SOLUTION: The position of an object moving in a straight line is given by s=2tsquared - 3t, where s is in meters and t is the time in secoonds the object has been in motion. How long (to the

Question 122844This question is from textbook college algebra
: The position of an object moving in a straight line is given by s=2tsquared - 3t, where s is in meters and t is the time in secoonds the object has been in motion. How long (to the nearest tenth) will it take the object to move 19 meters? This question is from textbook college algebra

Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The position of an object moving in a straight line is given by s=2tsquared - 3t, where s is in meters and t is the time in secoonds the object has been in motion. How long (to the nearest tenth) will it take the object to move 19 meters?
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s(t) = 2t^2-3t
19 = 2t^2-3t
2t^2-3t-19=0
t = [3 +- sqrt(9-4*2*-19)]/4
t = [3 +- sqrt(161)]/4
t = [3 +- 12.6886]/4
Positive answer:
t = [3+12.6886]/4
t = 3.9221 seconds
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Cheers,
Stan H.

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
The problem can be solved by letting s (the distance in meters) = 19 and solving for t (the time in seconds):
Substitute s = 19:
Subtract 19 from both sides.
Solve this quadratic equation using the quadratic formula: and the a, b, and c, come from
So here, a = 2, b = -3, and c = -19.
Make the substitutions:

or
or
Only the positive answer is meaningful here, so...
It will take 3.9 seconds for the object to move 19 meters.
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