are you sure that is 2y^3 and not 2 t^3 ?
2 y^3 is all right, but unusual and probably not intended.
The equation of motion of a particle is s= 2y^3 - 7t^2 + 4t +1 where s is in meters and t is in seconds
a) Find the velocity and acceleration as functions of t.
b) Find the acceleration after 1 second.
3 answers
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
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
Thanks