Answer
Relative to the hitchhiker, the magnitude of the acceleration at the top of the tire is $1500~m/s^2$
Work Step by Step
We can convert the speed of the car into units of $m/s$:
$v = (80~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 22~m/s$
The velocity of the tire's center of mass is equal to the velocity of the car.
Since the car is moving at a constant velocity, the car's acceleration is zero, and the acceleration at the center of the tire is zero.
Since the hitchhiker is stationary, the hitchhiker's acceleration is zero.
At the top of the tire, there is a centripetal acceleration directed toward the center of the tire. We can find the magnitude of this acceleration:
$a = \frac{v^2}{r}$
$a = \frac{(22~m/s)^2}{0.33~m}$
$a = 1500~m/s^2$
The magnitude of the acceleration at the top of the tire relative to the hitchhiker is $(1500~m/s^2-0)$ which is $1500~m/s^2$
Relative to the hitchhiker, the magnitude of the acceleration at the top of the tire is $1500~m/s^2$.
