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.
The car is traveling East at 30 m/s with respect to the road but is orbiting the center of the Earth at
Answers & Comments
Answer:
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.
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