# Non-Uniform Circular Motion: Locomotive Rounding a Curve

1. Oct 28, 2011

### Becca93

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

As a locomotive rounds a circular curve of radius 2.10 km (which would be 2100 m to keep all the units the same), its speed is increasing at a rate of 0.440 m/s2. An instrument in the cab (an accelerometer) indicates that the magnitude of the locomotive's total acceleration at a particular instant is 0.760 m/s2. What is the locomotive's speed at that instant?

After I got it wrong the first few times, I was also given the hint: "The total acceleration is the VECTOR sum of the centripetal acceleration and the tangential acceleration."

2. Relevant equations

The equations I have in my notes regarding non-uniform circular motion are:

Tangential acceleration: At = d|v|/dt

and Total Acceleration: Atot = √(Ar^2 + At^2)

3. The attempt at a solution

To solve, would it be correct to do the following:

Ar = √(Atot^2 - At^2)

And then sub that number into

V = √((RAr)/(-m)

But, if I were to do that, I would get the square root of a negative number, which is an irrational number, which I can't have as a velocity?

2. Oct 28, 2011

### Staff: Mentor

The negative sign just indicates that the acceleration is towards the center. Just worry about the magnitude.

3. Oct 28, 2011

### Becca93

But there is no 'm' given in the question, and I don't know how to get radial acceleration any other way.

4. Oct 28, 2011

### Staff: Mentor

Oops, I didn't see that. Your equation is not quite right:
That's the centripetal force. The radial acceleration is just v^2/R.

5. Oct 28, 2011

### Becca93

I must have copied it incorrectly during class. Thank you!