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**UNIT 1: 1-D KINEMATICS Lesson 4:**

CENTRE HIGH: PHYSICS 20 UNIT 1: 1-D KINEMATICS Lesson 4: Graphical Analysis of Accelerating Motion

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Reading Segment #1: Average and Instantaneous Velocity on Position-Time Graphs To prepare for this section, please read: Unit 1: p.14

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**D1. Position-Time Graphs (Accelerated Motion)**

Recall, when the motion of an object is uniform, - the velocity is constant - since velocity is the slope of a position-time graph, the graph is a straight line (constant slope). But now, the velocity is going to change. As a result, the line is no longer going to be straight.

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**Consider the following position-time graph**

for an accelerating object: (Forward is positive) d (m) t (s) Can you describe the motion?

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d (m) Zero slope t (s) At the start, the slope is zero. Thus, the object starts from rest.

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d (m) Slopes are increasing t (s) But, over time, the slope increases. Thus, the object's velocity increases, which in this case means that the object is speeding up.

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d (m) Slopes are increasing t (s) Since the velocity is changing, the object is accelerating.

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**For non-linear position-time graphs, there are two types**

of velocities (i.e. slopes) that could be calculated: 1. Average velocity - from one time to another 2. Instantaneous velocity - at one point in time

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1. Average velocity To find the average velocity from t = t1 to t = t2, Find the initial position (at t1) and the final position (at t2) on the graph Draw a straight line from the two positions (called a secant line) Find the slope of this secant line

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**e.g. How would we find the average velocity from t = 3.0 s**

to t = 5.0 s on the graph below? d (m) t (s)

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d (m) t (s) First, find the initial and final position on the line.

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d (m) secant line t (s) Draw a straight line between the two points. This is called the secant line.

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d (m) slope = vavg t (s) Find the slope of the secant line. The slope is the average velocity between 3.0 and 5.0 s.

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**2. Instantaneous velocity**

To find the instantaneous velocity at t = t1 : Find the position (at t1) on the graph Draw a tangent line at this point Choose 2 points on the tangent line and find its slope

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**e.g. How would we find the instantaneous velocity at t = 4.0 s ?**

d (m) 4.0 t (s)

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d (m) 4.0 t (s) First, find the position at t = 4.0 s on the line.

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d (m) tangent line 4.0 t (s) Next, draw a tangent line at this point. A tangent line touches the curve only at this point, and if done correctly, it should show the slope at that point.

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d (m) slope of tangent = velocity 4.0 t (s) Finally, locate two points on the tangent line and find its slope. The slope is the instantaneous velocity.

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**Ex. 1 For the graph below, find: (North is positive)**

a) the average velocity from 0.25 s to 5.0 s b) the instantaneous velocity at 1.0 s

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**a) Locate the positions at t = 0.25 s and t = 5.0 s**

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secant line a) Draw a straight line between the two points. This the secant line.

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(5.0 s, 50.0 m) secant line (0.25 s, 11.0 m) a) Choose two points on the secant line and find the slope.

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**a) Average velocity between 0.25 s and 5.0 s:**

vavg = slope of secant line = y2 - y1 = (50.0 m) - (11.0 m) x2 - x1 (5.0 s) - (0.25 s) = m/s North Be sure to include the direction for velocity.

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**b) Locate the position at t = 1.0 s**

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tangent line b) Draw the tangent line at this point. It should touch the curve only at this point.

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(3.25 s, 48.0 m) tangent line (1.0 s, 22.0 m) b) Find two points on the tangent line. Calculate the slope.

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**b) Instantaneous velocity between 1.0 s:**

v1.0 = slope of tangent line = y2 - y1 = (48.0 m) - (22.0 m) x2 - x1 (3.25 s) - (1.0 s) = m/s North Be sure to include the direction for velocity.

