On a displacement-time graph, what does the gradient of the graph tell you?

The gradient of a displacement-time graph is the velocity of the object. A positive gradient means the object is moving forwards , and a negative gradient means it is moving backwards . The steeper the line, the greater the speed.

True or False? A horizontal section of a displacement-time graph shows that the object is moving at a constant velocity.

False. A horizontal section of a displacement-time graph means the displacement is not changing, so the object is stationary . Strictly, a stationary object does have a constant velocity , of zero, so the word doing the work in the statement is moving . A constant non-zero velocity gives a straight sloping line on a displacement-time graph, not a horizontal one; it is on a velocity -time graph that a horizontal line means constant velocity.

On a displacement-time graph, what does a straight section tell you, and what does a curved section tell you?

On a displacement-time graph, a straight section shows the object moving at a constant velocity , because the gradient, and so the velocity, is the same all the way along it. A curved section shows the velocity changing , so the object is accelerating or decelerating . The gradient of a curve is different at every point.

What does it mean if a displacement-time graph touches the time axis?

Where a displacement-time graph touches the time axis the displacement is zero , so the object is at the fixed origin that its displacement is measured from. This does not mean the object has stopped. An object is stationary only where the graph is horizontal .

On a displacement-time graph, what does the value where the graph meets the vertical axis represent?

Where a displacement-time graph meets the vertical axis, the value is the object's initial displacement : how far it is from the fixed origin at time zero. An object does not have to start at the origin, so this value is not always zero. A graph starting at 100 m, for example, could show someone setting out on a walk from a point 100 m from their house.

On a velocity-time graph, what does the gradient of the graph tell you?

The gradient of a velocity-time graph is the acceleration of the object. A straight line means the acceleration is constant , and a horizontal line means the acceleration is zero, so the object is moving at a constant velocity .

On a velocity-time graph, what does the area between the graph and the time axis represent?

The area between a velocity-time graph and the time axis gives the change in displacement of the object over that time. Where the graph is above the axis the object is moving forwards, so that area is travel in the positive direction. Where the graph is below the axis the object is moving backwards.

What does it mean if a velocity-time graph touches the time axis?

Where a velocity-time graph touches the time axis the velocity is zero , so the object is instantaneously at rest . If the graph crosses the axis rather than only touching it, the velocity changes sign, so the object changes direction at that moment.

True or False? A section of a velocity-time graph that lies below the time axis always shows an object that is slowing down.

False. Below the time axis on a velocity-time graph the velocity is negative, so the object is moving backwards . Whether it is speeding up or slowing down is decided by the gradient , not by which side of the axis it is on. Below the axis, a negative gradient means the object is speeding up as it moves backwards, and a positive gradient means it is slowing down while still moving backwards.

Why can a velocity-time graph go below the time axis when a speed-time graph cannot?

Velocity is a vector , so it can be negative, and a negative velocity is drawn below the time axis to show motion in the backward direction. Speed is a scalar : it is the size of the velocity and can never be negative, so a speed-time graph never goes below the axis. On a speed-time graph, motion backwards still appears as a positive speed, with the graph coming down to touch the axis at the moment the direction changes.

A journey has a period of uniform acceleration, followed by a period at constant velocity. What shape is each of those sections on a velocity-time graph?

Uniform means constant , so a constant acceleration is a constant gradient and the uniformly accelerating section is a straight line sloping upwards . A section at constant velocity has zero acceleration, so it is a horizontal line. A uniform deceleration would also be a straight line, sloping downwards .

A LevelMathsEdexcelMechanicsFlashcardsKinematicsGraphs in Kinematics

Graphs in Kinematics (Edexcel A Level Maths: Mechanics): Flashcards

Exam code: 9MA0

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FrontDisplacement-Time Graphs
On a displacement-time graph, what does the gradient of the graph tell you?
BackDisplacement-Time Graphs
False.
A horizontal section of a displacement-time graph means the displacement is not changing, so the object is stationary.
Strictly, a stationary object does have a constant velocity, of zero, so the word doing the work in the statement is moving.
A constant non-zero velocity gives a straight sloping line on a displacement-time graph, not a horizontal one; it is on a velocity-time graph that a horizontal line means constant velocity.

