Question

System of equations (?)
Find the position equation s=(1/2)at^2+(vo)t+(so) for the object. vo and so are both variables. Solving as a system, a=-32, but it is equal to -16 in the book. The answer should end up being s=-16t^2+144.

t=1, s=128ft
t=2, s=80 ft
t=3, s=0ft

Everything and everywhere I check, a=-32, but the book is never wrong, and I refuse to believe that it is.

Answer 1

128 = 0.5A + 1Vo + So
80 = 2A + 2Vo + So
0 = 4.5A + 3Vo + So

80 - 128 = (2A + 2Vo + So) - (0.5A + 1Vo + So)
-48 = 1.5A + Vo
0 - 80 = (4.5A + 3Vo + So) - (2A + 2Vo + So)
-80 = 2.5A + Vo
(-80) - (-48) = (2.5A + Vo) - (1.5A + Vo)
-32 = A

:(

Answer 2

Be careful. You have the right answer, you're just not taking the last step.

Remember the equation is:\[s=\frac{1}{2}at^2+v_0t+s_0\]When you substitute in a, you get...\[s=\frac{1}{2}(-32)t^2+\ ...\]

Answer 3

I know how to solve for the other variables (one is 1) but the answer shouldn't be -32
for a

Answer 4

Are you sure? You said the final form of the equation was:\[s=-16t^2+144\]Which implies that the acceleration, a, SHOULD be -32 (it gets multiplied by the 0.5).

Answer 5

oh

Answer 6

No. Just think of the general form of the equation for parabolic motion as:\[s=\frac{1}{2}at^2+v_0t+s_0\]All you did was solve for the three variables, and then you can plug them back into this equation to get the full equation of motion.

Answer 7

I spend an hour on this, and overlooked such a thing

Answer 8

I originally was using substitution, so I have about 8 equations on my paper XD

Answer 9

Heh. It's OK. I did the same thing a lot when I was first working with these equations of motion. I remember getting frustrated over the exact same thing (missing the 0.5) :P.