# Constant Acceleration With Two Objects

• eltavo809
In summary, the passenger arrives at the track 6 seconds after the train has departed and needs to find the slowest constant speed to catch the train. Using the equations for position and velocity, it is determined that setting the positions of the passenger and train to be equal will give a function of time in terms of the initial velocity of the passenger. By taking the derivative of this function and setting it equal to 0, the minimum value of the function can be found, giving the time at which the passenger catches the train.
eltavo809

## 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:
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 ;).

## 1. What is constant acceleration with two objects?

Constant acceleration with two objects refers to the situation where two objects are moving together in a straight line, and both are experiencing the same acceleration at all times.

## 2. How is constant acceleration calculated for two objects?

Constant acceleration for two objects is calculated by dividing the change in velocity of the objects by the time it takes for the change to occur. This can be represented by the equation a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

## 3. What factors can affect the constant acceleration of two objects?

The constant acceleration of two objects can be affected by factors such as the mass of the objects, the force applied, and the surface on which the objects are moving. Other factors like air resistance and friction can also play a role in altering the acceleration.

## 4. How does constant acceleration with two objects differ from constant velocity?

Constant acceleration with two objects is different from constant velocity in that the objects are not moving at a constant speed but rather experiencing a change in velocity over time. In contrast, constant velocity means the objects are moving at a constant speed in a straight line without any changes in velocity.

## 5. What are some real-life examples of constant acceleration with two objects?

Some real-life examples of constant acceleration with two objects include a car towing a trailer, two rollercoasters moving together on parallel tracks, and two objects falling towards the ground due to gravity.

• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
12
Views
4K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
34
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
6K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
4K
• Introductory Physics Homework Help
Replies
3
Views
2K