SOLUTION: The speed of a car was tracked by radar in a police helicopter over a period of 10 minutes. The police found that the car's speed could be represented by the equation: s = 3t squar

Click here to see ALL problems on Travel Word Problems

Question 166462: The speed of a car was tracked by radar in a police helicopter over a period of 10 minutes. The police found that the car's speed could be represented by the equation: s = 3t squared - 30t + 135
where s is the speed of the car in kilometers per hour after t minutes have passed.
a. When was the car traveling at 80km/h?
b. When was the car traveling at 120km/h?
Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website!
equation states that s = 3t^2 - 30t + 135
s is the speed in km/h after t minutes have passed.
the constant of 135 is presumed to be in km/h.
-----
for the first equation,
80 = 3t^2 -30t + 135
subtract 80 from both sides and equation becomes:
0 = 3t^2 - 30t + 55
use the quadratic equation to solve:
t = 7.58198... minutes
or
t = 2.41801... minutes
-----
either value of t in the equation satisfies it.
car was traveling at 80km/h in either 2.418... minutes, or 7.58... minutes
-----
for the second equation,
120 = 3t^2 - 30t + 135
subtract 120 from both sides and equation becomes:
0 = 3t^2 - 30t + 15
use the quadratic equation to solve:
t = 9.47213... minutes
or
t = 0.52786... minutes
-----
either value of t in the second equation satisfied it.
-----
the answer rounded to the nearest hundredth of a minute appears to be:
car was traveling at 80kmph after 2.42 minutes were passed and after 7.58 minutes were passed.
car was traveling at 120kmph after .53 minutes were passed and after 9.47 minutes were passed.
-----
since both values of t are within the 10 minute window for the 80kmph equation and the 120kmph equation, then both values of t appear to be valid for each of those equations.
-----
a graph of this equation looks like the following: