Any point along the circular line at the center of the ring of the doughnut shaped mass will still undergo gravitational pressure from the fact that it is not centered on the total mass of the ring.
https://www.quora.com/Where-is-the-center-of-gravity-of-a-dense-and-massive-doughnut-located-And-what-shape-would-it-have-Would-it-be-a-point-at-the-center-of-its-empty-hole-or-a-gravity-line-in-the-center-of-the-doughnuts-ring. answerd byJeffery Werbock - musician and lecturer
T= ~32 hours
Car A = 1hr/50km
Car B = 1hr/60km
Distance= 550 miles = 885.1km
<550 miles >
(1/50 + 1/60) hr/km = Xhr/885.1km
((~0.03667) hr/km ) / 885.1km = T
Because a thermometer only measures temperatures
Let d represent as displacement.
Let R represent as resultant.
Let Φ represents as the angle.
d1 = 5 km, 90°
d2 = 3 km, 0°
d3 = 1 km, 90°
For horizontal x-axis:
For vertical y-axis:
For the resultant (or the magnitude displacement):
R = √((x)^2 + (y)^2)
For the angle (althought unnecessary):
Φ = arctan (y/x)
d1x = (5 km)(cos90) = 0 km
d1y = (5 km)(sin90) = 5 km
d2x = (3 km)(cos0) = 3 km
d2y = (3 km)(sin0) = 0 km
d3x = (1 km)(cos90) = 0 km
d3y = (1 km)(sin90) = 1 km
Summation of dx = (d1 + d2 + d3)x = 3 km
Summation of dy = (d1 + d2 + d3)y = 6 km
R = 3√5 km or 6.708203932 km
Φ = 63.43494882°
The displacement is approximately 6.71 km (or 7 km) at an angle of 63.43°.