The diameter of a car's tire is 52 cm. While the car is being driven, the tire picks up a nail. How high above the ground will the nail be after the car has travelled 0.1 km?

Question

It also asked me to graph this information. So I created a periodic graph that starts at 0 and then after half the circumference it is at 52 cm and then after the complete circumference it is back at 0.
But in the answers section of my textbook it started at -26 and after a complete rotation went back to -26. I was confused by this.

Answer 1

This looks like the bicycle wheel question I answered for you before.
I agree with you that the -26 and +26 might be confusing.
I think "they" considered the axle as the zero of elevation, so the lowest point in a rotation would be 26 cm below, and the highest point would be 26 above.
However the wording referred to the height above ground level, so I would have put it as 0 to 52.

Circumf of car's tire = 52π cm
.1 km = 10 000 cm
number of rotations at .1 km
= 10 000/52π = 61.213

So it is 0.213 into the 62nd rotation.
Can you find how high it is?

Answer 2

Thank you
I found that at 0.213 of a cycle the distance travelled would be 34.8 cm in terms of the first cycle. And I found the height at that distance to be 22.15 cm
Is that correct?