An object moving vertically is at the given heights at the specified times. Find the position equation s = 12 at^2 + v (small o) t + s (small o) for the object.

Question

At t = 1 second, s = 161 feet At t = 2 seconds, s = 98 feet At t = 3 seconds, s = 3 feet
a. s=−8t^2−t−192
b. s=−32t^2 −15t+161 c. s=−16t2 +15t+161 d. s=−16t^2 −15t−192 e. s=−16t2 −15t+192

Answer 1

You mean
s = (1/2) at^2 + v (small o) t + s (small o)
so
161 = .5 a + Vo + So
98 = .5 a(4) + 2 Vo + So
3 = .5 a(9) + 3 Vo + So
or

1 a + 2 Vo + 2 So = 322
2 a + 2 Vo + 1 So = 98
4.5a+ 3 Vo + 1 So = 3

solve three linear equations for a, Vo So

Answer 2

I wonder if a comes out -32 ft/second^2 :)

Answer 3

which is the acceleration of gravity on earth in ft/s^2
and indeed it works ;)

Answer 4

but I cheated. [ used a = dV/dt = d/dt (dx/dt) ]

Answer 5

so is the answer s=−32t^2 −15t+161?