Motion · MYP 4 Physics

Coordinates and position

Start with one dimension
We will need some space to describe relative placement of objects. Let’s start with one dimension, only left and right. A straight route tells us where motion can happen.
Choose an origin
We will need to start somewhere. Let’s pick one point and assign it x = 0. This reference point is called the origin.
Set the scale
Next move is the scale. Space the ticks evenly and label at least one.
Place two objects
Here each gap between neighboring ticks represents 1 meter.
Read signed positions
Count intervals from zero and assign a sign. Three intervals left is −3 m. Four intervals right is +4 m.
Find the separation
The tree and bus stop are 7 m apart. Subtract the positions in either order. The bars keep the size and drop the sign. This is the definition of the distance between two points: the absolute value of the difference of their positions.
$| x_{B} - x_{A} |$ $= | 4 - (- 3) |$ $= 7 m$ $| x_{A} - x_{B} |$ $= | - 3 - 4 |$ $= 7 m$
Move the origin
The distance between two points stays the same wherever you put the origin.

Describing a journey

Add a traveller and clock

Use the axis to record where a journey starts and ends. Use a clock to record when those two events happen.

Time: 7 s
Position: +4 m

Δx = +7 m · Δt = 7 s

Equation

Δx

= x_{f} - x_{0}

Δt

= t_{f} - t_{i}

Set the route. A and B are stationary markers. The traveller will go from A at −3 m to B at +4 m.
Place the traveller at A. Record the starting position x₀ = −3 m.
Set the clock to zero. Choose departure from A as t = 0 s. The traveller and clock begin at the same event.
Travel to B. As the clock runs, the traveller moves right. The clock reads 7 s on arrival at B.
Mark the endpoints. Keep the start and finish visible so you can compare their coordinates.
Calculate the changes. Displacement is final position minus starting position: (+4 m) − (−3 m) = +7 m. Duration is the elapsed time: 7 s − 0 s = 7 s.

The traveller ends at +4 m after moving +7 m from A in 7 s. Final position, displacement and duration answer different questions.

Describing a journey

Constant speed and velocity

Take a snapshot at the same time interval again and again. If the traveller’s signed displacements are equal in equal time intervals, the motion is uniform.

Time: 4 s
Position: +5 m

Equation

At constant velocity, v

= \frac{Δx}{Δt}

Follow equal time intervals. Start at x = −3 m when t = 0 s. The traveller moves 2 m to the right in the first second.
Mark the first second. The first two snapshots are at −3 m and −1 m. Their times are one second apart.
Compare the gaps. Keep taking snapshots every second. The position increases by 2 m each time, so every one-second displacement is +2 m.
Find the speed. Speed is the rate at which distance is travelled. The traveller covers 8 m in 4 s at a steady rate, so the speed is 2 m/s.
Include direction in the velocity. Velocity is the rate of change of position. The traveller’s displacement is +8 m in 4 s, so the constant velocity is +2 m/s. The plus sign means rightward.

Each second, the traveller moves 2 m to the right. The speed is 2 m/s and the velocity is +2 m/s.

Describing a journey

Turn around

When a traveller reverses, part of the route is covered twice. Distance counts the length of each leg. Displacement measures the change between the starting and ending positions.

Time: 5 s
Position: +6 m

Distance = 8 m + 2 m = 10 m

Equation

d_{total}

= d_{1} + d_{2} + \dots

Δx

= x_{f} - x_{0}

Start at zero. Follow the traveller from 0 m to +8 m and then partway back. Watch how the start, turn and finish are used.
Travel to +8 m. The first leg covers 8 m. Highlight that stretch of the path.
Turn and travel back to +6 m. The second leg is 2 m long. It covers part of the same route again.
Add the lengths. The total distance is 8 m + 2 m = 10 m. Each leg contributes its non-negative length.
Draw one displacement arrow. Draw the arrow directly from 0 m to +6 m. Δx = (+6 m) − 0 m = +6 m, so it points right.

From 0 m to +8 m and back to +6 m, distance is 10 m and displacement is +6 m.

Worked example

Distance and displacement after a turn

A traveller goes from −2 m to +5 m, then turns and ends at +1 m. Find the distance travelled and displacement.

