Calculate tangential acceleration knowing velocity and time....

In summary, the conversation discusses a train moving along a circle with a radius of 2 km. The train accelerates from 30 km/h to 100 km/h and crosses a distance of 1200 m during the acceleration. The time and tangential acceleration of the train are also discussed, with the formula aT = Δv/Δt being used to calculate the tangential acceleration. However, there is confusion over the units used in the calculation, with the correct units being km/(3600h) instead of m/s.
  • #1
JulienB
408
12

Homework Statement



A train moves along a circle with radius R = 2 km. The train accelerates with a constant rate, from v1 = 30 km/h to v2 = 100 km/h. It crosses SF = 1200 m during the acceleration.
a) Find the time tF of the acceleration.
b) What is the tangential acceleration aT of the train?

Homework Equations



aT = Δv/Δt
and anything relevant to non-constant displacement along a circle.

The Attempt at a Solution


[/B]
Although it seems pretty easy, I am unsure on how to use the formulas I have. For the first question, I calculated the average speed during the acceleration ((100-30)/2 + 30 = 65 km/h) and then the time it takes at this speed to cross 1.2 km (1 min 6s).

Then I have a lot of different formulas for how to calculate the tangential acceleration, especially that one:
aT = Δv/Δt

Can someone please help me with how to use this formula in a problem? I would guess Δd is (100 - 30)⋅602 (in order to get a result in m/s) and Δt is 66s, am I right?

Thank you very much in advance for your answers.J.
 
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  • #2
JulienB said:
It crosses SF = 1200 m during the acceleration.
During the acceleration? While it is still accelerating?

Then I have a lot of different formulas for how to calculate the tangential acceleration, especially that one:
aT = Δv/Δt

Can someone please help me with how to use this formula in a problem? I would guess Δd is (100 - 30)⋅602 (in order to get a result in m/s) and Δt is 66s, am I right?
Why do you need a Δd here (and why should it be different from 1200m?), and where do the factors of 60 come from?
 
  • #3
Yes it is still accelerating for 1200m. The factors of 60 are for the velocity to be expressed in m/s, I meant Δv and not Δd of course.
 
  • #4
That gives me a tangential acceleration of 3818.2 m/s2 btw.
 
  • #5
JulienB said:
... The factors of 60 are for the velocity to be expressed in m/s, ...
That will give km/s, not m/s.
 
  • #6
It should give km/(3600h).

JulienB said:
That gives me a tangential acceleration of 3818.2 m/s2 btw.
That would be lethal.
 
  • #7
mfb said:
It should give km/(3600h).

That would be lethal.
Yes, of course!

I stand corrected. DUH!
 

FAQ: Calculate tangential acceleration knowing velocity and time....

What is tangential acceleration?

Tangential acceleration is the rate of change of an object's tangential velocity, which is the speed and direction of its motion along a curved path.

How is tangential acceleration calculated?

Tangential acceleration can be calculated by dividing the change in tangential velocity by the change in time. It can also be calculated by multiplying the tangential velocity by the angular acceleration, which is the rate of change of an object's angular velocity as it moves along a curved path.

Can tangential acceleration be negative?

Yes, tangential acceleration can be negative. A negative tangential acceleration means that the object is slowing down or changing direction, while a positive tangential acceleration means that the object is speeding up or maintaining a constant speed along a curved path.

What units are used to measure tangential acceleration?

Tangential acceleration is typically measured in meters per second squared (m/s^2) or radians per second squared (rad/s^2).

How does tangential acceleration relate to centripetal acceleration?

Tangential acceleration and centripetal acceleration are related to each other through the equation a = v^2/r, where a is centripetal acceleration, v is tangential velocity, and r is the radius of the curved path. This equation shows that tangential acceleration is directly proportional to centripetal acceleration and inversely proportional to the radius of the curved path.

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