## Calling all physicists, especially those who are an expert i

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### Calling all physicists, especially those who are an expert i

Ok, xkcd, I hate to do this but this is my last resort. I am calling in the big guns to help me answer a friend's physics question homework. I haven't taken kinematics in awhile so I am a bit rusty so I will detail the question out, there is no diagram. I spent literally like 40 minutes on this question and this is as far as I got. See question below.

An automobile enters a constant 90 meter radius of curvature turn traveling at 25 m/s north and exits the curve traveling east. The car completes the turn in 5.4 seconds. Assume the speed of the car can be modeled as a quadratic function of time.

Ok so here is what I did.

I modeled the velocity equation, (the professor told us that the function of speed and velocity can be used interchangably). We are required to figure out initial
and final values of arc length, velocity, acceleration, angular velocity, and angular acceleration.

v(t) = a(t)^2 + bt + c

plugged in initial conditions t(0) and simplified the velocity function to

v(t) = a(t)^2 + bt + 25

so for the final value of velocity equation I have

v(5.4) = a(5.4)^2 + (5.4)b + 25

I used derivation to figure out the function for acceleration which I got to be: (ac is acceleration since a is already used)

ac(t) = 2at+b

and using initial conditions I found the initial acceleration equal to:

ac(0) = b

and final value of acceleration to be

ac(5.4) = 10.8+b

and finally I used integration of the velocity function to get the function or arc length equal to (I used the equation s = r(x) where x is equal to theta, s = arc length and r = radius, in radians to figure out arc length)

90s(t) = a(t)^3 / 3 + b^2 / 2 +25t.

Using initial conditions I simplified initial arc length to 0 and final arc length to be equal to

52.488a + 14.58b = 70.686 after simplifying by figuring out final arc length to be 70.686 by using trigonomtery to figure out that the car did an exact quarter of a circle arc.

And that is as far as I got, I have 3 equations with 4 unknowns and no idea how to figure out the rest of the data.
Any help would be greatly appreciated, time is a issue here, it is due in about 6 hours.
Last edited by BirdMav on Mon Apr 16, 2012 8:39 am UTC, edited 1 time in total.
Bipolar : The maximum of this function you can't begin to comprehend as well as the minimum of this function you cannot begin to end.
BirdMav

Posts: 52
Joined: Sun Oct 24, 2010 4:03 pm UTC

### Re: Calling all physicists, especially those who are an expe

BirdMav wrote:See question below.

An automobile enters a constant 90 meter radius of curvature turn traveling at 25 m/s north and exits the curve traveling east. The car completes the turn in 5.4 seconds. Assume the speed of the car can be modeled as a quadratic function of time.

What's the question? You state some facts and an assumption, but I really don't see what you're trying to calculate.
tooyoo

Posts: 96
Joined: Sat Jan 22, 2011 5:39 pm UTC

### Re: Calling all physicists, especially those who are an expe

Question is find the initial values and final values of linear (tangential velocity), linear acceleration, angular velocities and accelerations.

Initial value is at time = 0 seconds and final value is at time = 5.4 seconds.

EDIT: For the record, I did not write the question, I think it is very poorly worded and constructed with piss poor vocabulary.
Bipolar : The maximum of this function you can't begin to comprehend as well as the minimum of this function you cannot begin to end.
BirdMav

Posts: 52
Joined: Sun Oct 24, 2010 4:03 pm UTC

### Re: Calling all physicists, especially those who are an expe

Well, it looks like you're doing more or less the right thing - possibly with some smaller errors.

Now, to get a start on the problem: The angular velocity and acceleration are trivial to compute once you have the tangential ones. The tangential acceleration is also the derivative of the tangential velocity. So it's sufficient to compute that.

\dot{x}(t) = \alpha t^2 + \beta t + \gamma

with three constants. As you correctly point out, we have
\gamma = \dot{x}(0) = 25m/s

To get the position, you can simply integrate. So after 5.4 seconds, the car has traveled
x(5.4s) = \frac{\pi}{2} \times 90m = \frac{\alpha}{3} (5.4s)^3 + \frac{\beta}{2} (5.4s)^2 + 25m \times 5.4

You then proceed to solve this for one of the two remaining constants. The point is - and this is what seems to be confusing you - that the system is not fully determined. You're more or less able to choose one of the constants freely, the other will depend on it.

This should not be surprising. If you want to drive a distance in a given time, you can either drive at a constant speed or drive slow first and later fast, or first fast later slow, or ... - Here you don't have this full freedom, since you're restricted to a quadratic dependence of velocity on time.
tooyoo

Posts: 96
Joined: Sat Jan 22, 2011 5:39 pm UTC