Show solution

Draw a number line with right positive. Mark start (−2 m), turn (+5 m) and finish (+1 m). Follow −2 → +5 → +1.
Distance adds both leg lengths. The first is |(+5 m) − (−2 m)| = 7 m. The second is |(+1 m) − (+5 m)| = 4 m. The total is 7 m + 4 m = 11 m.
Displacement uses the start and finish. Δx = (+1 m) − (−2 m) = +3 m, so the finish is 3 m right of the start.
Check with the signed legs: +7 m + (−4 m) = +3 m. The return adds 4 m to the distance and reduces the net rightward change by 4 m.

Describing a journey

Average speed and average velocity

A traveller can change speed or turn around during a journey. To average the whole journey, divide the appropriate total by the total elapsed time.

Time: 7 s
Position: +10 m

Average speed ≈ 3.14 m/s · Average velocity ≈ +1.43 m/s

Equation

Average speed

= \frac{d_{total}}{{Δt}_{total}}

v_{avg}

= \frac{{Δx}_{total}}{{Δt}_{total}}

Begin the two-leg journey. Start at 0 m. Move right at +4 m/s for 4 s, then turn and move left at −2 m/s for 3 s.
Time the outward leg. At 4 s the traveller is 16 m to the right of the start, since 4 m/s × 4 s = 16 m.
Finish and add the totals. The return leg covers 2 m/s × 3 s = 6 m to the left. Total distance is 16 m + 6 m = 22 m. Total elapsed time is 4 s + 3 s = 7 s.
Calculate average speed. Divide the full path length by the full time: 22 m ÷ 7 s ≈ 3.14 m/s. Average speed has no direction.
Calculate average velocity. The net change from 0 m to +10 m is +10 m. Divide by the same 7 s to get an average velocity of about +1.43 m/s.
Account for unequal times. The simple mean (4 + 2) ÷ 2 = 3 m/s would give both legs equal weight. The traveller spends 4 s at the higher speed and 3 s at the lower speed, so the whole-journey average is closer to 4 m/s.

This traveller covers 22 m of path and changes position by +10 m in 7 s. The average speed is about 3.14 m/s and the average velocity is about +1.43 m/s.

Worked example

Two averages for a two-leg journey

A traveller goes at +4 m/s for 4 s, then at −2 m/s for 3 s. Find the total distance, displacement, average speed and average velocity.

Show solution

Choose 0 m as the start and right as positive. The route runs start (0 m) → turn (+16 m at 4 s) → finish (+10 m at 7 s).
Each leg has constant velocity. Multiply signed velocity by duration: (+4 m/s)(4 s) = +16 m and (−2 m/s)(3 s) = −6 m.
Add lengths for distance: 16 m + 6 m = 22 m. Add signed changes for displacement: +16 m − 6 m = +10 m.
Divide by 4 s + 3 s = 7 s. Average speed = 22 m ÷ 7 s ≈ 3.14 m/s. Average velocity = +10 m ÷ 7 s ≈ +1.43 m/s.
Check that the average speed falls between the leg speeds of 2 and 4 m/s and is closer to 4 m/s because the traveller spent longer at that speed. The positive average velocity agrees with the finish being right of the start.

Describing a journey

What an average speed tells you

Measure distance in metres and elapsed time in seconds. Their ratio gives average speed in metres per second.

Time: 30 s
Position: +600 m

Average across this stretch: 20 m/s

Equation

Average speed

= \frac{d_{total}}{{Δt}_{total}}

Measure ten metres. The traveller starts at 0 m and the finish is at +10 m. Five intervals of 2 m give a path length of 10 m.
Measure two seconds. Set the clock to 0 s as the traveller leaves the start. It reads 2 s at the finish.
Calculate the rate. Divide 10 m by 2 s to get an average speed of 5 m/s.
Measure a road journey. The camera at entry records when the vehicle passes. A second camera records the exit time. Across 600 m in 30 s, the average speed is 20 m/s.
Ask what happened at one instant. Instantaneous velocity describes how fast and in which direction an object moves at a particular instant. The two camera records give the average speed across the stretch. A vehicle could slow down and speed up between them.

An average-speed camera uses the distance and duration between two points. Its result describes the whole stretch.

Worked example

What two cameras can measure

A vehicle covers a measured 600 m stretch between two average-speed cameras in 30 s. Find the average speed in m/s. What can you infer about its speed at any particular point?

Show solution

Label the route entry (0 m, 0 s) → exit (+600 m, 30 s). The measured path is 600 m and the elapsed time is 30 s.
Use the distance and time for the whole stretch: average speed = 600 m ÷ 30 s = 20 m/s.
Check the rate: at a steady 20 m/s, travelling for 30 s would cover 600 m.
The two times give an average for the stretch. Many different patterns of speeding up and slowing down could give the same average, so an individual instant would require another measurement.

