One car is going 88km/h and the other is going 115 km/h, suppose they are both going in the same direction with the same starting position. How far must the faster car travel to arrive 16 mins before the slower car? Answer in units of km.
2007-09-09 16:00:53 · 1 answers · asked by Skyler H 1 in Science & Mathematics ➔ Physics
Each car travels the same distance "d" and the 115km/h car gets there 16 mins (16/60 hours) before the other car. If "t" is the time it takes the slower car then the time "T" it takes the faster car is T = t - 16/60. So:
d = 88t or t = d/88
d = 115T and T = t - 16/60
d = 115(t - 16/60)
d = 115[(d/88) - 16/60]
d = 1.3068d - 30.6667
0.3068d = 30.6667
d = 99.951 km
check:
t = d/88 = 1.136 hours
T = t - 16/60 =
115T = 99.951 km
So the faster car travels the same distance as the slower car and does it in a time 16 minutes less than the time of the slower car.
2007-09-09 17:17:02 · answer #1 · answered by Captain Mephisto 7 · 0⤊ 0⤋