Algebra
posted by Megan on .
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^23t, 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?

17 = 2 t^2 3 t (started out moving backwards but accelerates forwards  tricky :)

Thanks, I think I have it.

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
(t3/4)^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 
2 t^2 3 t  14.75 = 0
t = 3/4 +/(1/4)sqrt (9 + 118)
= .75 +/ 2.82
= 3.57