# Setting up problem.

• Oct 1st 2009, 09:09 AM
Brndo4u
Setting up problem.
A railroad track is laid along the arc of a circle of radius 1800. The circular part of the track subtends a central angle 40 degrees. How long in Sec will it take s point on the front of the train moving at 30 mph to go around this portion.
I'm given: Radius-1800 Theta-40 Velocity-30. So, would I use V=r(theta)/time. Then, divide by 60 time to convert to seconds.
• Oct 1st 2009, 09:48 AM
Wilmer
Hint: length of arc = 40/360 * pi(2*1800) = pi(3600) / 9 = ?
• Oct 1st 2009, 10:26 AM
Brndo4u
Still does not really help me out.
• Oct 1st 2009, 03:17 PM
aidan
Quote:

Originally Posted by Brndo4u
A railroad track is laid along the arc of a circle of radius 1800. The circular part of the track subtends a central angle 40 degrees. How long in Sec will it take s point on the front of the train moving at 30 mph to go around this portion.
I'm given: Radius-1800 Theta-40 Velocity-30. So, would I use V=r(theta)/time. Then, divide by 60 time to convert to seconds.

Your speed is given in miles per HOUR -- 3600 seconds in a hour.

You can determine the arc length: $\displaystyle 400 \pi$ feet.
(Wilmer gave you that -- I just did the division of 3600 by 9 to leave 400.)

The speed: 30 mph is equivalent to:

$\displaystyle \dfrac{ 30 miles/hour \cdot 5280 feet/mile}{3600 seconds/hour}$ = $\displaystyle \dfrac{44 feet}{second}$

Then divide the arc length by the speed:

$\displaystyle \dfrac{ 400 \pi \, feet }{44 feet / second}$ to get the time in seconds to traverse the arc.
• Oct 1st 2009, 03:58 PM
Wilmer
OK; rounding out:

Arc length = 400*pi = 1257 MILES (if the given 1800 is in miles; IS IT?)

1257 / 30 mph = 42 HOURS

42 * 60 * 60 = 151,200 SECONDS.

In other words, your problem makes little sense.
Are you sure you entered it correctly?