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Cards in this collection (19)

On a displacement-time graph, what does the gradient of the graph tell you?
The gradient of a displacement-time graph is the velocity of the object.
A positive gradient means the object is moving forwards, and a negative gradient means it is moving backwards.
The steeper the line, the greater the speed.
True or False?
A horizontal section of a displacement-time graph shows that the object is moving at a constant velocity.
False.
A horizontal section of a displacement-time graph means the displacement is not changing, so the object is stationary.
Strictly, a stationary object does have a constant velocity, of zero, so the word doing the work in the statement is moving.
A constant non-zero velocity gives a straight sloping line on a displacement-time graph, not a horizontal one; it is on a velocity-time graph that a horizontal line means constant velocity.
On a displacement-time graph, what does a straight section tell you, and what does a curved section tell you?
On a displacement-time graph, a straight section shows the object moving at a constant velocity, because the gradient, and so the velocity, is the same all the way along it.
A curved section shows the velocity changing, so the object is accelerating or decelerating. The gradient of a curve is different at every point.
What does it mean if a displacement-time graph touches the time axis?
Where a displacement-time graph touches the time axis the displacement is zero, so the object is at the fixed origin that its displacement is measured from.
This does not mean the object has stopped. An object is stationary only where the graph is horizontal.
Fill in the blanks to complete the two formulas for a journey:
$average speed = \frac{total______travelled}{time taken}$
$average velocity = \frac{______from the starting point}{time taken}$
The completed formulas for a journey are:
$average speed = \frac{total distance travelled}{time taken}$
$average velocity = \frac{displacement from the starting point}{time taken}$
Average speed uses distance, so it can never be negative. Average velocity uses displacement, so it can be positive, negative or zero: an object that finishes where it started has an average velocity of zero, however far it has travelled.
How do you find the total distance travelled from a displacement-time graph?
To find the total distance travelled from a displacement-time graph, add together the size of every rise and every fall, ignoring whether it goes up or down.
So, for example, a graph that rises by 11 m and then falls by 11 m back to where it started shows a total distance travelled of $11 + 11 = 22$ m, even though the displacement at the end is zero.
On a displacement-time graph, what does the value where the graph meets the vertical axis represent?
Where a displacement-time graph meets the vertical axis, the value is the object's initial displacement: how far it is from the fixed origin at time zero.
An object does not have to start at the origin, so this value is not always zero. A graph starting at 100 m, for example, could show someone setting out on a walk from a point 100 m from their house.
On a velocity-time graph, what does the gradient of the graph tell you?
The gradient of a velocity-time graph is the acceleration of the object.
A straight line means the acceleration is constant, and a horizontal line means the acceleration is zero, so the object is moving at a constant velocity.
On a velocity-time graph, what does the area between the graph and the time axis represent?
The area between a velocity-time graph and the time axis gives the change in displacement of the object over that time.
Where the graph is above the axis the object is moving forwards, so that area is travel in the positive direction. Where the graph is below the axis the object is moving backwards.
A velocity-time graph goes both above and below the time axis. Fill in the blanks:
total displacement $=$ (sum of the areas above the axis) $______$ (sum of the areas below the axis)
total distance travelled $=$ (sum of the areas above the axis) $______$ (sum of the areas below the axis)
The completed rules for a velocity-time graph are:
total displacement $=$ (sum of the areas above the axis) $-$ (sum of the areas below the axis)
total distance travelled $=$ (sum of the areas above the axis) $+$ (sum of the areas below the axis)
An area below the axis is travel backwards. It brings the object back towards its starting point, so it reduces the displacement, but it still adds to how far the object has actually moved.
What does it mean if a velocity-time graph touches the time axis?
