00:01
So essentially, most of the analysis is done in sample problem, a car in a banked circular turn.
00:21
And the analysis leads to equation 626, which gives us the theta equaling arc tan of v squared over the acceleration due to gravity times the radius of curvature.
00:40
Now, we know that here v is equaling 60 kilometers per hour.
00:45
We're going to multiply this by 1 ,000, rather we can say 1 meter per second for every 3 .6 kilometers per hour, giving us approximately 17 meters per second.
01:00
And we know that r is 200 meters.
01:04
So given this, we can then solve for theta.
01:08
And theta would be arc tan of 17 meters per second, quantity squared, divided by 9 .80.
01:20
Meters per second squared multiplied by r of 200 meters and this is going to give us 8 .1 degrees.
01:31
Now if we were to consider a vehicle taking this banked curve at 40 kilometers per hour, we can say that v prime would be equal to 40 kilometers per hour.
01:44
Let's convert one meter per second for every 3 .6 kilometers per hour.
01:52
This is going to equal 11 meters per second.
01:57
And we can then say that it's horizontal acceleration here would be considered a prime.
02:03
This would be again v prime squared divided by r and it has components parallel to the incline and perpendicular to the incline.
02:13
So we can say that a parallel would be equal to a prime cosine of theta where a perpendicular would be equal to a sign a prime sine of theta...