Uniform motion

From movement to x(t)

Record the traveller’s position as the clock runs, then use those records to draw a position–time graph.

Time: 4 s
Position: +10 m
Velocity: +3 m/s

Position–time

Motion readings
t (s)	x (m)
0	−2
1	+1
2	+4
3	+7
4	+10

Mark the starting position. The traveller starts 2 m to the left of the origin, at x₀ = −2 m.
Start the clock. Set t = 0 s at the starting event. The first row records t = 0 s and x = −2 m.
Move for one second. At +3 m/s the traveller moves 3 m right each second. After 1 s, the position is −2 m + 3 m = +1 m.
Record each second. Keep the velocity at +3 m/s. At 2, 3 and 4 s, the traveller reaches +4, +7 and +10 m. Each new row adds the same displacement.
Transfer the rows to axes. Use time horizontally and position vertically. The first point is (0 s, −2 m). The last is (4 s, +10 m).
Join the points. The velocity also stays constant between whole seconds, so the points lie on one straight line. At 2.5 s the line gives x = +5.5 m.

Plot each pair as (time, position). A traveller moving at constant velocity makes a straight x(t) line. The height at t = 0 gives the starting position.

Uniform motion

Interpreting the x(t) gradient

Use the change in position and the time it takes to read velocity from a position–time graph.

Time: 4 s
Position: +10 m
Velocity: +3 m/s

Position–time

Equation

v

= \frac{Δx}{Δt}

x

= x_{0} + vt (constant velocity)

Read the line. This is the line from the previous scene. Its height at t = 0 s is −2 m, the starting coordinate.
Choose two points. Read (1 s, +1 m) and (3 s, +7 m). Use points you can read clearly and take each change in time order.
Draw the gradient triangle. Move 2 s to the right and then +6 m up. These horizontal and vertical changes are Δt and Δx.
Divide rise by run. The gradient is (7 − 1) m / (3 − 1) s = +3 m/s. The positive sign means rightward velocity.
Verify on the route. As the clock advances from 1 s to 3 s, the traveller moves from +1 m to +7 m. The +6 m displacement agrees with the graph.
Predict a later position. At t = 4 s, x = x₀ + vt = −2 m + (3 m/s)(4 s) = +10 m. The first term sets the starting coordinate.

The vertical coordinate gives position. The gradient gives velocity. For constant velocity, add vt to the starting position x₀ to predict x.

Worked example

Predict a position

An object starts at x₀ = −2 m and moves at +3 m/s for 4 s. What is its final position?

Show solution

On the axis, label the start at −2 m and point the velocity arrow to the right. On x(t), the line starts at (0 s, −2 m). We want its height at 4 s.
The velocity is constant, so Δx = vt = (+3 m/s)(4 s) = +12 m.
Add the displacement to the starting position. x = x₀ + vt = −2 m + 12 m = +10 m.
Check the direction and scale. Four rightward steps of 3 m from −2 m finish at +10 m, which agrees with the graph.

Uniform motion

Four shapes of uniform motion

Compare lines on identical axes to read direction and speed from their gradients.

Rightward

Time: 3 s
Position: +6 m
Velocity: +2 m/s

Position–time

Leftward

Time: 3 s
Position: −6 m
Velocity: −2 m/s

Position–time

Stationary

Time: 3 s
Position: 0 m
Velocity: 0 m/s

Position–time

Faster rightward

Time: 3 s
Position: +9 m
Velocity: +3 m/s

Position–time

Equation

Velocity

= gradient of x (t)

Speed

= | velocity |

A rising line. Position rises from 0 m to +6 m in 3 s. The gradient is +2 m/s and the traveller moves right.
A falling line. Position falls from 0 m to −6 m in 3 s. The gradient is −2 m/s, so the traveller moves left with the same speed of 2 m/s.
A horizontal line. Position stays at 0 m throughout the 3 s interval. Its gradient is zero, so the traveller is stationary.
A steeper line. Position rises to +9 m in the same 3 s. The gradient of +3 m/s is steeper than the first line, and the traveller covers more metres each second.

On x(t), rising means moving right, falling means moving left and horizontal means stationary. On identical axes, a steeper line in either direction means a larger speed. The graph records position over time along a straight route.

