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Chapter overview
Learning goals
Learning Goals
Maths
Summary
Like distance-time graphs, velocity-time graphs give information about journeys. They tell you about the velocity at different points of time. They can also be used to calculate acceleration by finding the gradient of line segment.
There is a subtle difference between speed and velocity. Speed is an example of a scalar, meaning it only has information about magnitude (size). Velocity is an example of a vector, meaning it has information on magnitude and direction. So you can't technically have a negative speed, but you can have a negative velocity (direction of travel is backwards).
The unit for velocity is distance per time unit. For example, kilometres per hour ($\text{km}/\text{h}$).
The velocity of a section of a journey is $-50$ kilometres per hour. What is the speed and direction of this journey?
The magnitude of a vector is just the numerical value without the negative sign. So the speed is fifty kilometres per hour. The negative in the velocity tells you that the direction is backwards toward the starting point of the whole journey.
Note: At the beginning of a problem or scenario, pick a direction to be positive and thus the opposite direction to be negative. Stay consistent to this choice within the problem or scenario.
By taking any point on a velocity-time graph, you will find the velocity ($y$-coordinate) at a point in time ($x$-coordinate).
Using the velocity-time graph below, find the velocity when:
a. time = one hour;
b. time = two hours.
Note: Time in hours is on the $x$-axis and velocity in kilometres per hour is on the $y$-axis.
a. When $x=1$, the $y$-coordinate is $30$ so velocity is thirty kilometres per hour.
b. When $x=2$, the $y$-coordinate is $20$ so velocity is twenty kilometres per hour.
As long as your axes' units are compatible (using the same time unit), the gradient of a velocity-time graph represents the acceleration of that segment's motion. Acceleration is the rate of change of velocity, so is a measure of how velocity is changing. Recall that
$\text{gradient}=\frac{\text{change in }y}{\text{change in }x}$
Distance is on the $y$-axis and time is on the $x$-axis, so it follows from
$\text{acceleration}=\frac{\text{change in velocity}}{\text{change in time}}$
that gradient gives acceleration.
Note: A negative gradient represents that the motion is slowing down, not that motion is in the direction back towards the beginning. Velocity can still be positive even if the gradient at that point is negative.
A flat line on a velocity-time graph represents no change in velocity. In other words, motion occurs at a constant speed. Don't get this confused with a flat line on distance-time graph which means motion has stopped.
In the velocity-time graph below, find the acceleration in the first minute.
Note: Time in hours is on the $x$-axis and velocity in kilometres per hour is on the $y$-axis.
You look for the gradient of the line segment in the first minute of the graph. You can use the start and end coordinates: $(0,0)$ and $(1,60)$. Hence the gradient is
$\text{gradient}=\frac{60-0}{1-0}=60$
Thus, the acceleration here is sixty kilometres per hour squared. Note: The unit used for acceleration is distance per time unit squared (here it is $\text{km}/\text{h}^2$)
As long as your axes' units are compatible, you can find distance travelled in a time-period from a velocity-time graph by using the area under the graph (the area between the $x$-axis and the graph's line) in that time-period.
Note: If the line goes below the $x$-axis, then the direction of motion is back towards the beginning.
Using this velocity time graph (the same as above), find the distance travelled between hour one and hour three.
You want the area under between the graph and the $x$-axis between $x=1$ and $x=3$. This is a compound shape made up of a rectangle and a triangle. The area of the rectangle between $x=1$ and $x=2$ is $60$, so in this time period, sixty kilometres are covered. Between $x=2$ and $x=3$, the triangle has an area of $30$. Hence, in total, in the two hour window between hour one and hour three, ninety kilometres have been covered. Note: The unit has come from the fact that the velocity is in kilometres per hour.
Like distance-time graphs, velocity-time graphs give information about journeys. They tell you about the velocity at different points of time. They can also be used to calculate acceleration by finding the gradient of line segment.
