I have difficulty conceptualizing a displacement as an area. The math makes sense. I just can't "see" it.

For me, I looked at it by drawing a flat line between the times on the x axis starting at the y of where the diagonal's going up📈 starts or going down📉 ends. Then if the triangle doesn't connect to the bottom you add a rectangle above or below the triangle so that it touches making sure to still keep it within the x range that the triangle above it is in. If there is a flat line, then draw a rectangle above or under it until it touches the x-axis and remains under the max time on the x-axis.

How do you you find the Instantaneous velocity speed from graphs review?

The slope of a given point is the instantaneous velocity. For example if you had a position vs. time graph, then if you were to find the instantaneous velocity of a moment, point P, then the slope of point P is the instantaneous velocity. Was this helpful?

can any one please explain to me the part, At what time does the penguin have the same position as t=0s? why we need to calculate the two triangles and cancel them out then find the rest displacement ? can anyone help and thank you guys.

The first triangle has an area of 5 from the line. The second triangle has an area of -2.5 from the line. That means you need to find at what point past 3 seconds will remove another 2.5. The answer is 3.5. That's because at 3.5 seconds a rectangle from the -2.5 triangle to the end of 3.5 is formed. It's area is -2.5 because 5 * 0.5 = 2.5, and it's under the line so negative.

Can someone tell me why the area under the slope is the displacement?

Displacement is basically the vector form of distance, and we know distance is speed x time. So displacement is velocity (vector form of speed) x time, which is what’s plotted on the graph. The area under the graph will thus be velocity x time, giving you the displacement.

What does it mean when it says that instantaneous speed is the magnitude of instantaneous velocity?

It's the quantity without the sign. So if the instantaneous velocity is, for example, -2m/s, the instantaneous speed is 2m/s.

what is the formula for velocity

Velocity is equal to displacement over time. V=d/t

I don't really get the concept of displacement. What is the easiest way to figure out displacement problems?

say you walk 1 meter right, away from your house, then you walk 1 meter back to your house. you travelled 2 meters, so your distance is 2, but you ended up 0 meters from your house, so your displacement is 0. If walk 1 meter to the left of your house, your distance is 1 meter and your displacement is either 1 or -1, depending on what you arbitrarily designate as forward (positive) or backward (negative) (the convention is to use right as forwards). On a graph, a line going forwards is like you walking to the right, and a line going down is like going to the left. if the y is positive, you are to the right of your house, if it is negative, you are at the left of your house.

I dont understand the concept of area found by the displacement on graph

On a Velocity-Time graph, you can find the displacement of a time interval by finding the area under the curve. Usually this involves triangles and rectrangles.

Main content

Course: AP®︎/College Physics 1 > Unit 1

Lesson 4: Velocity and speed from graphs

Instantaneous velocity and speed from graphs review

Google Classroom

Review the key terms and skills related to analyzing motion graphs, such as finding velocity from position vs. time graphs and displacement from velocity vs. time graphs.

Key terms

Term	Meaning
Instantaneous velocity	Velocity at a given moment in time. Has SI units of $\frac{m}{s}$ ‍.
Instantaneous speed	Speed at a given moment in time. Equal to the magnitude of the instantaneous velocity. Has SI units of $\frac{m}{s}$ ‍.

Equation	Symbol breakdown	Meaning in words
$\bar{v} = \frac{Δ x}{Δ t}$ ‍	$\bar{v}$ ‍ is average velocity, $Δ x$ ‍ is displacement, and $Δ t$ ‍ is change in time.	Average velocity is displacement divided by time interval of displacement.
$v_{avg} = \frac{d}{Δ t}$ ‍	$v_{avg}$ ‍ is average speed, $d$ ‍ is distance, and $Δ t$ ‍ is change in time .	Average speed is distance divided by time interval for the distance traveled.

Analyzing motion graphs

Velocity is the slope of position vs. time graph

The equation for the slope of a position vs. time graph matches the definition of velocity exactly.

slope = velocity = \frac{Δ x}{Δ t}

To calculate the average velocity between two points

P_{1}

and

P_{2}

, we divide the change of position

Δ x

by the change in time

Δ t

The instantaneous velocity at point

P_{1}

is equal to the slope of the position graph at point

P_{1}

The simplest non-calculus approach uses the fact that, within a region of constant slope, the average slope between any two points

P_{1}

and

P_{2}

in that region will be equal to the instantaneous slope at every point

P

in that region.

In other words, to calculate the instantaneous velocity at point

P

shown above, just find the average velocity between any two points

P_{1}

and

P_{2}

within the constant slope section of the graph that contains point

P

Note that finding the average slope between points

P_{1}

and

P_{3}

shown above would NOT give you the correct instantaneous velocity at point

P

even though it would correctly give the average velocity between points

P_{1}

and

P_{3}

Displacement is the area under the curve on a velocity vs. time graph

To find the displacement between two points

P_{1}

and

P_{2}

on a velocity vs. time graph, we find the area under the curve between the two points.

Area = displacement = v Δ t

The change in time

Δ t

will be the width of the area, and the height

v

is on the vertical axis.

Common mistakes and misconceptions

Some people think instantaneous velocity and speed are the same as average velocity and speed. When people use the words speed or velocity, they usually mean instantaneous velocity or instantaneous speed. Average velocity and speed account for motion occurring over a time period, and instantaneous velocity and speed describe motion at a given moment in time.

Learn more

For deeper explanations of velocity and speed see the videos on instantaneous speed and velocity and position vs. time graphs.

