Jan 31 8:19 PM. Chapter 9: Uniform Rectilinear Motion

Transcription

1 Unit 3: Kinematics Uniform Rectilinear Motion (velocity is constant) Uniform Accelerated Rectilinear Motion The Motion of Projectiles Jan 31 8:19 PM Chapter 9: Uniform Rectilinear Motion Position: point of an object relative to something else. involves direction therefore it is a vector except that it is one dimensional. The direction is along the x axis Frame of reference. We talk about position as positive or negative. We often place the origin at the origional position and when there is movement we can then talk about displacement relative to the original position. Displacement: Change in position Also one dimension vector : Positive or negative Velocity: rate of change in position v = change in positon/change in time Feb 4 8:03 PM 1

2 Graphical Representation of position as a function of time. We always place time on the x axis (independent) position on the y axis (dependent) Positive displacement will give a positive slope Negative displacement will give a negative slope. Example position (km) John starts at position A. He moves forward 10km in 2 minutes He returns to his original position in 4 minutes. He then moves in the other direction and travels 20 km in 5 minutes. time (min) Graph his position as a function of time. Feb 4 8:17 PM position (m) Describe Sylvie's position as she travels along a straight road. time (min) Feb 4 8:36 PM 2

3 position (m) What can you say about how fast Sylvie is moving? time (min) Feb 4 8:36 PM Graphical representation of position as a function of time Uniform rectilinear motion is motion in a straight line at a constant velocity. We can break down a car trip into intervals and graph the relationship displacement (m) 1 5 time (s) The tram backs up 2.5 m then moves forward 4.5 m for a final displacement of 2m. How long did it take? 25 s. The displacement is initially negative and then positive. It has direction! How fast was it travelling? Part 1 2.5m/5s = 0.5 m/s Jan 9 3:29 PM 3

4 Velocity on a position time graph The change in y values divided by the change in x values Δy/Δx = slope In the case of distance over time, change in distance/change in time is m/s which is velocity Velocity can be positive or negative Velocity is negative that means that displacement is negative: remember it is relative and we use the x axis to describe displacement, (+) is to the right (-) is left. Example: a car drives forward, then turns around a returns to the initial position (or think of it as backing up). Dec 9 5:50 PM We can graph change in position versus time... Where does Thomas start? Where does he finish? Position in meters 5 A 1 B C D time in minutes When is the velocity (+)? When is the velocity (-)? What can we say about the slope? (what does it represent?) Dec 9 5:35 PM 4

5 We can graph change in position versus time... How fast is he travelling? Position in meters 5 A 1 B C D time in minutes Interval A: Interval B: Interval C: Interval D: Dec 9 5:35 PM Graphical representation of velocity as a function of time Represents motion The slope will by change in velocity over time: acceleration The displacement can be found by finding the area under the curve I Feb 4 8:50 PM 5

6 y 10 9 What is this graph telling us? Velocity m/s x time (s) What is happening in terms of displacement? Dec 5 8:36 AM Velocity m/s y What is the acceleration? x time (s) Dec 5 8:36 AM 6

7 graphs: slope of a line Pay attention to the variables What does the y axis represent? What does the x axis represent? If the function is linear then the slope = rise/run If the graph is displacement versus time: slope = Δd/Δt = d2 - d1/ t2 - t1 = velocity If the graph is velocity versus time: slope = Δv / Δt = v 2 - v1/ t2 - t1 = acceleration Jan 31 8:22 PM Velocity (m/s) A B C D E time(s) What is the initial velocity? Find the Velocity for each interval A B B C C D D E Feb 4 9:06 PM 7

8 Velocity (m/s) B C Find the displacement for each interval AB BC A D E time(s) CD DE Feb 4 9:06 PM constant acceleration A car is at rest A c A ca car is The car accelerates at 2 m/s 2 means every second the car will go 2m/s faster Time (s) Velocity (m/s) Jan 24 10:11 AM 8

