Asked by john daniel
an automobile moving at constant velocity of 15m/sec passes a gasoline station. two seconds later another automobile leaves the gasoline station and accelerates at a constant rate of 2m/ sec squared. how soon will the second automobile overtake the first ?
Answers
Answered by
drwls
They will pass when the distances traveled are equal.
15 t = (a/2)(t-2)^2 = (t-2)^2
Solve for t, which is measured from the time the 15 m/s car goes by the station.
t^2 -4t +4 = 15 t
t^2 -19t +4 = 0
t = (1/2)[19 +sqrt345) = 18.79 s
Both cars will have travelled 282 m at that time
15 t = (a/2)(t-2)^2 = (t-2)^2
Solve for t, which is measured from the time the 15 m/s car goes by the station.
t^2 -4t +4 = 15 t
t^2 -19t +4 = 0
t = (1/2)[19 +sqrt345) = 18.79 s
Both cars will have travelled 282 m at that time
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.