The velocity equation is the integral of the acceleration equation.
So v(t) = ∫-32 dt
This is an indefinite integral so when we integrate, we have to add "+C": v(t) = -32t + C
This allows us to use the information that when t = 0, v(0) = 80 ft/sec: v(0) = -32 (0) + C = 80 or 0 + C = 80 or C = 80.
So now we have our velocity equation: v(t) = -32t + 80 ft/sec.
The position equation is the integral of the velocity equation.
So s(t) = ∫(-32t + 80) dt
Again, this is an indefinite integral, so we integrate to get: s(t) = -32 (1/2) t2 + 80t + C or s(t) = -16t2 + 80t + C
Again, we use the information given in the problem that when t = 0, s(0) = 10 ft
So s(0) = 16 (0)2 + 80 (0) + C = 10 or C = 10.
So now we have our position equation: s(t) = -16t2 +80t + 10
You just have to memorize the relationship between position, velocity and acceleration.
position (derivative) -> velocity (derivative) -> acceleration
acceleration (integral) -> velocity (integral) -> position