Quick question about a train/car problem

  • Context: High School 
  • Thread starter Thread starter imagiro1
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Discussion Overview

The discussion revolves around a kinematics problem involving a car and a truck, specifically focusing on how to calculate the time it takes for the car to catch up to the truck given their speeds and the initial distance between them. The scope includes mathematical reasoning and conceptual clarification of the equations of motion.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant presents a kinematic equation but struggles with incorporating the initial distance of 100m between the car and the truck into their calculations.
  • Another participant corrects the initial equation format and points out that acceleration is zero for both vehicles.
  • A participant expresses confusion over the signs in their equations and questions whether changing the distance to a negative value accurately represents the car being behind the truck.
  • There is a suggestion to delve deeper into the mathematics of physics, emphasizing the importance of understanding calculus in solving such problems.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation of the equations and the interpretation of the initial conditions. There is no consensus on the best approach to solve the problem, and some confusion remains regarding the application of the equations.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in the problem, particularly in how to accurately represent the initial distance and the implications of using negative values in their equations.

Who May Find This Useful

This discussion may be useful for students working on kinematics problems, particularly those involving relative motion and the application of equations of motion in physics.

imagiro1
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I got a few problems like this one. The only thing I can't figure out is where to put the km. Here's what I got. A car traveling 95km/hr is 100m behind a truck traveling 75km/h. How long will it take the car to reach the truck. My 2 formulas are:

X=Xo+Vo+(1/2)at2
Car: X=95t2
Truck: X=75t2

Put both formulas equal to each other and solve for t. But where do I plug in the 100m that the car is behind? Thanks.
 
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Welcome to PF!

Hi imagiro1! Welcome to PF! :smile:
imagiro1 said:
X=Xo+Vo+(1/2)at2
Car: X=95t2
Truck: X=75t2

Nooo :cry: …

i] it's X=Xo+volt+(1/2)at2

ii] a (the acceleration) is obviously zero, isn't it?
Put both formulas equal to each other and solve for t. But where do I plug in the 100m that the car is behind? Thanks.

iii] that would be the Xo (a different one for each vehicle, of course) :wink:
 
I forgot that t, I swear I got it on paper.

I ended up with Xc=.110+95t and Xt=75t, which gave me the right answer, but negative. I tried it again with Xc=95t and Xt=.110-75t and it gave me the 20 sec which is the answer.

If I change it to -.110 would that be the same thing as saying the car is .110km behind the truck?
 
imagiro1 said:
I ended up with Xc=.110+95t and Xt=75t, which gave me the right answer, but negative. …

The car is 100m behind the truck …

so if the truck is at position 0 at time 0 (consistent with Xt = 75t), then the car is at position -100 at time 0, so Xc = … ? :smile:
 
-.110+95t. Awesome. Thanks for the help. I'm sure I'll be back later.
 
Imagiro, all tiny tim said is perfect but for further progresses I may suggest to try to see deeper into the methematics behind physics. I mean x=x0+v0t+1/2at^2 is not just a recipt where you input some data and it gives you another, certainly it can do so, but it is just its most superficial use. Cinematics, at the level that you are working (about the one the problem is), can be fully understood by learning and catching the basic calculus (functions, derivatives and integrals, in general).

Then, once you regard space and time as variables in a "calculus" way, you'll have no problem in solving problems like this and a lot more complicated that you never expected, even trying, before.

That's my advice, HOPE it's usefull.

Good night and good science :biggrin:
 

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