HELP!
an object moves in a straight line given by s=2x^2-3t where s is in meters and t is the time in seconds the object is in motion. how long to nearest tenth will it take to move 17 meters?
my attempt
s=2t^2-3t
17=2t^2-3t
-17=-17
0=2t^2-3x-17
t=-b +or- sqrtb^2-4ac/2a
t=-(-3)+or-sqrt-3^2-4(2)(-17)/2(2)
Did I do all of the correct steps? I think I did but this is really confusing
I have worked it more.....
t=3+or-sqrt9+136/4
t=3+or-sqrt145/4 (is this my answer?)
yes, you are right, but when typing on here you should use brackets to establish the correct order of operation
t = (3 ± √145)/4
I got the ± by holding down the Alt key and at the same time punching in 241 on the number key pad, then releasing the Alt key
I got the √ by holding down the Alt key and at the same time punching in 251 on the number key pad, then releasing the Alt key
I am using a PC with Firefox on XP
It did not work on my Apple IBook laptop
Your attempt is on the right track, but there are a few mistakes in your calculations. Let's go through the steps again to find the correct answer.
1. Start with the given equation: s = 2x^2 - 3t.
2. Substitute the value of s with 17 meters: 17 = 2x^2 - 3t.
3. Rearrange the equation to be in quadratic form: 2x^2 - 3t - 17 = 0.
4. Now we can use the quadratic formula to find the values of t. The quadratic formula is t = (-b ± √(b^2 - 4ac)) / 2a.
a = 2, b = -3, c = -17.
5. Plug in the values into the formula: t = (-(-3) ± √((-3)^2 - 4(2)(-17))) / (2(2)).
6. Simplify the equation: t = (3 ± √(9 + 136)) / 4.
7. Further simplification gives us: t = (3 ± √145) / 4.
8. Now we have two possible values for t. To find the time it will take to move 17 meters to the nearest tenth, we need to choose the appropriate value based on the context of the problem.
t ≈ (3 + √145) / 4 ≈ 2.88 seconds (rounded to the nearest tenth).
Therefore, it will take approximately 2.88 seconds to move 17 meters along the given straight line.