There is a subtle difference between speed and velocity. Speed is an example of a scalar, meaning it only has information about magnitude (size). Velocity is an example of a vector, meaning it has information on magnitude and direction. So you can't technically have a negative speed, but you can have a negative velocity (direction of travel is backwards).
The unit for velocity is distance per time unit. For example, kilometres per hour ($\text{km}/\text{h}$).
The velocity of a section of a journey is $-50$ kilometres per hour. What is the speed and direction of this journey?
The magnitude of a vector is just the numerical value without the negative sign. So the speed is fifty kilometres per hour. The negative in the velocity tells you that the direction is backwards toward the starting point of the whole journey.
Note: At the beginning of a problem or scenario, pick a direction to be positive and thus the opposite direction to be negative. Stay consistent to this choice within the problem or scenario.
By taking any point on a velocity-time graph, you will find the velocity ($y$-coordinate) at a point in time ($x$-coordinate).
Using the velocity-time graph below, find the velocity when:
a. time = one hour;
b. time = two hours.
Note: Time in hours is on the $x$-axis and velocity in kilometres per hour is on the $y$-axis.
a. When $x=1$, the $y$-coordinate is $30$ so velocity is thirty kilometres per hour.
b. When $x=2$, the $y$-coordinate is $20$ so velocity is twenty kilometres per hour.
As long as your axes' units are compatible (using the same time unit), the gradient of a velocity-time graph represents the acceleration of that segment's motion. Acceleration is the rate of change of velocity, so is a measure of how velocity is changing. Recall that
$\text{gradient}=\frac{\text{change in }y}{\text{change in }x}$
Distance is on the $y$-axis and time is on the $x$-axis, so it follows from
$\text{acceleration}=\frac{\text{change in velocity}}{\text{change in time}}$
that gradient gives acceleration.
Note: A negative gradient represents that the motion is slowing down, not that motion is in the direction back towards the beginning. Velocity can still be positive even if the gradient at that point is negative.
A flat line on a velocity-time graph represents no change in velocity. In other words, motion occurs at a constant speed. Don't get this confused with a flat line on distance-time graph which means motion has stopped.
In the velocity-time graph below, find the acceleration in the first minute.
Note: Time in hours is on the $x$-axis and velocity in kilometres per hour is on the $y$-axis.
You look for the gradient of the line segment in the first minute of the graph. You can use the start and end coordinates: $(0,0)$ and $(1,60)$. Hence the gradient is
$\text{gradient}=\frac{60-0}{1-0}=60$
Thus, the acceleration here is sixty kilometres per hour squared. Note: The unit used for acceleration is distance per time unit squared (here it is $\text{km}/\text{h}^2$)
As long as your axes' units are compatible, you can find distance travelled in a time-period from a velocity-time graph by using the area under the graph (the area between the $x$-axis and the graph's line) in that time-period.
Note: If the line goes below the $x$-axis, then the direction of motion is back towards the beginning.
Using this velocity time graph (the same as above), find the distance travelled between hour one and hour three.
You want the area under between the graph and the $x$-axis between $x=1$ and $x=3$. This is a compound shape made up of a rectangle and a triangle. The area of the rectangle between $x=1$ and $x=2$ is $60$, so in this time period, sixty kilometres are covered. Between $x=2$ and $x=3$, the triangle has an area of $30$. Hence, in total, in the two hour window between hour one and hour three, ninety kilometres have been covered. Note: The unit has come from the fact that the velocity is in kilometres per hour.
Distance-time graphs
FAQs
Question: What does the area under a velocity-time graph tell you?
Answer: As long as your axes' units are compatible, you can find distance travelled in a time-period from a velocity-time graph by using the area under the graph (the area between the x-axis and the graph's line) in that time-period.
Question: What does the gradient tell you on a velocity-time graph?
Answer: As long as your axes' units are compatible (using the same time unit), the gradient of a velocity-time graph represents the acceleration of that segment's motion.
Theory
Exercises
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