Uniform motion

Two objects and a meeting

Follow A and B on the same route and clock. Their graphs meet exactly when the two objects share a position.

Time: 5 s
Position A: +4 m
Position B: +4 m
Velocity A: +2 m/s
Velocity B: −1 m/s

Position–time

A · solid lineB · dashed line

Equation

x_{A}

= x_{A0} + v_{A} t

x_{B}

= x_{B0} + v_{B} t

at a meeting, x_{A}

= x_{B}

Mark both starting positions. At t = 0 s, A is at −6 m and B is at +9 m. Both coordinates use the same origin and positive direction.
Start together. A moves right at +2 m/s and B moves left at −1 m/s. After 2 s, A is at −2 m and B is at +7 m.
Plot both positions on one graph. A’s line has gradient +2 m/s and B’s has gradient −1 m/s. A vertical read at any time gives both positions at that instant.
Reach the same coordinate. At 5 s, A has moved 10 m right and B has moved 5 m left. Both are at +4 m.
Read the intersection. The full lines cross at (5 s, +4 m). They then continue with their original velocities.
Verify with both equations. At 5 s, x_A = −6 + 2(5) = +4 m and x_B = 9 − 5 = +4 m. Both equations confirm the meeting point.

The intersection gives the common time and position. Each line retains its own gradient, which gives that object’s velocity.

Worked example

Find a meeting time and place

A starts at −6 m and moves at +2 m/s. B starts at +9 m and moves at −1 m/s. They start together. When and where do they meet?

Show solution

Mark A at −6 m with a rightward arrow and B at +9 m with a leftward arrow. On x(t), A’s line rises from −6 m and B’s falls from +9 m.
Each velocity is constant. With t in seconds and coordinates in metres, x_A = −6 + 2t and x_B = 9 − t.
At the intersection, they are at the same coordinate at the same time. Set −6 + 2t = 9 − t, giving 3t = 15 and t = 5 s.
Substitute 5 s into both equations. A reaches −6 + 2(5) = +4 m and B reaches 9 − 5 = +4 m.
Check that the 15 m starting separation closes at 2 + 1 = 3 m each second. It takes 5 s, and the meeting coordinate lies between the starting positions.

Velocity–time graphs

Same motion, different graph

Use the same constant-velocity journey to connect the gradient of x(t) with the height and area of v(t).

Time: 4 s
Position: +10 m
Velocity: +3 m/s

Position–time

Velocity–time

Equation

Signed area under v (t)

= Δx

for constant v, Δx

= vΔt

Replay the journey. Start at −2 m and move right at +3 m/s. In 4 s the traveller reaches +10 m.
Show the position graph. The x(t) line rises by 3 m each second. Its gradient stays at +3 m/s.
Plot velocity on the same clock. Plot +3 m/s at every instant. This produces a horizontal line above v = 0 while the traveller keeps moving right.
Shade the rectangle. Shade between the velocity line and the time axis from 0 s to 4 s. This entire area is positive because it is above v = 0.
Multiply base by height. The rectangle has base 4 s and signed height +3 m/s. Its signed area is (4 s)(+3 m/s) = +12 m. Seconds cancel, leaving metres.
Check against the route. Subtract the starting position from the final position. Δx = +10 m − (−2 m) = +12 m, the same displacement given by the velocity graph.

On x(t), the height is position and the gradient is velocity. On v(t), the height is velocity and the signed area gives displacement. A horizontal v(t) line shows constant velocity. Only a line at v = 0 shows a stop.

Velocity–time graphs

A multi-stage velocity graph

Split a journey into constant-velocity stages. Track signed displacement and total distance using the same graph.

Time: 6 s
Position: +8 m
Velocity: −2 m/s

Position–time

Velocity–time

Equation

Δx

= sum of signed areas

d

= sum of area magnitudes

Go right. The traveller leaves x = 0 m and moves at +4 m/s for 3 s, reaching +12 m. This first segment has constant velocity.
Pause. From 3 s until 4 s the traveller stays at +12 m, so v = 0 m/s.
Go left. From 4 s to 6 s, the traveller moves left at −2 m/s, returning 4 m to reach +8 m.
Plot three velocity stages. Read +4 m/s from 0 to 3 s, 0 m/s from 3 to 4 s, and −2 m/s from 4 to 6 s. The model switches velocity at the stage boundaries.
Shade the signed areas. The rightward rectangle gives (+4)(3) = +12 m. The stop lies on the time axis. The leftward rectangle gives (−2)(2) = −4 m.
Add in two ways. Add signed areas for Δx = +12 − 4 = +8 m. Add their magnitudes for d = 12 + 4 = 16 m.
Build the position graph. On x(t), the slopes are +4, 0 and −2 m/s. The three segments join at (3 s, +12 m) and (4 s, +12 m), then the line ends at (6 s, +8 m).

