A train slows down at a constant rate as it
rounds a sharp circular horizontal turn. Its
initial speed is not known. It takes 18.1 s to
slow down from 80 km/h to 33 km/h. The
radius of the curve is 163 m.
As the train goes around the turn, what is
the magnitude of the tangential component of
the acceleration?
Answer in units of m/s2

Question

anonymous · Answer

So its represented by a change in distance over time.... therefore 80-(33km/hr)/18.1 s

thn convert because its asking for m/s^2

anonymous · Answer

(13.06m/s)/18.1s = .721 ms^-2

anonymous · Answer

Then As the train goes around the turn, what is
the sign of the tangential component of the
acceleration? Take the direction in which the
train is moving to be the positive direction. 

I just use the sign of the previous answer which is positive? right?

turingtest · Answer

1) this is not a question of distance
2) you can't just divide by time again to get m/s^2, you need to follow the physics
all acceleration is in m/s^2 in SI units
the formula you need here seems to be$$a_{tan}=\frac{\Delta v}{\Delta t}$$

turingtest · Answer

...not sure why the radius of the curve is given

turingtest · Answer

note: you will need to convert km/hr into m/s

anonymous · Answer

isnt the velocity 80km/hr - 33km/hr ????

anonymous · Answer

47 km/hr converts to 13.06 m/s

turingtest · Answer

that is delta v, but you said that was distance in your answer
and yes to your conversion, so I guess you had the right answer, you just phrased it strangely for me
looks good now though :D

anonymous · Answer

I just spoke incorrectly... thankyou

anonymous · Answer

what about the second part of the question

anonymous · Answer

is it positive?

anonymous · Answer

It will be negative because it slows the train.