Train turning, tangential acceleration and total acceleration

  • Thread starter gap0063
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  • #1
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1. (a)A train slows down at a constant rate as it
rounds a sharp circular horizontal turn. Its
initial speed is not known. It takes 16.6 s to
slow down from 86 km/h to 30 km/h. The
radius of the curve is 162 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.

(b) 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.
1. +
2. Not enough information
3. 0
4. −

(c)At the moment the train’s speed is 50 km/h,
what is the magnitude of the total accelera-
tion?
Answer in units of m/s2.




Homework Equations


atangential = d|v|/dt
not sure on others


The Attempt at a Solution


I tried part (a) with atangential = d|v|/dt
where atangential = |30-86|/16.6=3.37349 km/h^2
then converting to m/s^s by times 3600s and dividing 1000m and getting 12.1446 m/s^2
wrong
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,160
504
You are messing up your units and conversions....first change km/hr to m/s, find the change in v, then divide by t (in seconds) to get the tangential acceleration.
 

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