MOTION ALONG A STRAIGHT LINE I. DISPLACEMENT AND AVERAGE VELOCITY: A. DISPLACEMENT: Distance travelled, in this case along a straight line, x-axis. 1) x1- Initial point x2- Terminal point Δx = x2 – x1 --- displacement 2) If time is factored in, then if t1 – time at point x1 and t2 – time at point x2 Δt = change in time during displacement = t2 – t1 B. AVERAGE VELOCITY: The displacement, change in x – component, divided by the time interval. 1) Average velocity = Vav – x = Δx / Δt = (x2 – x1) / (t2 – t1) 2) Keep in mind the direction matters. This is a vector, so the opposing distances must be subtracted to get the net displacement. C. AVERAGE SPEED: Direction is taken out of the equation. So we take sum of all distances instead of net displacement. D. EXAMPLE 1: You decide to go camping up north. You drive 65 miles at 55 mph. you realize that you travelled 22 miles too far so you travel due south at 75 mph. 1) What was the total time travelled? Speed = Distance/Time North: 55 mph = 65 miles/TN TN = 65 miles/ 55 mph = 1.18 hours. South: 75 mph = 22 miles/TS TS = 22 miles/ 75 mph = 0.29 hours. Total time: TN + TS = 1.18 + 0.29 = 1.47 hours. 2) What is the average velocity? Because the trips are in the opposite directions, the net displacement must be found. Net displacement = DistN – DistS = 65 – 22 = 43 miles. Total time of the journey was found in the previous part to be as 1.47 hours. Vav-x = 43 miles/1.47 hours = 29.25 mph. 3) What is the average speed? Average speed = Total displacement (independent of direction) Total time Total displacement = DistN + DistS = 65 + 22 = 87 Average speed = 87 miles / 1.47 hours = 59.2 mph. E. INSTANTANEOUS VELOCITY: When you want the direction and how fast at any given time during an interval. 1) Formal definition: The limit of the average velocity as the time interval approaches zero; the instantaneous rate of change of position with time. (the derivative with respect to time) Vx = lim Δx/Δt = dx/dt Δt -> 0 = lim Δx/Δt t -> t0 F. EXAMPLE 2: A car 50m ahead is accelerating at a rate of 10 m/s2. The position of the vehicle is described by the equation: 2 2 Xt = 50 + 10 m/s T 1) Find displacement for to = 2s, tt = 6s? 2 2 X1 = 50m + 10m/s (2s) = 50 + 40 = 90m 2 2 X2 = 50m + 10m/s (6s) = 50 + 360 = 410m ΔX = 410 – 90 = 320m. 2) Find the average velocity during this time? ΔT = tt – to = 6s – 2s = 4s. Vav-x = ΔX/ΔT = 320m/4s = 80 m/s. 3) Find the instantaneous velocity at t = 5s? Find the lim using Δt = 0.1, 0.01, 0.001, 0.0001. t -> 5 Plug in the t + Δt into Xt and see what number Xt approaches. t + Δt 5.1 5.01 5.001 5.0001 Xt 310.1 301.001 300.10001 300.0100001 So as t 5s, Xt approaches 300m. Thus, the instantaneous velocity = 300m/5s = 60m/s. G. Find the following using the graph below. Tangent line Secant line 400 __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 300 __ Δx = 400 – 100 = 300m 200 __ 100 __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Δt = 4 – 2 = 2s 1 2 3 4 5 6 7 1) Average velocity: Vax – slope of secant line = (400 -100)/(4 - 2) = 150 m/s 2) Instantaneous velocity at t = 4s: V – slope of tangent line = (400 - 0)/(4-3) = 400m/s H. AVERAGE ANG INSTANTANEOUS ACCELERATION: 1) Average acceleration: The change in the velocity in the x- component, ΔVx divided by the time interval, ΔT. Aav-x = ΔVx/ ΔT = (V2x – V1x)/(t2 – t1) 2) Concepts of velocity apply to acceleration. Practice both types. 