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Practice Problems Try these problems in the Physics 20 Workbook: Unit 1 p. 16 #1

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Reading Segment #2: Velocity-Time Graphs To prepare for this section, please read: Unit 1: p.15

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**D2. Velocity-Time Graphs (Accelerated Motion)**

There are two major calculations we find for velocity-time graphs: 1. Acceleration - this will be the slope of a v-t graph 2. Displacement - this will be the area "under" the v-t graph

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**1. Acceleration on a velocity-time graph**

The slope of a v-t graph is the acceleration Thus, if the object has uniform acceleration: - the acceleration is constant - the slope is constant on the v-t graph - this produces a straight line

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**e.g. Ref: Forward is positive**

Backward is negative If the graph has a positive slope (i.e. a positive acceleration), it means the object's velocity is increasing. v v t

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**Ref: Forward is positive**

Backward is negative If the graph has a negative slope (i.e. a negative acceleration), it means the object's velocity is decreasing. v v t

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**Ref: Forward is positive**

Backward is negative If the graph has a zero slope (i.e. a zero acceleration), it means the object has a constant velocity. v v t

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**Ex. 2 For the graph below, find the acceleration at t = 3.0 s.**

(East is positive) Velocity 8.0 ( 103 m/s) 2.0 4.0 t (s)

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Velocity 8.0 ( 103 m/s) 2.0 4.0 t (s) Acceleration is the slope of a v-t graph. Since this is a straight line, you can choose any 2 points on the line to calculate the slope.

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Velocity (0 s, 8.0 103 m/s) ( 103 m/s) (4.0 s, 2.0 103 m/s) 2.0 4.0 t (s) Choose 2 points on the line.

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Acceleration at 3.0 s: a3.0 = slope of line = y2 - y1 = (2.0 103 m/s) - (8.0 103 m/s) x2 - x (3.0 s) - (0 s) = 103 m/s2 = 103 m/s2 West Be sure to include the direction for acceleration.

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**Ex. 3 Describe the motion showed by the following**

velocity-time graph: (Forward is positive) v (m/s) 10 t (s) -6

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**v (m/s) Forward is positive**

t (s) -6 At the very start of the motion (A), the object is moving forward at 10 m/s.

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**v (m/s) Forward is positive**

B t (s) -6 Along AB, the object is slowing down at a constant acceleration (constant slope), from 10 m/s to 0. It is still moving forward, since the velocity is positive. At B, the object is at rest.

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**v (m/s) Forward is positive**

B t (s) C Along BC, the object is speeding up at a constant acceleration (constant slope), from 0 to 6 m/s. It is now moving backward, since the velocity is negative. At C, the object is at moving backwards at 6 m/s.

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**v (m/s) Forward is positive**

B t (s) C D Along CD, the object is at a constant velocity of -6.0 m/s. Acceleration (slope) is zero. At D, the object is at moving backwards at 6 m/s.

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**v (m/s) Forward is positive**

B E t (s) C D Along DE, the object is slowing down at a constant acceleration (slope), from 6 m/s to zero. It is moving backwards, since the velocities are negative. At E, the object has come to rest.

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**2. Displacement on a velocity-time graph**

The area between the line and the t-axis on a v-t graph is the displacement

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**Ref: Forward is positive**

Backward is negative e.g. When the line is above the t-axis, displacement (area) is positive. i.e. A forward displacement v v t t

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**Ref: Forward is positive**

Backward is negative e.g. When the line is below the t-axis, displacement (area) is negative. i.e. A backward displacement v v t t

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**Ex. 4 Calculate the displacement for the first 8.0 seconds.**

(North is positive) v (m/s) 33 14 8.0 t (s)

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v (m/s) 33 14 8.0 t (s) Displacement is the area between the line and the t-axis

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v (m/s) 33 1 14 2 8.0 t (s) To calculate the area, find the area of the triangle (1) and the rectangle (2).

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v (m/s) 33 m/s 14 m/s 8.0 s t (s) Show the dimensions on the diagram.

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**Displacement for the first 8.0 seconds:**

A1 = Area of triangle = b h = (8.0 s) (19 m/s) = m A2 = Area of rectangle = L w = (8.0 s) (14 m/s) = 112 m Thus, d = A1 + A2 = m North

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Practice Problems Try these problems in the Physics 20 Workbook: Unit 1 pp #2 - 5

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