Train Acceleration on Sharp Turn

In summary, the train rounds a sharp curve and slows from 80.7 km/h to 58.9 km/h in 12.4 seconds. The magnitude is -.712 and the direction is 89 degrees.
  • #1
chanv1
30
0

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?
 
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  • #2
chanv1 said:

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:

[tex] v^2 = v_0^2 + 2 a \Delta x [/tex]

Careful with your units. You have km/h.
 
  • #3
when I plugged it into that equation, I got ..

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

is that correct?
 
  • #4
chanv1 said:
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.
 
  • #5
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:
  • #6
chanv1 said:
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
 
  • #7
hm, how did you get v = 16.36 m/s? I'm not getting that number ..
would you mind showing me your steps please?
 
  • #8
chanv1 said:
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.
 
  • #9
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
chanv1 said:
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.
 
  • #11
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:
  • #12
chanv1 said:
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
 
  • #13
thank you so much for your help LP !
 

1. What is magnitude?

Magnitude refers to the size or amount of a physical quantity, such as force or velocity.

2. How is magnitude measured?

Magnitude is typically measured using a numerical scale, such as meters or newtons, depending on the specific quantity being measured.

3. What is direction?

Direction refers to the path or orientation of a physical quantity, such as the direction of a force or the direction of motion.

4. How is direction represented?

Direction can be represented using words, symbols, or vectors. Vectors are often used in physics to represent both magnitude and direction, with the direction indicated by the angle and the magnitude by the length of the vector.

5. Why is it important to consider both magnitude and direction?

In physics, the magnitude and direction of a physical quantity are both essential components in fully describing a system. For example, the force acting on an object may be the same in magnitude, but its direction can greatly impact its effect on the object's motion.

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