# Oke Descriptive Essay

879 Words4 Pages
Session 3: The Derivatives 1 2 f(x) = x2 – 1 3 • If a ball is thrown vertically upward with an initial velocity of 128 ft/sec, the ball's height after t seconds is s(t) = 128t - 16t2 . – Sketch graphs of position, velocity as well as the acceleration function of the problem discussed. 4 Example • If a ball is thrown vertically upward with an initial velocity of 128 ft/sec, the ball's height after t seconds is s(t) = 128t 16t2 a) What is the velocity function? Solution: v(t) = 128 - 32t b) What is the velocity when t = 2, 4, 6 Solution: v(2) = 128 - 32(2) = 64 v(4) = 128 - 32(4) = 32 v(6) = 128 - 32(6) = -64 5 c) When is the velocity zero? Solution: Set the velocity function to zero and solve 0 = 128 - 32t -128 = -32t 4=t d) What is the height of the ball at the time the velocity is 0? Solution: let t = 4 in the position function s(4) = 128(4) - 16(16) = 512 - 256 = 256' 6 e) When does the ball hit the ground? Solution: Set the position function to zero and solve! 128t - 16t2 = 0 16t(8 - t) = 0 At t = 0, or t = 8. Think. At t = 0, you haven't thrown the ball yet! The ball will hit the ground in 8 seconds. You could have reasoned that it took 4 seconds to reach maximum height and another 4 secs to come back down. f) What is the velocity when the ball hits the ground? Solution: Find v(8) v(8) = 128 - 32(8) = -128 ft/sec. It is negative because the ball is coming back down! Notice it has the same velocity as when it was thrown upward! 7 g) What is the acceleration function? Solution: Find a(t) by finding the derivative of v(t)! a(t) = -32 ft/sec2 h) What is the acceleration at t = 3, 5, 8? Solution: a(3) = -32 a(5) = -32 a(8) = -32 No matter what time it will always be -32. 8 Your Task  Sketch graphs of position, velocity as well as the acceleration function of the problem discussed. 9