An automobile moving at a constant velocity of 14 m/s passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at a constant rate of 1.8 m/s². How soon will the second automobile overtake the first?
2 answers
14 * t = (1.8 / 2) t^2
d1 = r*T = 14*2 = 28 m.
d2 = 0.5*a*T^2.
0.5*a*T^2 = 28 +14T
0.5*1.8T^2 = 28+14T.
0.9T^2-14T-28 = 0
T = ( -B+-sqrt(B^2-4AC))2A
T = (14+-sqrt(196+100.8))/1.8 = (14+-17.2)/1.8 = 17.3 s, and -1.78 s.
T = 17.3 s.
d2 = 0.5*a*T^2.
0.5*a*T^2 = 28 +14T
0.5*1.8T^2 = 28+14T.
0.9T^2-14T-28 = 0
T = ( -B+-sqrt(B^2-4AC))2A
T = (14+-sqrt(196+100.8))/1.8 = (14+-17.2)/1.8 = 17.3 s, and -1.78 s.
T = 17.3 s.
3 answers