Constant Acceleration With Two Objects

Click For Summary

Homework Help Overview

The problem involves a train accelerating from a station and a passenger attempting to catch up to it after a delay. The subject area includes kinematics and equations of motion under constant acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the positions of the train and the passenger, questioning the setup of equations and the variables involved. There are attempts to derive a function for the passenger's speed and the time to catch the train.

Discussion Status

Participants are actively engaging with the problem, offering corrections and suggestions for deriving functions. Some have noted the importance of calculus in finding minimum values, while others have expressed confusion regarding the relationship between distance and velocity.

Contextual Notes

There is a mention of the passenger arriving 6 seconds after the train has passed a point, which introduces a time delay into the equations. Participants are also navigating the implications of using the quadratic formula in this context.

eltavo809
Messages
5
Reaction score
0

Homework Statement



A train pulls away from a station with a constant
acceleration of 0.40 m/s2. A passenger arrives at the
track 6.0 s after the end of the train has passed the
very same point. What is the slowest constant speed at
which she can run and catch the train?

Homework Equations



X = X0 + V0 * t + 0.5 * a * t2
V = V0 + a * t

The Attempt at a Solution



The passenger arrieves 6 sec. latter, so: tp = tt - 6sec

Passenger

X = V0 * ( t - 6 sec)

Train

X = 0.5 * 0.4 m/s2 * t2

Both, train and passenger, have to be at the same place at the same time in order to achieve an encounter, so:

xtrain = x passenger

V0 * ( t - 6 sec) = 0.2 m/s2 * t2So here is the problem. I know I have to find out V0 , but I have 3 variables and 2 equations. Any ideas? This is supose to be easy... :(
 
Last edited:
Physics news on Phys.org
For the passenger, X is Vo*(t-6) not +, careful!

An easy way to check this is plug in t=6s, he should be at x=0.

Then your strategy of setting X1=X2 is correct.

You should first get a function of time as a function of Vo. Then try to remember how you can find the minimum value of a function!
 
After the 6 s head start, the train is moving at 2.4 m/s and has gone 7.2 m.
Easier to let t be the time after this point.

(edited - I had a mistake earlier)
 
Last edited:
We know the position of the person is X1 = V0 * ( t - 6 )

The position of the train is X2 = 0.2 * t^2

All he needs to do is set X1=X2 and find a function of the time taken to catch up in terms of Vo, he can then use calculus to determine the best choice of Vo :)
 
Okay, I see it - had confused distance and velocity and got a v function with no minimum.
How interesting, at speed 5 the person catches the train in 9 seconds. At speed 4, she never catches it.
 
So, after some basic algebra, I got this:

v0 * (t - 6) = 0.2 * t2

-0.2 t2 + v0 t - v0 6 = 0

From the quadratic formula, I get:

(- v0 t +/- √ v02 + 4 * 0.2 * 6 v0) / 0.4

This should give me the time at which she catches the train, right?
 
Using the quadratic formula would be meaningless here, you don't want the values of Vo such that t=0 (which is what the quadratic formula would give you)!

Do you know how to take the derivative of a function and find its minimum value? :) If not I can just show you a way to find it =-).
 
Hey, dontdisturbmycircles, first of all let me thank you all your time and dedication.

Yes, i know how to derivate and find a minimun of a function. I just was not expecting to come to that kind of solution. Having said that, thanks again for the help. I got it now!
 
You are welcome ;).
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
Replies
12
Views
5K
  • · Replies 34 ·
2
Replies
34
Views
3K
Replies
2
Views
4K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
9
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 32 ·
2
Replies
32
Views
5K