# Linear Acceleration in a Circular Path

1. Oct 28, 2007

### pecosbill

1. The problem statement, all variables and given/known data

A schoolbus travels around a circular path with acceleration a(t)=0.5t m/s/s with t in seconds.
At some point it has a velocity of 8 m/s.
What are the magnitudes of its velocity and acceleration when it has travelled a fourth of the circular track from the point at which it had v = 8 m/s?

Radius of the track is 250m

2. Relevant equations

ds=vdt

3. The attempt at a solution

dv=0.5tdt

I integrate both sides, but get confused as to where I should integrate from since the time the bus has a velocity of 8 m/s is not provided. After consideration, I choose:

v initial is 0, v final is 8 and t initial is 0, t final is just t

After integrating, I solve for t to find

t=6.32 at v=10 and v(t)=0.25t^2

then if a fourth of the circle is travelled, the distance travelled is a fourth of the circumference:

(2*pi*250)/4=125*pi

then ds=vdt=(0.03t^2)dt
We integrate again to find distance travelled as a function of t
integrate ds from 0 to 125*pi and vdt from 6.32 to t, yielding

125*pi=0.083(t)^3-0.083(6.32)^3

From this equation, t=17.08 seconds or it will take 17 seconds to travel a fourth of the way around from this point. From here, we can plug into the a(t) and v(t) equations to answer the question.

I think this is right; however, the first time I solved the problem, I integrated the dv=adt expression from 10 to v on the dv side and 0 to t on the adt side to get v(t)=0.25*t^2+10
I then integrated it again and found the distance as a function of time, solving the time it took to travel from the point at 10 m/s to a point a fourth of the way around the track. What is wrong with this approach?

2. Oct 29, 2007

### Mindscrape

Your bounds for your first approach are strange. You are saying that

$$\int_{10}^{v(t)} dv = \int_{10}^{t} a dt$$

but you don't know v(10).

3. Oct 29, 2007

### pecosbill

no, that's not correct.
first, the acceleration integral is from starting point t=0.
secondly, the velocity integral is going from 0 to 8 (or as I mistyped it, 10)

4. Oct 29, 2007

### pecosbill

also, it will take 17-6.32 seconds to travel that 200*pi

5. Oct 30, 2007

### Mindscrape

I don't know if I confused you, or you're confusing me, but your second approach, the first half of your post is right.

For your other method are you saying you wanted to do (?)

$$\int_{v(0)}^{v(t)} dv = \int_0^t \frac{1}{2}t dt$$

that is what you did. If I take you literally (what was the typo again?), you are not considering that the bounds have to be the same on both sides of the integral

$$\int_{v(t_1)}^{v(t_2)} dv = \int_{t_1}^{t_2} .5 t dt$$

You can't integrate velocity from 0 8, and time from t(2) to t(10). I don't know if this is what were trying to do. On the other hand, you could solve for time when velocity is 8 this way.

$$\int_{0}^{8} dv = \int_{t(0)}^{t(8)} .5 t' dt'$$

If this doesn't clear things up, could you reformulate the other method you tried?

Last edited: Oct 30, 2007
6. Oct 30, 2007

### pecosbill

I think that's what I did. The confusion is in part because of the 8-10 typo, but also because of the two methods, and the fact that I don't have the capability to make integrals easily. If you know a program that will do this, I'd appreciate it.

The bounds and integrals that I used ended up being:

0 to the final velocity, say 8....integrated dv
equal to
t=0 to t=x, x being the unknown time at which v has reached 8 for the integral of adt

this gave me the time it initial took to go from rest to the point it was at v=8,
at which point i could do the integral of

d to d+125*pi the integral of ds
equal to
x to y(y being the time it took to travel 125*pi from the point v=8) the integral of vdt

does this make sense? i'm doing my best to explain the method clearly, but it's hard wihout integral notation here.

7. Oct 30, 2007