Above-axis area counts positive for displacement, below-axis area counts negative and a stop contributes zero. For distance, add the lengths from all stages. On x(t), join the positions with a rising, flat and then falling segment.

Worked example

Read displacement and distance

A velocity–time graph shows +4 m/s for 3 s, then immediately −2 m/s for 2 s. Find the displacement and total distance.

Show solution

Velocity–time

On v(t), draw a rectangle above zero from 0 to 3 s at +4 m/s and a rectangle below zero from 3 to 5 s at −2 m/s. This version of the journey omits the stop.
Each stage has constant velocity, so signed area is vΔt. The first stage gives (+4 m/s)(3 s) = +12 m. The second gives (−2 m/s)(2 s) = −4 m.
Add the signed areas for displacement. Δx = +12 m − 4 m = +8 m. The final coordinate is 8 m to the right of the start.
Add the area magnitudes for distance. d = 12 m + 4 m = 16 m. This includes all 4 m of the return leg.
Check that 16 m is greater than the displacement magnitude of 8 m because the traveller reverses direction.

Acceleration

Motion need not be uniform

Take snapshots at equal time intervals to see how the motion of a cart changes after it starts from rest.

Time: 4 s
Position: +16 m
Velocity: +8 m/s
Acceleration: +2 m/s²

Motion readings
t (s)	x (m)
0	0
1	+1
2	+4
3	+9
4	+16

Equation

Acceleration

= \frac{change in velocity}{elapsed time}

Release the cart from rest. The cart begins at the origin. Its initial velocity is u = 0 m/s at t = 0 s.
Mark the first second. After one second the cart is at +1 m. Use the same one-second interval for every snapshot.
Compare the gaps. The cart reaches +4 m at 2 s and +9 m at 3 s. Successive one-second gaps grow from 1 m to 3 m to 5 m.
Track the velocity arrow. The velocity arrow points right and grows longer as the cart moves. At 4 s the cart is at +16 m and its velocity is +8 m/s.
Name the change. Acceleration measures how much velocity changes per unit time. Here the velocity increases by +2 m/s each second, so a = +2 m/s².

At equal time intervals, larger gaps in the same direction indicate greater average speed over those intervals. In this model the cart’s rightward velocity increases by 2 m/s every second.

Acceleration

Acceleration from v(t)

Read the gradient of a velocity–time graph to find the acceleration, then represent it on a third graph.

Time: 2 s
Position: +8 m
Velocity: +2 m/s
Acceleration: −2 m/s²

Velocity–time

Acceleration–time

Motion readings
t (s)	v (m/s)
0	+6
1	+4
2	+2

Equation

Δv

= v - u

a

= \frac{Δv}{Δt}

signed area under a (t)

= Δv

Plot the velocity readings. Velocity starts at +6 m/s. The model gives +4 m/s at 1 s and +2 m/s at 2 s. Plot t in seconds and v in metres per second.
Join the points. Velocity changes by −2 m/s each second. With constant acceleration, the points join in one straight line.
Draw the gradient triangle. From t = 0 s to t = 2 s, the run is Δt = 2 s and the rise is Δv = +2 − (+6) = −4 m/s.
Calculate acceleration. The gradient is a = (−4 m/s)/(2 s) = −2 m/s². The velocity is still positive, so the traveller is moving right.
Read the unit. Metres per second per second is written m/s². Here the signed velocity decreases by 2 m/s every second.
Add the acceleration graph. The a(t) line stays at −2 m/s². Its signed area over 2 s is (−2 m/s²)(2 s) = −4 m/s, the same change in velocity as before.

The gradient of v(t) gives acceleration. Equal changes in velocity over equal times give a straight v(t) line and a flat a(t) line. Signed area under a(t) gives the change in velocity.

Worked example

Find a negative acceleration

Velocity changes from +6 m/s to +2 m/s in 2 s. What is the average acceleration? How does the object move in the straight-line model?

