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Description of motion

The purpose of this chapter is to describe motion, and now that we understand the definitions of displacement, distance, velocity, speed and acceleration, we are ready to start using these ideas to describe how an object is moving. There are many ways of describing motion:

  1. words
  2. diagrams
  3. graphs

These methods will be described in this section.

We will consider three types of motion: when the object is not moving (stationary object), when the object is moving at a constant velocity (uniform motion) and when the object is moving at a constant acceleration (motion at constant acceleration).

Stationary object

The simplest motion that we can come across is that of a stationary object. A stationary object does not move and so its position does not change, for as long as it is standing still. An example of this situation is when someone is waiting for something without moving.The person remains in the same position.

Lesedi is waiting for a taxi. He is standing two metres from a stop street at t = 0 s. After one minute, at t = 60 s , he is still 2 metres from the stop street and after two minutes, at t  = 120  s , also 2 metres from the stop street. His position has not changed. His displacement is zero (because his position is the same), his velocity is zero (because his displacement is zero) and his acceleration is also zero (because his velocity is not changing).

We can now draw graphs of position vs. time ( x vs. t ), velocity vs. time ( v vs. t ) and acceleration vs. time ( a vs. t ) for a stationary object. The graphs are shown in [link] . Lesedi's position is 2 metres from the stop street. If the stop street is taken as the reference point, his position remains at 2 metres for 120 seconds. The graph is a horizontal line at 2 m.The velocity and acceleration graphs are also shown. They are both horizontal lines on the x -axis. Since his position is not changing, his velocity is 0 m · s - 1 and since velocity is not changing, acceleration is 0 m · s - 2 .

Graphs for a stationary object (a) position vs. time (b) velocity vs. time (c) acceleration vs. time.
Gradient

The gradient of a line can be calculated by dividing the change in the y -value by the change in the x -value.

m = Δ y Δ x

Since we know that velocity is the rate of change of position, we can confirm the value for the velocity vs. time graph, by calculating the gradient of the x vs. t graph.

The gradient of a position vs. time graph gives the velocity.

If we calculate the gradient of the x vs. t graph for a stationary object we get:

v = Δ x Δ t = x f - x i t f - t i = 2 m - 2 m 120 s - 60 s ( initial position = final position ) = 0 m · s - 1 ( for the time that Lesedi is stationary )

Similarly, we can confirm the value of the acceleration by calculating the gradient of the velocity vs. time graph.

The gradient of a velocity vs. time graph gives the acceleration.

If we calculate the gradient of the v vs. t graph for a stationary object we get:

a = Δ v Δ t = v f - v i t f - t i = 0 m · s - 1 - 0 m · s - 1 120 s - 60 s = 0 m · s - 2

Additionally, because the velocity vs. time graph is related to the position vs. time graph, we can use the area under the velocity vs. time graph to calculate the displacement of an object.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
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emma Reply
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Magreth
progressive wave
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Siyavula textbooks: grade 10 physical science [caps]. OpenStax CNX. Sep 30, 2011 Download for free at http://cnx.org/content/col11305/1.7
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