9 Feb 4 9:52 AM If a car is slowing down the accelertation will be 3m/s 2 time (s) Velocity (m/s) Jan 24 10:19 AM 9

10 Feb 4 9:56 AM Feb 4 9:58 AM 10

11 Aristotle 4th century BC 2 kinds of motion violent (throw, push) natural (fire rised, stones fall down) Oresme 14 the century predicted that if the initial velocity was zero, distance was proportional to time squared Jan 24 10:29 AM Galileo Did experiments: marble on the inclined plane he noticed a marble on an inclined plane took more time than a marble dropped (free fall) "the gravitational force has been diluted" Jan 24 7:59 AM 11

12 Kinematics day notebook Jan 24 10:08 AM Feb 4 11:13 AM 12

13 Feb 4 1:57 PM Feb 4 11:21 AM 13

14 Feb 4 11:13 AM 450 years later Apollo 15 lands on the moon A feather falls at the same time as a hammer Jan 24 10:34 AM 14

15 Acceleration the change in velocity over time a = Δv/Δt = m/s 2 a truck's speed changes from 5 m/s to 50 m/s, in 60 seconds, what is its acceleration? 50m/s 5m/s / 60s = 45/60 m/s 2 =.75m/s 2 Jan 10 7:30 PM Uniformly Accelerated Rectilinear Motion velocity is no longer constant (more real) velocity varies moment to moment t i = initial time (s) t f = final time (s) x i = initial position (m) x f = final position (m) v i = initial velocity (m/s) v f = final velocity (m/s) a = acceleration (m/s 2 ) Jan 10 7:04 PM 15

16 UARM: constant acceleration (rate of change of velocity is linear) velocity vs time graphs Slope v Δv/Δt gives acceleration t Area Δv*Δt = displacement Jan 31 8:42 PM y d t = 2s t = 5s x UARM: position vs time Velocity is changing so the curve for position vs time is quadratic. We can still find instantaneous velocity take the tangent of the curve tangent: a straight line (linear) with the same slope as the curve at that moment in time. to t find the slope: take 2 points on the tangent line and find it using the slope formula a = Jan 31 8:39 PM 16

17 a = y2 - y1 x 2 - x 1 Jan 10 7:19 PM d t Jan 10 7:15 PM 17

18 Equations of Uniform Acceleration Equation 1: v 2 = v 1 + at Equation 1: a = Δv/Δt = v 2 - v 1/ Δt a * Δt = v 2 - v 1 a * Δt + v 1 = v 2 v 2 = v 1 + at Equation 2: x f = x i + ½(v i + v f )Δt Equation 3: x f =x i + v 1 t + ½at 2 Equation 4: v f 2 = v i 2 + 2aΔx Jan 31 8:17 PM How to solve kinematic problems: textbox p232 draw a diagram identify your point of reference ex: origin of x axis, at starting point, x i = 0 id known parameters (be sure to indicate signs (+ or -) Find one of the 4 equations where the quantity sought is the only unknown Jan 10 7:41 PM 18

19 Example: The driver of a car which is moving east at 25m/s applies the brakes and begins to decelerat at 2.0m/s 2 How far does the car travel in 8.0s? a = -2.0 m/s 2 v i = 25 m/s t = 8.0 s d =? which formula? Jan 31 9:14 PM d = v i t + 1/2 a t 2 d = 25(8.0) + 1/2(-2.0)(8.0 2 ) d = d = 136 m (+) Jan 31 9:18 PM 19

20 Feb 15 12:08 PM Examples in text book p 232 A: a cyclist accelerates from rest to a velocity of 10m/s in 12 s. with a constant acceleration a) what is the acceleration? b)what is the distance travelled during this phase of acceleration? Jan 31 9:21 PM 20

