Calculate tangential acceleration knowing velocity and time....

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JulienB
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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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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?
 
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.
 
That gives me a tangential acceleration of 3818.2 m/s2 btw.
 
It should give km/(3600h).

JulienB said:
That gives me a tangential acceleration of 3818.2 m/s2 btw.
That would be lethal.