To check your understanding and work toward mastering these concepts, check out these exercises:

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aqbar135
Posted 6 years ago. Direct link to aqbar135's post “can you show example and ...”
can you show example and exercise how to solve instantaneous velocity or speed involving curve graph.. i very like this website to learn things that i missed in class bcs of the way of explanation process very clear, fun, and easy to understand.. i hope you can improve the 'instantaneous' topics and notify me please.. ^_^
Button navigates to signup pageComment on aqbar135's post “can you show example and ...”
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Diane Shawgo
Posted 4 years ago. Direct link to Diane Shawgo's post “I have difficulty concept...”
I have difficulty conceptualizing a displacement as an area. The math makes sense. I just can't "see" it.
Button navigates to signup pageComment on Diane Shawgo's post “I have difficulty concept...”
(10 votes)
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- 23jaivasamad
  Posted 9 months ago. Direct link to 23jaivasamad's post “For me, I looked at it by...”
  For me, I looked at it by drawing a flat line between the times on the x axis starting at the y of where the diagonal's going up📈 starts or going down📉 ends. Then if the triangle doesn't connect to the bottom you add a rectangle above or below the triangle so that it touches making sure to still keep it within the x range that the triangle above it is in. If there is a flat line, then draw a rectangle above or under it until it touches the x-axis and remains under the max time on the x-axis.
  Button navigates to signup page
  (3 votes)
Theresa Chaudhry
Posted 7 months ago. Direct link to Theresa Chaudhry's post “It would have been useful...”
It would have been useful if Sal had included the negative velocity aspect in his video.
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(9 votes)
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nayeliviera02
Posted 6 years ago. Direct link to nayeliviera02's post “How do you you find the I...”
How do you you find the Instantaneous velocity speed from graphs review?
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(4 votes)
Answer
- srikarve
  Posted 4 years ago. Direct link to srikarve's post “The slope of a given poin...”
  The slope of a given point is the instantaneous velocity. For example if you had a position vs. time graph, then if you were to find the instantaneous velocity of a moment, point P, then the slope of point P is the instantaneous velocity. Was this helpful?
  Comment on srikarve's post “The slope of a given poin...”
  (9 votes)
yikian2021
Posted a year ago. Direct link to yikian2021's post “Can someone tell me why t...”
Can someone tell me why the area under the slope is the displacement?
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(3 votes)
Answer
- Koline
  Posted a year ago. Direct link to Koline's post “Displacement is basically...”
  Displacement is basically the vector form of distance, and we know distance is speed x time. So displacement is velocity (vector form of speed) x time, which is what’s plotted on the graph. The area under the graph will thus be velocity x time, giving you the displacement.
  Button navigates to signup page
  (6 votes)
shehab shaar
Posted 8 months ago. Direct link to shehab shaar's post “can any one please explai...”
can any one please explain to me the part,

At what time does the penguin have the same position as
t=0s?

why we need to calculate the two triangles and cancel them out then find the rest displacement ? can anyone help and thank you guys.
Button navigates to signup pageComment on shehab shaar's post “can any one please explai...”
(4 votes)
Answer
- Gabe
  Posted 8 months ago. Direct link to Gabe's post “The first triangle has an...”
  The first triangle has an area of 5 from the line.
  The second triangle has an area of -2.5 from the line.
  That means you need to find at what point past 3 seconds will remove another 2.5.
  The answer is 3.5.
  That's because at 3.5 seconds a rectangle from the -2.5 triangle to the end of 3.5 is formed.
  It's area is -2.5 because 5 * 0.5 = 2.5, and it's under the line so negative.
  Button navigates to signup page
  (4 votes)
Lee Mommsen
Posted a year ago. Direct link to Lee Mommsen's post “I don't really get the co...”
I don't really get the concept of displacement.
What is the easiest way to figure out displacement problems?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- rylan.wetsell
  Posted a year ago. Direct link to rylan.wetsell's post “say you walk 1 meter righ...”
  say you walk 1 meter right, away from your house, then you walk 1 meter back to your house. you travelled 2 meters, so your distance is 2, but you ended up 0 meters from your house, so your displacement is 0.
  If walk 1 meter to the left of your house, your distance is 1 meter and your displacement is either 1 or -1, depending on what you arbitrarily designate as forward (positive) or backward (negative) (the convention is to use right as forwards).
  On a graph, a line going forwards is like you walking to the right, and a line going down is like going to the left. if the y is positive, you are to the right of your house, if it is negative, you are at the left of your house.
  Button navigates to signup page
  (8 votes)
srikarve
Posted 4 years ago. Direct link to srikarve's post “What does it mean when it...”
What does it mean when it says that instantaneous speed is the magnitude of instantaneous velocity?
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(3 votes)
Answer
- J.A. Brown
  Posted 4 years ago. Direct link to J.A. Brown's post “It's the quantity without...”
  It's the quantity without the sign. So if the instantaneous velocity is, for example, -2m/s, the instantaneous speed is 2m/s.
  Comment on J.A. Brown's post “It's the quantity without...”
  (3 votes)
dan.kawadza
Posted 9 months ago. Direct link to dan.kawadza's post “what is the formula for v...”
what is the formula for velocity
Button navigates to signup pageComment on dan.kawadza's post “what is the formula for v...”
(3 votes)
Answer
- 23jaivasamad
  Posted 9 months ago. Direct link to 23jaivasamad's post “Velocity is equal to disp...”
  Velocity is equal to displacement over time.
  V=d/t
  Button navigates to signup page
  (2 votes)
saraa1202
Posted 9 months ago. Direct link to saraa1202's post “I dont understand the con...”
I dont understand the concept of area found by the displacement on graph
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(2 votes)
Answer
- Cameron J. Hopkins
  Posted 9 months ago. Direct link to Cameron J. Hopkins's post “On a Velocity-Time graph,...”
  On a Velocity-Time graph, you can find the displacement of a time interval by finding the area under the curve. Usually this involves triangles and rectrangles.
  Button navigates to signup page
  (3 votes)