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
Well, you got it
4t^2 - 20t - 150 = 0
by the formula
t = (20 ± √2800)/8
=9.11 or a non-admissible negative value
To solve this problem algebraically, we need to set up a system of equations.
Let's assume the time at which the cars are side by side as "t" seconds. At this time, the distance traveled by the first car should be equal to the distance traveled by the second car.
For the first car:
C(t) = 4t^2
For the second car:
C(t) = 20t + 150
Setting these two equations equal to each other, we get:
4t^2 = 20t + 150
Now, to solve this quadratic equation, we want to bring all terms to one side and set it equal to zero:
4t^2 - 20t - 150 = 0
Next, we can divide the entire equation by 2 to simplify:
2t^2 - 10t - 75 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula. The quadratic formula states:
t = (-b ± √(b^2 - 4ac)) / (2a)
Using the coefficients from the quadratic equation above, the formula becomes:
t = (10 ± √(10^2 - 4(2)(-75))) / (2(2))
Simplifying further:
t = (10 ± √(100 + 600)) / 4
t = (10 ± √700) / 4
Now, we can simplify the value under the square root:
t = (10 ± √(100 * 7)) / 4
t = (10 ± 10√7) / 4
We can further simplify by dividing both numerator and denominator by 2:
t = (5 ± 5√7) / 2
So, there are two possible solutions for "t":
t = (5 + 5√7) / 2
t = (5 - 5√7) / 2
These are the two solutions for the time at which the cars will be side by side.