Displacement-Time Graphs | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Displacement-Time Graphs

A displacement-time graph records position relative to a chosen origin. Read the axes before reading the line: its height is displacement, while its is velocity.

Reading straight sections

Gradient means rise over run. Here the rise is displacement and the run is time, so gradient = $\Delta s/\Delta t$ = velocity. That is exactly the definition from the last lesson.
A steeper line has a greater speed. A flat line means zero velocity, even when the displacement is large. A falling line means negative velocity, not negative speed.

Read the slope, not the height. Rising: moving forward. Flat: not moving. Falling: moving back toward the start.

Worked example

Warm-up: velocity from a straight section

On an s-t graph, a straight section rises from 0 m to 24 m between t = 0 and t = 8.0 s. Find the velocity.

Gradient: $v = \dfrac{\Delta s}{\Delta t} = \dfrac{24}{8.0} = 3.0 \text{ m s}^{-1}$ .

Answer

v = 3.0 \text{ m s}^{-1}

On an s-t graph, read the slope, never the height. A high flat line means far away and not moving.

Distance-time graph	Displacement-time graph
height = total path length	height = position from the origin
gradient = speed	gradient = velocity
cannot slope downward	may slope downward when motion reverses
cannot have a negative value	may have a negative value

A curved graph: use a tangent

A curved s-t graph means the velocity is changing: the gradient is different at every point.
To find the velocity at one instant, draw a at that point and take the tangent's gradient. That gradient is the instantaneous velocity from lesson 2.1.

On a curve, draw a tangent at the point you care about. Its gradient is the velocity at that instant: rise ÷ run = 16 m ÷ 4 s = 4 m s⁻¹.

Worked example

A harder step: instantaneous velocity from a tangent

On the curve above, a tangent is drawn at t = 4.0 s. It passes through (2.0 s, 0 m) and (6.0 s, 16 m). Find the velocity at t = 4.0 s.

Rise and run of the tangent: $\Delta s = 16 - 0 = 16 \text{ m}$ , $\Delta t = 6.0 - 2.0 = 4.0 \text{ s}$ .
$v = \dfrac{16}{4.0} = 4.0 \text{ m s}^{-1}$ .

Answer

v = 4.0 \text{ m s}^{-1}

When two displacement-time lines cross, the objects have the same displacement at that time. They are at the same position only if both graphs use the same origin and direction.

Common mistake

Use two points that are far apart on the tangent line, not two points on the curve. Points on the curve give an average velocity, not the velocity at that instant.

Your turnCAIE-style · original · 3 marks

A curved displacement-time graph becomes progressively steeper. State what happens to the velocity, describe how to find the velocity at 4.0 s, and state what a horizontal tangent would mean.

Show worked answer

The velocity increases because the gradient increases.
Draw a tangent at 4.0 s and calculate its gradient.
A horizontal tangent has zero gradient, so the object is instantaneously at rest.

Displacement-Time Graphs

A displacement-time graph records position relative to a chosen origin. Read the axes before reading the line: its height is displacement, while its is velocity.

Reading straight sections

Gradient means rise over run. Here the rise is displacement and the run is time, so gradient = $\Delta s/\Delta t$ = velocity. That is exactly the definition from the last lesson.
A steeper line has a greater speed. A flat line means zero velocity, even when the displacement is large. A falling line means negative velocity, not negative speed.

Read the slope, not the height. Rising: moving forward. Flat: not moving. Falling: moving back toward the start.

Worked example

Warm-up: velocity from a straight section

On an s-t graph, a straight section rises from 0 m to 24 m between t = 0 and t = 8.0 s. Find the velocity.

Gradient: $v = \dfrac{\Delta s}{\Delta t} = \dfrac{24}{8.0} = 3.0 \text{ m s}^{-1}$ .

Answer

v = 3.0 \text{ m s}^{-1}

On an s-t graph, read the slope, never the height. A high flat line means far away and not moving.

Distance-time graph	Displacement-time graph
height = total path length	height = position from the origin
gradient = speed	gradient = velocity
cannot slope downward	may slope downward when motion reverses
cannot have a negative value	may have a negative value

A curved graph: use a tangent

A curved s-t graph means the velocity is changing: the gradient is different at every point.
To find the velocity at one instant, draw a at that point and take the tangent's gradient. That gradient is the instantaneous velocity from lesson 2.1.

On a curve, draw a tangent at the point you care about. Its gradient is the velocity at that instant: rise ÷ run = 16 m ÷ 4 s = 4 m s⁻¹.

Worked example

A harder step: instantaneous velocity from a tangent

On the curve above, a tangent is drawn at t = 4.0 s. It passes through (2.0 s, 0 m) and (6.0 s, 16 m). Find the velocity at t = 4.0 s.

Rise and run of the tangent: $\Delta s = 16 - 0 = 16 \text{ m}$ , $\Delta t = 6.0 - 2.0 = 4.0 \text{ s}$ .
$v = \dfrac{16}{4.0} = 4.0 \text{ m s}^{-1}$ .

Answer

v = 4.0 \text{ m s}^{-1}

When two displacement-time lines cross, the objects have the same displacement at that time. They are at the same position only if both graphs use the same origin and direction.

Common mistake

Use two points that are far apart on the tangent line, not two points on the curve. Points on the curve give an average velocity, not the velocity at that instant.

Your turnCAIE-style · original · 3 marks

A curved displacement-time graph becomes progressively steeper. State what happens to the velocity, describe how to find the velocity at 4.0 s, and state what a horizontal tangent would mean.

Show worked answer

The velocity increases because the gradient increases.
Draw a tangent at 4.0 s and calculate its gradient.
A horizontal tangent has zero gradient, so the object is instantaneously at rest.