21 Sample Problems 1. A ball rolling down a hill at 4.0 m/s accelerates at 2.0 m/s 2 What is its velocity 5.0s later Given: v 1 = 4.0 m/s a = 2.0 m/s 2 t = 5.0s We can use equation 1 Equation 1: v 2 = v 1 + at v 2 = 4.0m/s m/s 2 *5.0s v 2 = 4.0 m/s m/s = 14.0 m/s The ball reaches a velocity of 14 m/s in 5.0 s. Jan 31 9:22 PM 2. A car travelling at 10 m/s (2 Sig figs) accelerates at 4.0 m/s 2 for 8.0s. What is its displacement during this interval? Given v 1 = 10 m/s a = 4.0 m/s 2 t = 8.0s find Δd We can use equation # 3 Equation 3: x f =x i + v 1 t + ½at 2 Δx =vit + 1/2 at 2 =10 m/s* 8.0s + 1/2 * 4.0m/s 2 * 8.0s 2 = 80m + 128m = 208 m = 2.1 x 10 2 m The cars displacement for 8.0s is 2.0 x 10 2 m Jan 24 7:24 AM 21

22 3. A car accelerating at 5.0 m/s 2 has a displacement of 114m in 6.0s. What was its velocity at the beginning of the interval? given: a = 5.0 m/s 2 t = 6.0s Δd = 114 find v 1 We can use equation 3 Equation 3: x f =x i + v 1 t + ½at 2 Δd = x f - x i Δ d = v i t + 1/2at 2 114m = vi(6.0s) + 1/2 (5.0 m/s 2 )(6.0s) 2 114m = 6.0s(vi) + 90 m combine like terms 114m - 90m = 6.0s (vi) 24m = 6.0s (vi) vi = 4.0 m/s Jan 24 7:32 AM 4. A ball rolls at an initial velocity of 4.0 m/s up a hill. five seconds later it is rolling down the hill at 6.0 m/s 2. Equation 1: v 2 = v 1 + at Find the following: Equation 2: x f = x i + ½(v i + v f )Δt a) acceleration Equation 3: xf =xi + v1t + ½at 2 b) displacement at 5.0 s. a) assuming up the hill is positive and down the hill is negative (in terms of displacement and therefor speed) given: v1 = 4.0 m/s v2 = 6.0 m/s t = 5.0s Equation 4: v f 2 = v i 2 + 2aΔx find a a = (v 2 v 1 )/Δt = ( 6.0 m/s 4.0 m/s)/5.0s = 10m/s/5s = 2m/s 2 b) Use a to find displacement Equation #2 or # 4 Equation 2: x f = x i + ½(v i + v f )Δt Equation 4: v f 2 = v i 2 + 2aΔx Using equation 2 Δx =( 1/2)(v1 +v2)*t 0.5* (-6.0m/s +4.0m/s)(5.0s) = 0.5(-2.0m/s)(5.0s) =-5m The ball is 5.0 m down the hill from its starting pint after 5.0s Using equation 4 (-6.0m/s) 2 = (4.0m/s) 2 + 2(-2m/s 2 )(Δx) 36 = x 20 = -4x x = -5m Feb 2 3:22 PM 22

23 Practice: Section 10.2 p. 234 p What do we know? Δd = 402m v i = 0m/s Δt = 6.0s x i = 0m x f = 402m?a? v f (km/h) Look for formulas with only one unknown... xf = xi + viδt + 1/2(aΔt 2 ) Δd = viδt + 1/2(aΔt 2 ) find a 402 = 0 + (0*6 + 1/2(a*6^2) 402 = 1/2(a*36) 804 = 36a a = 22.3m/s 2 = 22m/s 2 vf vf 2 = vi 2 +2aΔx vf 2 = 0 + 2* 22.3 * 402 vf = vf = m/s = 130 m/s Feb 2 3:46 PM A car is travelling at 27 m/s and then decelerates at a a constant rate of 9 m/s 2. The car comes to a stop 40 m from where it began decelerating. How long did it take? Given: vi = 27m/s a = 9 m/s 2 Δd = 40 m t =? d= vi*t + 1/2*a*t 2 0 = 40m + (27m/s)t + 0.5( 9m/s 2 )t 2 0 = 40m + 27m/s*t + 4.5m/s 2 *t 2 a = 4.5 b = 27 c = 40 t = [ b + (b 2-4ac) ]/ 2a t = [ 27 + (27 2-4(-4.5)(-40))]/2(-4.5) Feb 1 3:46 PM 23

24 Mar 10 1:35 PM 24