Asked by Anna
A car stops for a red ligth. The light turns green and the car moves for 3 seconds at a steadily increasing speed. During this time, it travels 20 meters. The car then travels at a constant speed for another 3 seconds for a distance of 30 meters. Finally, when approaching another red light, the car steadily slows to a stop during the next 3 seconds in 15 meters. Create equations with respect to time for the position and velocity of the car at each segment.
Answers
Answered by
Henry
d1 = 0.5a*t^2 = 20 m.
0.5a*3^2 = 20,
4.5a = 20,
a = 4.44 m/s^2.
V = at = 4.44 m/s^2 * 3 s=13.32 m/s.
V = d2/t = 30/3 = 10 m/s.
a = (Vf-Vo)/t = (0-10) / 3=-3.33 m/s^2
d3 = Vo*t + 0.5a*t^2 = 15 m.
10*3 + 0.5a*3^2 = 15,
30 + 4.5a = 15
4.5a = 15-30 = -15,
a = --3.33 m/s^2.
0.5a*3^2 = 20,
4.5a = 20,
a = 4.44 m/s^2.
V = at = 4.44 m/s^2 * 3 s=13.32 m/s.
V = d2/t = 30/3 = 10 m/s.
a = (Vf-Vo)/t = (0-10) / 3=-3.33 m/s^2
d3 = Vo*t + 0.5a*t^2 = 15 m.
10*3 + 0.5a*3^2 = 15,
30 + 4.5a = 15
4.5a = 15-30 = -15,
a = --3.33 m/s^2.
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