Asked by Beka
Use the given information to write a systems of equations, and then solve the system algebraically to answer the question:
A car begins at rest and accelerates. Its distance in meters, t, by the formula C(t) =4t^2. A second car, 150 meters ahead, is traveling at a constant speed of 20 meters per second. Its distance, C(t), in meters can be determined as a function of time, t, in seconds by the formula C(t)=20t+150. How long after the first car accelerates will the cars be side by side?
What I've done so far: 4t^2=20t+150
A car begins at rest and accelerates. Its distance in meters, t, by the formula C(t) =4t^2. A second car, 150 meters ahead, is traveling at a constant speed of 20 meters per second. Its distance, C(t), in meters can be determined as a function of time, t, in seconds by the formula C(t)=20t+150. How long after the first car accelerates will the cars be side by side?
What I've done so far: 4t^2=20t+150
Answers
Answered by
Reiny
Well, you got it
4t^2 - 20t - 150 = 0
by the formula
t = (20 ± √2800)/8
=9.11 or a non-admissible negative value
4t^2 - 20t - 150 = 0
by the formula
t = (20 ± √2800)/8
=9.11 or a non-admissible negative value
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