Show solution

Label (0 s, +6 m/s) and (2 s, +2 m/s) on v(t). Both endpoints lie above v = 0, and a straight segment between them shows constant acceleration.
Subtract initial velocity from final velocity. Δv = (+2 m/s) − (+6 m/s) = −4 m/s.
Divide by the elapsed time. a = Δv/Δt = (−4 m/s)/(2 s) = −2 m/s². For the straight segment this is the acceleration throughout.
Check that the line falls by 2 m/s each second while staying above zero. The object moves right throughout and its speed decreases.

Acceleration

Three signed acceleration cases

Use velocity to tell which way an object moves. Use velocity together with acceleration to tell what happens to its speed.

Rightward and speeding up

Time: 3 s
Position: +15 m
Velocity: +8 m/s
Acceleration: +2 m/s²

Velocity–time

Acceleration–time

Rightward and slowing down

Time: 3 s
Position: +15 m
Velocity: +2 m/s
Acceleration: −2 m/s²

Velocity–time

Acceleration–time

Leftward and speeding up

Time: 3 s
Position: −15 m
Velocity: −8 m/s
Acceleration: −2 m/s²

Velocity–time

Acceleration–time

Reversing direction

Time: 3 s
Position: −3 m
Velocity: −4 m/s
Acceleration: −2 m/s²

Velocity–time

Acceleration–time

Equation

Speed

= | v |

same signs of v and a

\to speeding up

opposite signs

\to slowing down

Positive v, positive a. Velocity rises from +2 to +8 m/s with a = +2 m/s². Both arrows point right. The object moves right and speeds up.
Positive v, negative a. Velocity falls from +8 to +2 m/s with a = −2 m/s². The velocity arrow points right and the acceleration arrow points left. The object moves right and slows down.
Negative v, negative a. Velocity changes from −2 to −8 m/s with a = −2 m/s². Both arrows point left. Its speed rises from 2 to 8 m/s while it moves left.
Crossing zero and reversing. With a = −2 m/s², velocity falls from +2 m/s to zero at 1 s, then reaches −4 m/s at 3 s. The object slows while moving right, stops momentarily and speeds up to the left.

The signs of v and a together determine whether speed increases or decreases. With a = −2 m/s², a positive velocity can decrease towards zero and a negative velocity can become more negative. A crossing of v = 0 marks a reversal.

Constant acceleration

From rest: triangle and half-final

A Smart Cart starts from rest on a ramp. Use the area under its velocity graph to explain the factor ½.

Time: 4 s
Position: +4 m
Velocity: +2 m/s
Acceleration: +0.5 m/s²

Velocity–time

Equation

From rest with constant a : v

= at

v_{avg}

= \frac{v}{2}

Δx

= \frac{1}{2} vt

= \frac{1}{2} a t^{2}

Start from rest. Release the Smart Cart at x₀ = 0 m with u = 0 m/s. Right is downhill in this example.
Build the velocity line. With a constant acceleration of +0.50 m/s², the cart reaches +2.0 m/s at 4.0 s. The v(t) line starts at the origin and rises steadily.
Shade the triangular area. The displacement is the signed area between the line and v = 0. The triangle has base 4.0 s and height +2.0 m/s.
Compare with a rectangle. A rectangle 4.0 s wide and +1.0 m/s high has the same area. Its height is half the final velocity. The area above the midline fills the missing area below it.
Find average velocity and displacement. The equivalent constant velocity is v_avg = v/2 = +1.0 m/s. The displacement is Δx = ½vt = ½(+2.0 m/s)(4.0 s) = +4.0 m.
Connect the two formulas. From rest, v = at. Substituting into ½vt gives Δx = ½at² = ½(+0.50 m/s²)(4.0 s)² = +4.0 m. The shortcut requires both constant a and u = 0.

With u = 0 and constant acceleration, the v(t) area is a triangle. A rectangle with the same base has the same area when its height is half the final velocity.

Constant acceleration

Non-zero initial velocity: trapezoid

Give the object a head start in velocity. The area from its initial motion adds a rectangle to the acceleration triangle.

