Finding the Speed of an Old-Fashioned Truck

[Amaroq Zev]

(I can't rightly find the right place to post this question, so if you could even point me in the right direction there that would be cool. I am new to this site, I saw the many other velocity questions, so I thought maybe this one could wiggle in there. If not sorry, please redirect me.)
Problem: old-fashioned trucks used a chain to transmit power from the engine to the wheels. Suppose that the drive sprocket had a diameter of 6 in and the wheel procket had a diameter of 2o in. If the drive sprocket goes 300 rpm:
A. Find the angular velocity of the drive sprocket in radians per minute(I have concluded that the answer is..uh..100 rpm but I am not certain. I divided the rpm by 3 inches, the driving sprocket's radius.)
B. find the linear velocity of the 20-in wheel sprocket in inches per minute.(18849.566in/m 300 rev/m*2(pie) rad/1 rev*10in/1 rad
C.)Find the angular velocity of the wheel in radians per minute(30 rpm) (got that by dividing the 300 rpm by the wheel sprocket's radius, since the wheel and sprocket share the same axel, the angular velocity is the same for the two of them, so 300 rpm divided by 10 in)
D.(this is my question) Find the speed of the truck to the nearest mile per hour.(I started setting this up but I don't know if I did it right so far, and I also forgotten how many ft ar in a mile so I can't continue really...anyway here is what I got so far. 30 revolutions/1 minute*60minutes/1 hr*2(pie)rad/1 revolution*19inches/1 radian*1mile/? from there I would have put the feet in there then 1 ft over 12 inches...or...I would have figured out how many inches are in a mile seperately and put the answer in the corresponding position in the equation...have I done this right so far? The picture has a small circle(drive sprocket)connected to another circle(wheel sprocket) by a chain, along the outside of the wheel sprocket is a axel sharing circle(wheel) of 38" diameter, if anyone knows if I am doing this right please tell meh...pwease?
alright I got it, now to continue the equation I still need to know if I am setting this up right or not.
30 revolutions/1 minute*60minutes/1 hr*2(pie)rad/1 revolution*19inches/1 radian*1mile/5280 feet*1 ft/1in
okay so then that would make...40.69790483 which rounds off to 40.698 mph...is this correct?

 

[BigJDubs]

Please tell me if I am doing this right or not...Yes, your setup and answer are correct. You correctly converted revolutions per minute to angular velocity (in radians), linear velocity (in inches), and then to miles per hour.

 

[Moetasim]

I would like to commend you on your efforts in solving this problem and using the correct equations and units. Your approach seems to be correct and your final answer of 40.698 mph is also correct. However, I would suggest using a more precise value for pi (3.14159) in your calculations to get a more accurate answer. Additionally, it is important to note that the linear velocity you calculated is for the outer edge of the wheel sprocket, not the actual speed of the truck. To find the speed of the truck, you would need to take into account the size of the wheel and the number of revolutions it makes in a given time period. Overall, your understanding and application of angular and linear velocity is impressive and I encourage you to continue exploring the world of physics and mathematics.