Posted by **Megan** on Tuesday, March 18, 2008 at 7:23pm.

Can someone help me? Do I subtract 17 on both sides and then............

The position of an object moving in a straight line is given by s=2t^2-3t, where s is in meters and t is the time in seconds the object has been in motion. How long (nearest tenth) will it take the object to move 17 meters?

- Algebra -
**Damon**, Tuesday, March 18, 2008 at 8:12pm
17 = 2 t^2 -3 t (started out moving backwards but accelerates forwards - tricky :)

- Algebra -
**Megan**, Tuesday, March 18, 2008 at 8:20pm
Thanks, I think I have it.

- Algebra -
**Damon**, Tuesday, March 18, 2008 at 8:32pm
so lets look at this parabola

2 t^2 - 3 t = s

t^2 - (3/2) t = s/2

t^2 - (3/2) t + 9/16 = s/2 + 9/16

(t-3/4)^2 = (1/2) (s+9/8)

opens up (t on x axis, s up) sketch it

vertex at t = .75 and s = -1.125

lets see when s = 0

2 t^2 - 3 t = 0

t = 0 and t = 1.5

SO:

after .75, s moved down 1.125 from zero

after 1.5, s is back at zero, total move so far = 2.25

17 - 2.25 = 14.75

so when s = 14.75, we have moved 17 meters total

14.75 = 2 t^2 - 3 t

- Algebra -
**Damon**, Tuesday, March 18, 2008 at 8:39pm
2 t^2 -3 t - 14.75 = 0

t = 3/4 +/-(1/4)sqrt (9 + 118)

= .75 +/- 2.82

= 3.57

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