Time: 4 s
Position: +12 m
Velocity: +4 m/s
Acceleration: +0.5 m/s²

Velocity–time

Equation

For constant a : v

= u + at

Δx

= ut + \frac{1}{2} a t^{2}

v_{avg}

= \frac{(u + v)}{2}

Begin with rightward velocity. At t = 0 s the object is already moving right at u = +2.0 m/s. Its v(t) line begins at +2.0 m/s.
Add constant acceleration. At +0.50 m/s², velocity rises by 0.50 m/s each second. After 4.0 s, v = u + at = +4.0 m/s.
Shade the trapezoid. The area now includes the velocity the object had at the start. Shade from v = 0 up to the line, between 0 s and 4.0 s.
Split it into two areas. The rectangle has height u and area ut = +8.0 m. The triangle has height v − u = at, so its area is ½t(at) = +4.0 m.
Add the areas. Δx = ut + ½at² = +8.0 m + 4.0 m = +12 m. The object began at the origin, so it finishes at x = +12 m.
Find the mean height. Because the velocity changes steadily, the mean graph height is v_avg = (u + v)/2 = +3.0 m/s. The equal-area rectangle gives (+3.0 m/s)(4.0 s) = +12 m.

With constant acceleration, the signed mean height of v(t) is (u + v)/2. The signed area equals Δx = [(u + v)/2]t = ut + ½at². If the object reverses direction, this gives displacement and signed average velocity. Add area magnitudes separately for distance.

Worked example

Predict motion with non-zero u

An object starts at u = +2.0 m/s and accelerates at +0.50 m/s² for 4.0 s. Find the final velocity, displacement and average velocity.

Show solution

Start the v(t) line at (0 s, +2.0 m/s). It rises steadily because the acceleration is constant. The 4.0 s interval is the trapezoid’s width.
Use v = u + at = +2.0 m/s + (+0.50 m/s²)(4.0 s) = +4.0 m/s.
For constant acceleration, Δx = ut + ½at² = (+2.0 m/s)(4.0 s) + ½(+0.50 m/s²)(4.0 s)² = +8.0 m + 4.0 m = +12 m.
The mean height is v_avg = (+2.0 + 4.0)/2 = +3.0 m/s. It lies between the initial and final velocities.
Check that (+3.0 m/s)(4.0 s) = +12 m. This agrees with the rectangle-plus-triangle area.

Constant acceleration

Check a motion prediction

Use a Smart Cart timing on a ramp to estimate acceleration, then predict what the cart will do over twice the distance.

Observed interval

Time: 2.2 s
Position: +0.6 m
Velocity: +0.545 m/s
Acceleration: +0.248 m/s²

Velocity–time

Motion data
Timed run	Value
Displacement	+0.60 m
Elapsed time	2.20 s

Predicted interval

Time: 3.098 s
Position: +1.2 m
Velocity: +0.775 m/s
Acceleration: +0.25 m/s²

Velocity–time

Motion data
Quantity	0.60 m timed run	1.20 m prediction
Displacement	+0.60 m measured	+1.20 m target
Time	2.20 s measured	3.10 s predicted
Final velocity	+0.545 m/s inferred	+0.775 m/s predicted
Acceleration	+0.248 m/s² inferred	+0.25 m/s² assumed

Equation

From rest with constant a : v_{avg}

= \frac{Δx}{t}

v

= 2 v_{avg}

t

= \sqrt{\frac{2 Δx}{a}}

Read the observed run. The cart starts from rest and passes the +0.60 m mark after 2.20 s. The stopwatch measures time and the marked ramp gives displacement.
Calculate average velocity. Divide the measured displacement by the elapsed time. v_avg = (+0.60 m)/(2.20 s) ≈ +0.273 m/s.
Infer final velocity and acceleration. Assume uniform acceleration. From rest the final velocity is twice the average, about +0.545 m/s. Divide the velocity change by 2.20 s to find a ≈ +0.248 m/s².
Set the new target. Release the cart from rest again. Use the rounded model value a = +0.25 m/s² and set the finish at +1.20 m on the same ramp.
Predict the longer run. Solve Δx = ½at² for time to get 3.10 s. Then v = at ≈ +0.775 m/s and v_avg = v/2 ≈ +0.387 m/s.
Compare measurement and prediction. The shorter run supplies measured time and displacement. Its graph is the fitted model. The longer run is a prediction that could be checked with a new timing under the same conditions.

The 0.60 m and 2.20 s values come from the timed run. The final velocity and acceleration are inferred using a model. The 1.20 m results predict a run from rest on the same ramp with the same approximately constant acceleration.

Worked example

Estimate acceleration from a timed run

A Smart Cart released from rest covers +0.60 m in 2.20 s on a ramp. Assuming constant acceleration, find its average velocity, final velocity and acceleration.

