Posted by Jilian on Sunday, February 24, 2013 at 5:15pm.
are you sure that is 2y^3 and not 2 t^3 ?
2 y^3 is all right, but unusual and probably not intended.
If as typed, the 2y^3 has nothing to do with the velocity and acceleration
v =ds/dt = -14 t + 4
a = dv/dt = -14
If as I kind of have a hunch you mean
v = 6 t^2 - 14 t + 4
a = 12 t - 14
it was 2 t^3
Thanks
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