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