Uniform Circular Motion of a train

In summary, the problem involves a train slowing down while rounding a sharp horizontal turn with a radius of 150m. The train slows down from 90.0km/h to 50.0km/h in 15.0s and the question asks for the acceleration at the moment it reaches 50.0km/h. There are two components of acceleration, one found using kinematic equations and the other using the equation for centripetal acceleration. The final answer is found using the formula a = [(a^2) + (a(centripital))^2]^1/2.
  • #1
darkmagicianoc
15
0
I was wondering if someone could help me figure out what formulas to use solving this problem. My professor gave me the answer (as a way to check our answers) : 1.48 m/s^2 inward and 29.9 degrees backward

The problem is: A train slows down as it rounds a sharp horizontal turn slowing from 90.0km/h to 50.0km/h in the 15.0s that it takes to round the bend. The radius of the curve is 150m. Compute the acceleration at the moment the train reaches 50.0km/her. Assume it continues to slow down at this rate.

Thank you in advance for helping me!
 
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  • #2
There are 2 components of acceleration. Find 1 with a kinematic and the other with the equation for centripetal acceleration. Then resolve the vectors to find magnitude and acceleration
 
  • #3
Ok. So do I use the kinematic equation: r(final) = r(initial) + Vi(t) +.5a(t^2)? When I do I get -2m/s^2.

I know centripital acceleration is (v^2)/r. But do I use the 13.89m/s (50km/h converted)?
 
  • #4
I am still very lost! Can someone please further help me!
 
  • #5
I figured it out! You use the formula Vf= Vi + at to find the acceleration, then use the formula a(centripital) = (v^2)/r. Then you find the acceleration by the formula: a = [(a^2) + (a(centripital))^2]^1/2
 
  • #6
darkmagicianoc said:
I am still very lost! Can someone please further help me!
The kinematic equation applies to the tangential deceleration along the curved bend. You are not given the distance traveled along that bend. But you are given the time and change in speed. Can you find another equation of kinematics that will give you the tangential deceleration without knowing the tangential displacement?
 
  • #7
Thank you for all of your help!
 
  • #8
darkmagicianoc said:
I figured it out! You use the formula Vf= Vi + at to find the acceleration, then use the formula a(centripital) = (v^2)/r. Then you find the acceleration by the formula: a = [(a^2) + (a(centripital))^2]^1/2
yes! That gives you the magnitude of the acceleration. What about its direction? BTW, this is non-uniform circular motion.
 

What is uniform circular motion?

Uniform circular motion is a type of motion in which an object moves in a circular path at a constant speed. This means that the object covers equal distances in equal time intervals and its velocity is always tangent to the circular path.

How does a train move in uniform circular motion?

In the case of a train, uniform circular motion occurs when the train travels along a curved track at a constant speed. The track provides a centripetal force, which is necessary to maintain the train's circular path.

What is the centripetal force acting on a train in uniform circular motion?

The centripetal force acting on a train in uniform circular motion is the force that pulls the train towards the center of the circular path. This force is provided by the track and is necessary to keep the train moving in a circular path.

How is the speed of a train in uniform circular motion related to its centripetal force?

The speed of a train in uniform circular motion is directly proportional to its centripetal force. This means that as the speed of the train increases, the centripetal force also increases in order to maintain the circular path.

Can a train experience uniform circular motion on a straight track?

No, a train cannot experience uniform circular motion on a straight track. In order for uniform circular motion to occur, the object must continuously change direction, which is not possible on a straight track.

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