A Mathematical Model of Motion Chapter 5

Position Time Graph Time t(s)Position x(m)

Describing Motion d(m) t(s) A B C D

Physics 1-8 Practice Problems:1-3 Page:85 Due: 9/25/01

Uniform Motion Uniform Motion means that equal changes occur during successive time intervals.

Slope d(m) t(s) rise Δy run Δx slope = rise run slope = Δy Δx

Slope of Distance vs Time Graph Velocity slope = Δy Δx v = Δd Δt v = d 1 – d 0 t 1 – t 0

assume: t 0 = 0s v = d 1 – d 0 t 1 – t 0 v = d 1 – d 0 t1 d 1 = d 0 + v t 1

d(m) t(s) v = d 1 – d 0 t 1 – t 0 v = 50m – 20m 5s – 2s v = 10m/s d 0 = 20m t 1 = 10s d 1 = d 0 + v t 1 d 1 = 20m + ( 10m/s)(10s) d 1 = 120m

Physics 1-8 Practice Problems:1-12 Pages:85, 87, 89 Section Review Page: 89 Due: 9/24/02

Problem 12 West East d 0 = 200 v = -15m/s d 0 = -400 d =d 0 + vt v = 12m/s d = td = t

d truck = d car t = t 27t =600 t =22s d = 200m + (-15m/s)(22s) d = 130m

Instantaneous Velocity

Velocity vs Time Curve Constant Faster Slower v(m/s) t(s)

v(m/s) t(s) v = Δd Δt Δd = vΔt Area underneath the v vs. t curve is Distance. A = l x w d = v x t { v vs t

Acceleration Acceleration is the rate of change of velocity. a = Δv = v 1 –v 0 Δt t 1 – t 0 Acceleration is the slope of the velocity vs. time curve.

Δv=5m/s Δt=1.5s Δv=1m/s Δt=8s

Find Acceleration from the Graph!! a = Δv Δt At: t = 1s At: t = 10s a = 1m/s 8s a = 3.3m/s² a = Δv Δt a = 5m/s 1.5s a = 0.13m/s²

Physics 1-8 Practice Probs:13-26 Pages:93,97,98 Section Review Page: 93 Due: 9/26/02

v t v0v0 d = v 0 t d =1/2(v- v 0 )t Finding d from V vs t curve d =1/2(v- v 0 )t + v 0 t

d =1/2(v+v 0 )t d = d 0 +1/2(v+v 0 )t Add Initial Displacement - d 0 d =(1/2v)-(1/2v 0 )t + v 0 t d =(1/2v)+(1/2v 0 )t

d = d 0 +1/2(v + v 0 )t v = v 0 + at d = d 0 +1/2(v 0 + at + v 0 )t d = d 0 +v 0 t + ½at 2 d = d 0 +1/2v 0 t + 1/2v 0 t + 1/2at 2

d = d 0 +1/2(v+v 0 )t Combine: v = v 0 + at t = (v-v 0 ) /a d = d 0 +1/2(v+v 0 ) (v-v 0 ) /a v 2 = v a(d-d 0 ) d = d 0 + (v 2 +v 0 2 ) 2a

d = d 0 +1/2(v+v 0 )t v 2 = v a(d-d 0 ) v = v 0 + at d = d 0 +v 0 t + ½at 2 *Basic Equations*

A motorcycle traveling at 16 m/s accelerates at a constant rate of 4.0 m/s 2 over 50 m. What is its final velocity? v 2 = v a(d-d 0 ) V 0 = 16m/s a = 4m/s 2 d = 50m v = ? Given:

v 2 = (16m/s) 2 +2(4m/s 2 )(50m) v 2 = v a(d-d 0 ) v = 25.6m/s 0 v = √ 656m 2 /s 2

Physics 3-3 Page:112 Problems: 52,54,57 Due: 10/3/06

Lab Results

Physics 1-10 Practice Probs:27-30 Pages:103 Section Review Page: 103 Due: 9/27/02

Falling Acceleration due to Gravity 9.8m/s² 32ft/s² a=g

t=0s,d=0m,v=0m/s t=1s,d=4.9m,v=9.80m/s t=2s,d=19.6m,v=19.6m/s t=3s,d=44.1m,v=29.4m/s

The Scream Ride at Six Flags falls freely for 31m(62m-205ft). How long does it drop and how fast is it going at the bottom? Known: a = -g = 9.8m/s² d 0 = 0m v 0 = 0m/s d = 55m Find: t = ? v = ? Equation: d = d 0 + v 0 t + ½at² d = ½at² t = √ 2d/a

t = √ 2(55m)/9.8m/s² t = 2.51s Equation: v = v 0 + at v = at v = (2.51s)(9.8m/s²) v = 24.6m/s = 55mph

Physics 3-4 Pages:112 Problems:66,67,70 Due: 10/10/06

Going straight Up and Down Slows down going up. Speeds up going down. Stops at the top. Acceleration is constant.

A ball is thrown up at a speed of 20m/s. How high does it go? How long does it take to go up and down? Use up as positive. Known: v 0 = 20m/s a = g = -9.8m/s² d 0 = 0m v = 0m/s Find: d = ? t =

Eq: v 2 = v a(d-d 0 ) 0 = v a(d) v 0 2 = -2a(d) = d v a

= d (20m/s) 2 -2(-9.8m/s 2 ) d = 20.4m

v = v 0 + at 0 = v 0 + at v 0 = -at v 0 -a = t 20m/s -(-9.8m/s 2 ) = t 2.04s = t The trip up! 4.08s = t

= d (20m/s) 2 -2(-9.8m/s 2 ) d = 20.4m

Physics 1-12 Ques: 3-5 Pages:107-8 Ques: Page:108 Due: 10/2/02

Physics 1-13 Ques: 6-11 Pages:10 Ques:39-43 Page:111 Due: 10/3/02 Labs Report:10/3/02

Physics 1-14 Ques:44-65 Page: Due: 10/7/02 Test: 10/8/02