# Unicycle straight line motion/time dependent acceleration Problem

1. Apr 21, 2014

### chiamocha

1. The problem statement, all variables and given/known data
Here is the problem I have been trying to figure out for the past hour.
A unicycle is traveling in a straight line, along the x axis. At t=0, the unicycle is at x = D. Initially the cyclist is accelerating backwards, in the minus x direction. Over time, the acceleration increases. The time-dependent acceleration is ax(t) = -2a + 6dt The quantities a and d are constants.

2. Relevant equations

All I know is the equations of constant acceleration vx(t) = v0x + ax(t)

position x(t) = x0 + v0xt + (1/2)axt2

vx2= v02x+2ax(x-x0)

3. The attempt at a solution

What I did to attempt this equation (sorry I am not very good at coding this properly so I will just write it out) is to integrate the equation for time dependent acceleration that was given to get equations for velocity [v = v(initial) +(from 0 to t) ∫ (a) dt'] and then for position[ x = x(initial) + (from 0 to t) ∫(v)dt']. And then since it is given that at time t=0 the object is at position D I plugged that in and since I assume (I am not sure I am correct in this assumption) that the initial position of the object is 0, I set the integral of v from 0 to t equal to D. And I am not sure where to go from there. And I am not sure how to account for the backwards acceleration in the minus x direction, and then change when over time the acceleration increases.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 22, 2014

### haruspex

That's a strange thing to do.
X(0)=D. X(t)=D+integral of v from 0 to t.
Please post what you get for those integrals.

3. Apr 22, 2014

### BiGyElLoWhAt

Hmmm... not sure what you're getting at.

If your looking for the position function, integrate it, plug in your initial conditions for your velocity (I don't recall you giving a magnitude, just a direction? ) then integrate again and use your initial conditions for position to solve for your constant if integration again.

From what you gave, I'm not sure that integrating from 0 to t should give you zero. Definitely not on your velocity (unless you're trying to solve for when it DOES equal 0), which is not t-0 as you stated that the cyclist was moving backwards.

Also I believe you said at t-0 the position was D, also not necessarily 0. If it was zero, likely your prof would use 0 and not D. Can't really do a whole lot other than limiting your first constant of integration with the given conditions, and thus limiting your second one (or getting it in terms of the first)

But again... you really didn't ask a question.

4. Apr 22, 2014

### BiGyElLoWhAt

Also, these numbers look really cooked, just pointing that out.

5. Apr 22, 2014

### chiamocha

Sorry, the question is to find the velocity and position of the unicycle as a function of time.

6. Apr 22, 2014

### SteamKing

Staff Emeritus
If you know the position of the unicycle as a function of time, what is the velocity of said unicycle as a function of time? Hint: there are no integrals involved. Look at the definition of velocity.

7. Apr 22, 2014

### chiamocha

Isn't it the derivative of the position?

8. Apr 22, 2014

### haruspex

Chiamocha, in the OP you mentioned integrating acceleration to get velocity and integrating that to get position. That was all correct. Please post what you got for those integrals and where you went from there.

9. Apr 22, 2014

### rcgldr

The problem statement defines acceleration versus time:

$$a_x(t) = (-2a \ + \ 6)\ \Delta t$$
Integrate this twice adding a constant of integration (v0 for the first integration, x0 for the second integration) for each integration step. The problem states that the unicycle is traveling in a straight line, but doesn't define the initial velocity v0. It does define the initial postion x0 as D.

Last edited: Apr 22, 2014
10. Apr 22, 2014

### SteamKing

Staff Emeritus
Yes, it is.

Given the information in the OP, write the position of the unicycle as a function of time. Use the acceleration (as a function of time) you were given.