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?
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 - 14.75 = 0
t = 3/4 +/-(1/4)sqrt (9 + 118)
= .75 +/- 2.82
= 3.57
Sure, I can help you with that! To find out how long it will take for the object to move 17 meters, we need to set up an equation using the given information.
The given equation is s=2t^2-3t, where s is the position of the object and t is the time in seconds.
To find the time it takes for the object to move 17 meters, we can set s equal to 17 and solve for t.
2t^2-3t = 17
To solve this quadratic equation, we can start by subtracting 17 from both sides:
2t^2 - 3t - 17 = 0
Now, we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 2, b = -3, and c = -17. To solve the equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
t = (-(-3) ± √((-3)^2 - 4(2)(-17))) / (2(2))
Simplifying further:
t = (3 ± √(9 + 136)) / (4)
t = (3 ± √(145)) / 4
Now, we have two possible solutions, t1 and t2:
t1 = (3 + √(145)) / 4
t2 = (3 - √(145)) / 4
To find the positive solution, since time cannot be negative, we can calculate t1:
t1 = (3 + √145) / 4
Evaluating t1 to the nearest tenth, we get:
t1 ≈ 2.8 seconds
So, it will take the object approximately 2.8 seconds to move 17 meters.