# Magnitude & Direction!

1. Oct 8, 2008

### chanv1

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

A train slows down as it rounds a sharp horizontal turn, slowing from 80.7 km/h to 58.9 km/h in the 12.4 s that it takes to round the bend. The radius of the curve is 165 m. Compute the acceleration at the moment the train speed reaches 58.9 km/h. Assume that it continues to slow down at this time at the same rate.

Magnitude?
Direction? ____° backward (behind the radial line pointing inward)

2. Relevant equations

a = change in V/ change in T

a = v^2/r

3. The attempt at a solution

I've attempted the equations above and got 42.37 and 0.489 for magnitude, which are both wrong. Can someone please help?

2. Oct 8, 2008

### LowlyPion

This might be a better way to determine a:

$$v^2 = v_0^2 + 2 a \Delta x$$

Careful with your units. You have km/h.

3. Oct 8, 2008

### chanv1

when I plugged it into that equation, I got ..

16.36^2 = 22.42^2 +2a (165)
a = -.712

is that correct?

4. Oct 8, 2008

### LowlyPion

Assuming your math is correct, now what is the centripetal acceleration?

Since these are both vectors, you can add them to determine total magnitude and direction.

5. Oct 8, 2008

### chanv1

so to find V, I used the eq. V = (2 pi r)/t and got V = 83.61

then plugged the V into V^2 / r and got centripetal acceleration of 42.37 m/s/s

then from what you said, I added the two vectors to determine magnitude

-.712^2 + 42.37^2 = 1795.72^2 so ultimately the magnitude would equal 42.38

right?

would the direction be inverse tan (42.37/.712) = 89 degrees?

Last edited: Oct 8, 2008
6. Oct 8, 2008

### LowlyPion

Whoa. Wait a minute. The tangential v is given as 58.9km/h or 16.36 m/s

v = 16.36 m/s
a = v2/R = 1.622 directed radially

7. Oct 8, 2008

### chanv1

hm, how did you get v = 16.36 m/s? I'm not getting that number ..

8. Oct 8, 2008

### LowlyPion

That's the problem statement.

9. Oct 8, 2008

### chanv1

So then do you add 1.622 and -.712 together to get the magnitude? How do I calculate the direction? inverse tan of the numbers?

10. Oct 8, 2008

### LowlyPion

Yes and yes.

Remember it is trailing the radius.

11. Oct 8, 2008

### chanv1

Okay, just to double check..

so it's 1.622 + -.712 = .91
and inverse tan (1.622/-.712) = -66.3 degrees? or do I add 180 degrees to that?

or do I have to do 1.62^2 + -.712^2 = c^2?

Last edited: Oct 8, 2008
12. Oct 8, 2008