Stephanie K. answered • 03/31/20

Calculus Instructor - AP, Dual Credit, College

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) t^{2} + 80t + C or s(t) = -16t^{2 }+ 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) = -16t^{2} +80t + 10

You just have to memorize the relationship between position, velocity and acceleration.

position (derivative) -> velocity (derivative) -> acceleration

acceleration (integral) -> velocity (integral) -> position