Question

A science project studying catapults sent a projectile into the air with an initial velocity of 30 m/s. The formula for distance (s) in meters with respect to time (t) in seconds is s(t) = -5t2 + 30t. Find an expression for the instantaneous rate of change (and thus a velocity function at any time) using s'(a)=limh→0s
((a+h)−s(a))/h where a is any value for t. Then, find the instantaneous rate of change at: a) t = 1 b) t = 3

Answer 1

s(t) = -5 t^2 + 30 t we want the time derivative which is velocity

s(t+h) = -5 (t+h)^2 + 30 (t+h)
= -5 (t^2 + 2 t h + h^2) + 30 t + 30 h
= -5 t^2 - 10 t h - 5 h^2 + 30 t + 30 h
now subtract s(t) = - 5 t^2 + 30 t
result = -10 t h -5 h^2 + 30 h
now divide by h
result = -10 t - 5 h + 30
now let h -----> 0
derivative = velocity = -10 t + 30
note that is if you shoot it straight up at 30 meters/second and g = -10 meters/second^2
now I think you can put in t = 1 and t = 3
note --- t = 3 seconds is the tippy top, when it stops and then falls