Question

A car that is moving at a constant velocity of 30 m/s, east for 5 seconds has an acceleration of what?

Answer 1

`a_c = 2.34*10^-2 m/s^2`

Explanation:

Let's assume we are talking about that car driving East in Columbus, Ohio, past the the Horseshoe Stadium. The surface of the Earth at that spot is traveling to the East at 356.49 m/s.

Reference: https://www.vcalc.com/wiki/MichaelBartmess/Rotational+Speed+at+Latitude

The car is traveling East at 30 m/s with respect to the road but is orbiting the center of the Earth at

`356.5 m/s + 30 m/s = 385.5 m/s`

The formula for centripetal acceleration is `a_c = v^2/r`.

So for the car, its centripetal acceleration is

`a_c = (3.855*10^2 m/s)^2/(6.36*10^6 m) = (14.86*10^4)/(6.36*10^6) m/s^2`

`a_c = v^2/r`

`a_c = 2.34*10^-2 m/s^2`

I hope this helps,
Steve