Train Acceleration on Sharp Turn

[chanv1]

Homework Statement

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)

Homework Equations

a = change in V/ change in T

a = v^2/r

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?

 

[LowlyPion]

Homework Statement

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)

Homework Equations

a = change in V/ change in T

a = v^2/r

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?

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.

 

[chanv1]

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

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

is that correct?

 

[LowlyPion]

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

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

is that correct?

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.

 

[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?

 

[LowlyPion]

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?

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

 

[chanv1]

hm, how did you get v = 16.36 m/s? I'm not getting that number ..
would you mind showing me your steps please?

 

[LowlyPion]

hm, how did you get v = 16.36 m/s? I'm not getting that number ..
would you mind showing me your steps please?

Compute the acceleration at the moment the train speed reaches 58.9 km/h

That's the problem statement.

 

[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?

 

[LowlyPion]

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?

Yes and yes.

Tangential acceleration/Radial acceleration

Remember it is trailing the radius.

 

[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?

 

[LowlyPion]

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?

No. Pythagoras is your guide.

And no. Tangential acceleration/Radial acceleration

 

[chanv1]

thank you so much for your help LP !