3) Instantaneous acceleration: Ax = lim (ΔVx/ ΔT) = dvx/dt Δt -> 0 4) Graphically: Avg. acceleration – secant line slope Instantaneous acceleration – tangent line slope *infer x-axis is still time but y-axis is now velocity II. MOTION WITH CONSTANT ACCELERATION: A. BECOMES A LINEAR RELATIONSHIP BETWEEN VELOCITY AND TIME vx ------------------------------------------------ Δv vo ------------------------------------------------ Δt 0 t1 t2 1) If t1 = 0, then a = Δv/ Δt can change to Vx = Vo + a*t 2) Vav = (Vo + Vx)/2 and Vx = Vo + a*t So, Vav = (Vo + Vo + a*t)/2 = Vo + 0.5*a*t Vav = (X2 – X1)/t = Vo + 0.5*a*t [*remember t2 = t, to = 0] 2 Xm – Xo = Vot + 0.5*a*t 3) If you can solve ax =(Vx – Vox)/t for t, You obtain t = (Vx – Vox)/ax and plug it into 2 Xm – Xo = Vot + 0.5*a*t Manipulate to get 2 2 V x = V ox + 2ax(Xm – Xo) Another important equation: X2 – X1 = [(Vox + Vx)/2]t *Know these equations well and when to apply! Plenty of practice problems in the text, main thing is what are given and what do you want. The pieces fit like a puzzle. III. FREE FALLING BODIES: A. FREE FALL: 1) All objects fall at the same rate of acceleration – acceleration due to gravity, g = 9.8 m/s2 or 32 ft/s2 2) So all equations in previous sections apply for free falling bodies except every time: Vy = Vo + ay*t 2 y = yo + Voy*t + 0.5*ay*t y – yo = [(Voy + Vy)/2]t 2 2 V y = V oy + 2ay(y – yo) *Remember g is a vector. B. EXAMPLE: A rocket is launched straight up at a speed of 8 m/s. When the rocket runs out of fuel and falls freely. 1) Find the position equation: 2 2 ym = yo + Voy*t + 0.5*g*t , yo = 0, Voy = 0, g = -9.8m/s 2 ym = - 4.9t 2) Find an equation for Vy? Vy = Voy + ay*t 2 Vy = 8m/s – 9.8m/s * t 3) Find y and Vy at t = 2s? Plug in t = 2s in both equations. 2 ym = -4.9*2 = -19.6m 2 Vy = 8 – 9.8*2 = -31.2m/s (Note: -ve sign implies motion opposite to acceleration due to gravity) 4) Find the time taken to reach max height and the max height? . From a perspective of time: We know that at max height Vy = 0. Vy = Voy + ay*t 0 = 8m/s – 9. t = -8/-9.8 t = 0.82s 2 Now plug into ym = - 4.9t 2 ym = -4.9*(0.82) = 3.28m . From point of view of velocity: Vy = 0m/s, yo = 0m 2 2 V y = V oy + 2ay(y – yo) 0 = (8m/s)2 + 2(-9.8m/s2)(y-0) -64m2/s2 = -19.6m/s2 * y Thus y = -64/-9.8 = 3.28m IV. VELOCITY AND ACCELERATION AS INTEGRALS A. EQUATIONS t 1) Vx = Vox + ∫axdt t 0 2) x = xo + ∫vxdt 0 Note: Given an ax function, integrate a couple of times and gt the position function. B. EXAMPLE 2 3 ax = 3.0 m/s – (0.2m/s )t, xo = 60m, Vox = 2m/s 1) Find the velocity equation as a function of time. t Vx = Vox + ∫axdt 0 2 3 = 2m/s + ∫(3m/s – 0.2m/s * t)dt 2 2 3 Vx = 2m/s + 3t m/s – 0.1t m/s 2) Find the position equation. xm = xo + ∫vxdt 2 2 3 xm = 60m + ∫(2m/s + 3t m/s – 0.1t m/s )dt 2 2 3 3 xm = 60m + 2t m/s + 3(t /2)m/s – 0.1(t /3)m/s Now you can plug in any t value and get the velocity or position at that time. .