Where a velocity-time graph touches the time axis the velocity is zero, so the object is instantaneously at rest.
If the graph crosses the axis rather than only touching it, the velocity changes sign, so the object changes direction at that moment.
True or False?
A section of a velocity-time graph that lies below the time axis always shows an object that is slowing down.
False.
Below the time axis on a velocity-time graph the velocity is negative, so the object is moving backwards. Whether it is speeding up or slowing down is decided by the gradient, not by which side of the axis it is on.
Below the axis, a negative gradient means the object is speeding up as it moves backwards, and a positive gradient means it is slowing down while still moving backwards.
How do you calculate the area between a velocity-time graph and the time axis when the graph is made up of straight sections?
To find the area under a velocity-time graph made of straight sections, split the region into triangles, rectangles and trapezia, work out each area separately, and add them.
So, for example, a section falling at a constant rate from $40 {m s}^{- 1}$ to $12 {m s}^{- 1}$ over 20 seconds encloses a trapezium of area $\frac{1}{2} (40 + 12) (20) = 520$ m.
Why can a velocity-time graph go below the time axis when a speed-time graph cannot?
Velocity is a vector, so it can be negative, and a negative velocity is drawn below the time axis to show motion in the backward direction.
Speed is a scalar: it is the size of the velocity and can never be negative, so a speed-time graph never goes below the axis.
On a speed-time graph, motion backwards still appears as a positive speed, with the graph coming down to touch the axis at the moment the direction changes.
A velocity-time graph is drawn with velocity in ${km h}^{- 1}$ on the vertical axis and time in hours on the horizontal axis. What are the units of its gradient, and of the area between the graph and the time axis?
With velocity in ${km h}^{- 1}$ and time in hours, the gradient is an acceleration in ${km h}^{- 2}$ and the area is a displacement in kilometres.
The units always come from the axes: the gradient is (vertical unit) divided by (time unit), and the area is (vertical unit) multiplied by (time unit). This is why both axes must be labelled with their units, and why values taken from a graph drawn in kilometres and hours have to be converted before they are used alongside values in metres and seconds.
A journey has a period of uniform acceleration, followed by a period at constant velocity. What shape is each of those sections on a velocity-time graph?
Uniform means constant, so a constant acceleration is a constant gradient and the uniformly accelerating section is a straight line sloping upwards.
A section at constant velocity has zero acceleration, so it is a horizontal line.
A uniform deceleration would also be a straight line, sloping downwards.
A velocity-time graph is being drawn from a description of a journey. Fill in the blanks:
If the object starts from rest, the line starts at $______$ .
If the object comes to rest, the line meets $______$ .
The completed sentences for a velocity-time graph are:
If the object starts from rest, the line starts at the origin.
If the object comes to rest, the line meets the time axis.
Being at rest means the velocity is zero, and zero velocity is the time axis, so these are the same fact at two different moments of the journey.
True or False?
A section of a velocity-time graph has gradient $- 1.4$ , so the deceleration over that section is $- 1.4 {m s}^{- 2}$ .
False.
The acceleration is $- 1.4 {m s}^{- 2}$ , because on a velocity-time graph the gradient is the acceleration, sign included.
A deceleration is the size of that rate of change, with the direction already carried by the word itself, so it is given as a positive number: the deceleration is $1.4 {m s}^{- 2}$ . A negative deceleration would mean the object was speeding up.
A velocity-time graph is made up of straight sections and the total distance travelled is known, but one of the times on the horizontal axis is unknown. How can you find that unknown time?
The total area under the graph can be written as an expression in the unknown time, then set equal to the known distance and solved, splitting the region into triangles, rectangles and trapezia as usual.
The section containing the unknown time $T$ gives an area in terms of $T$ : a triangle of height 12 and base $T - 28$ , for example, has area $\frac{1}{2} (12) (T - 28) = 6 (T - 28)$ .
Adding all the areas and setting the total equal to the known distance gives an equation in $T$ to solve.

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