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Module 1 1 Graph Analysis 1. Take note of the names and units of the axes. 2. Determine and interpret intercepts on the axes. 3. Determine and interpret the slope of the graph. The units will help. 4. Determine and interpret the area under a graph. The units will help. 5. Determine turning points maximum and minimum. 6. Determine and interpret any asymptotes.

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Module 1 1

Graph Analysis1. Take note of the names and units of the axes.

2. Determine and interpret intercepts on the axes.

3. Determine and interpret the slope of the graph. The units will help.

4. Determine and interpret the area under a graph. The units will help.

5. Determine turning points − maximum and minimum.

6. Determine and interpret any asymptotes.

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Example 1

The graph above is a velocity-vs-time graph. The value of the intercept on the y-axis, u stands for the initial velocity of the body. That is, the velocity when t = 0. The slope of the graph will have units ms–1/s which gives ms–2. This stands for acceleration. The area under the graph will have units ms–1 × s which gives m. This stands for a distance.

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Example 2

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Motion in a Straight Line (Linear Kinematics)

Displacement is the distance travelled in a specified direction. It is measured in metres (m) and is a vector quantity. Speed is the distance moved per unit time. It is measured in metres per second (ms–1) and is a scalar quantity. Velocity is the rate of change of displacement or the distance moved per unit time in a specified direction. It is measured in metres per second (ms–1) and is a vector quantity. Acceleration is the rate of change of velocity. It is measured in metres per square second (ms–2) and is a vector quantity. Average velocity is the total distance travelled divided by the total time taken.

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Note:

▪ The slope of a displacement/time graph represents velocity.

▪ The slope of a velocity/time graph represents acceleration.

▪ The area under a velocity/time graph represents displacement.

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Newton’s Equations of Motion

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For example, A car starts from rest and accelerates uniformly for 4 s at which time the velocity reaches 10 ms–1. It maintains this velocity for a further 6 s and is then uniformly brought to rest in a further 5 s. Sketch a velocity/time graph of the journey and use it to find:

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(i) the acceleration during the first 4 s.

(ii) the distance travelled during the first 4 s.

(iii) the distance travelled while the velocity was constant.

(iv) the distance travelled during the last 5 s.

(v) the deceleration during the last 5 s.

(vi) the average velocity over the entire journey.

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Motion Under Gravity Whenever a body is moving vertically, it will always be

under the influence of gravity which always acts

downwards and is constant. Gravity will lead to an

acceleration due to gravity which also acts downwards and

is constant. (In all the worked examples, g is taken as 10

ms–2 and air resistance is negligible)

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A Stone Released from Rest Velocity/time graph

Slope = g Area under graph = height

e.g. A stone is released from 20 m above the ground. Find:

(i) the time it takes to hit the ground. (ii) the velocity with which it hits the ground.

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A Ball Bumping Velocity/time graph

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A Stone Thrown Up in the Air Until it Returns to the Thrower’s Hand Velocity/time graph

Note: Slope is constant, so “g” is constant. Taking up as positive, “g” is negative.

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Distance/time graph

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Acceleration/time graph

Neglecting the time the ball is in contact with ground.