44.2 + Dg = 212 m
Dg = 212 - 44.2 = 167.8 m = Distance of green car.
Vor = 23km/h = 23,000m/3600s = 6.39 m/s.
Vr = 46km/h = 46,000m/3600s = 12.78 m/s.
Vor*T1 = 44.2
6.39T1 = 44.2
T1 = 6.92 s. = Time to pass.
Vr*T2 = 76.3
12.78*T2 = 76.3
T2 = 5.97 s. = Time to pass.
a. Vog*T1 = 167.8 m.
Vog*6.92 = 167.8
Vog = 24.2 m/s.
76.3 + Dg = 212
Dg = 212-76.3 = 135.7 m.
Vg*T2 = 135.7
Vg*5.97 = 135.7
Vg = 22.7 m/s
b. a = (Vg-Vog)/(T2-T1) =
Solve for a.
a red car and a green car move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 212 m. If the red car has a constant velocity of 23.0 km/h, the cars pass each other at x = 44.2 m. On the other hand, if the red car has a constant velocity of 46.0 km/h, they pass each other at x = 76.3 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs.
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