If a car is travelling at 60 mph down a straight road, what is the speed of the bottom, the top, & the dead center of the tires?
Speed with respect to what? The only part of this question that makes sense to my 15 watt brain is about the speed of the tire with respect to the pavement.
If you are coasting straight ahead, regardless of speed, the tire is going zero miles per hour with respect to the pavement. If the car is under power to maintain 60 MPH there is a small amount of slip between the tire and pavement. So the tire is going very slightly faster say 60.1 MPH.
In respect to the road at any given point!
I’d guess:
Bottom - stationary
Rearmost point 60
Top - forwards at 120
Front most point - 60
And which part of a train is always going backwards?
middle: 60mph
top: 60mph+linear velocity of the point at the top
bottom: same-extra velocity
Doesn’t seem that tricky… I’m with TheLoadedDog, with the addition that the centre is moving at 60mph 'cause he hasn’t specified. (Don’t know the answer to your riddle though).
chaoticdonkey would also seem to be technically right, if somewhat vague. 
I’m not so sure about the front most point. Wouldn’t it have a forward velocity component of 60 mph plus a downward velocity of 60 mph, adding up to 84.9 mph?
As for the train, the bottom edge of the wheel flange is always moving backwards.
The front “point” on the tire isn’t going forwards, just downwards.
Oh yeeeeah! 
Well that depends on how old the train is. Early trains were built with cylindrical wheels and flanged rails. So in this case, barring the case of the drunk careering backwards from the club car, no part of the train is moving backwards.
And who says I didn’t learn anything useful from my last job?
Doesn’t that depend which way the train’s going? And if it’s a wheeled train or a magnetic train?
OK, OK, we all knew what you meant, and someone got the answer…
Huh? Every point on the car has a translational velocity of 60mph. The wheel has an additional velocity due to the rotation around the axle. At any given moment, the front part of the tire has a rotational component that is completely downward, but this has to be added to the 60mph forward velocity that every point on the car has.
I think the trick part of the question is that the tire deforms due to the weight of the car, so it’s not a perfect circle. The bottom point has a zero velocity where it contacts the road (if we can assume no slip), meaning the rotational component is exactly 60mph backward to cancel the car’s 60mph forward. However, the radius of the wheel is slightly higher for the top part than the bottom part, so the top part of the wheel has a somewhat higher rotational velocity than the bottom, meaning the total velocity at the top of the wheel is somewhat more than 120mph.