimal places and be 21.2 seconds.
 
  • #20
mplayer
152
0


May want to double-check that conversion. 0.02 s is an awfully quick kilometer :P
 
  • #21
fphysicsclass
37
0


Haha, yes it is. That is why I did double check.

I submitted my answer for that one, and got it right. It was 21.2 seconds, and 0.53 km.

Now for #2!
 
  • #22
fphysicsclass
37
0


Thank God you can type in "1/170 hours to seconds" and get a perfect conversion.
 
  • #23
fphysicsclass
37
0


I'm lost again. I was trying to find distance when I should have been trying to find time. Any tips for starting number 2? I already drew a v-t graph, making sure my constant a was a straight line.
 
  • #24
fphysicsclass
37
0


I started number 2 using knowledge from 1. I set 2 displacement equations equal to each other. Was I wrong to say that: (1/2)(2.5)(t^2)=4.2t ?? I will add a second at the end to make up for the lost time.
 
  • #25
fphysicsclass
37
0


And I ended up on 2 getting 4.36 seconds. Did anyone else get this?
 
  • #26
azizlwl
1,065
10


From the graphs,

1a. Taking frontend as a point for intersection
1+80t=90t
1b. Taking backend of the train as origin and the point for intersection,
80t=-90t+1

2. 140/3600(t+1)=123/3600(t+1) + 0.5at2
 
Last edited:

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