Velocity & Other Rates of Change

3.4 Velocity & Other Rates of Change **Concepts** Vertical Motion s(t) = position v(t) = velocity a(t) = acceleration jerk where s'(t) = v(t) where v'(t) = a(t) a'(t) Speed = |velocity| Speed = never (-) **Example: Rock Blast** A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec (= 109 mph). It reaches a height of s = 160t - 16t^2 after t seconds. **a) How high does the rock go?** s'(t) = 160 - 32t 0 = 160 - 32t 32t = 160 t = 5 sec s(5) = 160(5) - 16(5)^2 = 800 - 400 = 400 ft **b) What is the velocity and speed of the rock when it is 256 ft above the ground on the way up? ...on the way down?** 256 = 160t - 16t^2 0 = 16t^2 - 160t + 256 0 = 16(t - 2)(t - 8) t = 2, t = 8 v(2) = 160 - 32(2) = 96 ft/sec Speed = 96 v(8) = 160 - 32(8) = -96 ft/sec Speed = 96 **c) What is the acceleration of the rock at any time t during its flight (after the blast)?** a(t) = v'(t) = -32 ft/sec^2 gravity pulling down **d) When does the rock hit the ground?** 0 = 160t - 16t^2 0 = 16t(10 - t) t = 0 and t = 10 sec 3.4 Particle Motion Suppose we are given the following function for the position of a particle on a coordinate axis. s(t) = (1/3)t³   - (9/2)t² + 14t + 2 0 ≤ t ≤ 9 so... v(t) = s'(t) = t² - 9t + 14 a(t) = v'(t) = 2t - 9 1) Find the average velocity on the interval [0, 9] Δ displacement / Δ time = (s(9) - s(0)) / 9 = (76.5 - 2) / 9 = 4.5 / 9 = 0.5 m/sec 2) What is the velocity of the particle at t = 3 seconds? (instantaneous velocity) v(t) = s'(t) = t² - 9t + 14 | t=3 = 3² - 9(3) + 14 = 9 - 27 + 14 = -4 m/sec 3) What is the speed of the particle at t = 3 seconds? speed = |velocity| = |-4| = 4 m/sec 4) Describe the motion of the particle (moving right, left, stopped) * velocity is positive? moving right * velocity is negative? moving left * velocity is zero? stopped (but doesn't change direction) [0, 2) U (7, 9] (2, 7) t = 2 t = 7 5) Find the acceleration of the particle at t = 3 seconds. a(t) = v'(t) = 2t - 9 = 2(3) - 9 = -3 m/sec^2 6) Find the velocity of the particle when acceleration is zero. 0 = 2t - 9 9 = 2t 4.5 = t v(4.5) = (4.5)² - 9(4.5) + 14 = 20.25 - 40.5 + 14 = -6.25 m/sec 3.4 Particle Motion - YOU A particle moves along a line so that is position at any time t ≥ 0 is given by the function s(t) = t² - 4t + 3, where s is measured in meters and t is measured in seconds. a) Find the displacement of the particle during the first 2 seconds. how far did object move? s(2) - s(0) [2² - 4(2) + 3] - [3] 4 - 8 + 3 - 3 = -4 meters b) Find the average velocity of the particle during the first 4 seconds. s(4) - s(0) 4 - 0 [16 - 16 + 3] - [3] 4 - 0 3 - 3 4 = 0 m/sec using V(4) - V(0) 4 - 0 is actually average acceleration c) Find the instantaneous velocity of the particle when t = 4. s'(t) = v(t) = 2t - 4 |t=4 = 8 - 4 = 4 m/sec d) What is the acceleration of the particle at t = 4? v'(t) = a(t) = 2 m/sec² it actually doesn't matter what time it is. e) Describe the motion of the particle. (at what values of t does the particle change directions?) if you look at v(t) = 0... that is when things are interesting 0 = 2t - 4 4 = 2t 2 = t Test these areas t = 1 2(1) - 4 2 - 4 -2 t = 3 2(3) - 4 6 - 4 2 V(3) = 2 indicates (2, ∞) indicates (-∞, 2)