Show solution

Mark the cart at x₀ = 0 m and the finish at +0.60 m. Draw v(t) starting from (0 s, 0 m/s) and rising to the unknown final velocity at 2.20 s.
The displacement is positive downhill. Use the measured displacement and time to find v_avg = +0.60 m / 2.20 s ≈ +0.273 m/s.
The model assumes constant acceleration and u = 0. The average velocity is half the final velocity, so v = 2(0.60/2.20) ≈ +0.545 m/s.
Use the unrounded velocity to estimate a = (v − u)/t ≈ (+0.545 m/s)/(2.20 s) ≈ +0.248 m/s².
Check that the average lies between 0 and +0.545 m/s and that the area ½vt gives approximately +0.60 m.

Worked example

Predict the motion over 1.20 m

Assume the same cart and ramp give a constant acceleration of +0.25 m/s² from rest. Predict the time, final velocity and average velocity over +1.20 m.

Show solution

Label the route from 0 to +1.20 m and sketch v(t) starting at zero with gradient +0.25 m/s². The target is the time at which the triangle’s area reaches +1.20 m.
With u = 0 and constant a, Δx = ½at². Use the positive elapsed time, t = √(2Δx/a) = √[2(1.20 m)/(0.25 m/s²)] ≈ 3.10 s.
Use v = at ≈ (+0.25 m/s²)(3.10 s) ≈ +0.775 m/s. Half the unrounded final velocity gives v_avg ≈ +0.387 m/s.
Check by dividing displacement by duration. (+1.20 m)/(3.10 s) ≈ +0.387 m/s. The longer run takes more time and reaches a higher final velocity.

Applications and evidence

Choose the right model

Identify what stays constant in each journey before deciding which equation or graph area to use.

Constant velocity

Time: 3 s
Position: +5 m
Velocity: +2 m/s

Position–time

Velocity–time

Piecewise velocity

Time: 5 s
Position: +3 m
Velocity: −1 m/s

Position–time

Velocity–time

Constant acceleration

Time: 3 s
Position: +12 m
Velocity: +7 m/s
Acceleration: +2 m/s²

Position–time

Velocity–time

Identify constant velocity. A traveller starts at −1 m and moves at +2 m/s for 3 s. The x(t) line has a constant gradient and the v(t) line is horizontal.
Use one uniform-motion equation. The velocity is constant for the full interval. Use x = x₀ + vt = −1 m + (+2 m/s)(3 s) = +5 m.
Identify separate velocity stages. A traveller goes right at +3 m/s for 2 s, then left at −1 m/s for 3 s. Each stage has its own constant gradient on x(t) and its own level on v(t).
Combine the areas stage by stage. Use vΔt separately. The signed areas give +6 m and −3 m, so Δx = +3 m. For distance, add 6 m and 3 m to get 9 m.
Identify constant acceleration. An object starts at +1 m/s and accelerates at +2 m/s² for 3 s. The v(t) line has a constant gradient, and the x(t) graph curves upward as the velocity increases.
Include the initial velocity. Using ½at² alone gives 9 m, the triangular part of the area. This object starts at u = +1 m/s. Include ut = +3 m to get Δx = +12 m. Check by multiplying the mean velocity of +4 m/s by 3 s.

Use x = x₀ + vt for constant velocity. When velocity is constant within stages, apply vΔt to each stage and combine signed displacements. Use x = x₀ + ut + ½at² when acceleration is constant. A falling ball released from rest can also fit a constant-acceleration model over a short fall when air resistance has little effect.

Applications and evidence

From evidence to conclusion

Process a small set of measurements, then use a velocity graph to judge how well a constant-acceleration model fits.

Velocity–time

Motion data
Time window (s)	Δx (m)	Midpoint (s)	Velocity estimate (m/s)
0.25–0.75	+0.055	0.50	+0.11
0.75–1.25	+0.125	1.00	+0.25
1.25–1.75	+0.190	1.50	+0.38
1.75–2.25	+0.245	2.00	+0.49

Equation

Mean time

= \frac{(t_{1} + t_{2} + \dots + t ₙ)}{n}

acceleration

= gradient of v (t)

Read the raw times. The three readings are for the same 0.60 m ramp interval. The run number identifies a repeat and each reading is an elapsed time.
Calculate the mean. Mean time = (2.18 + 2.20 + 2.22) s / 3 = 2.20 s. The mean summarizes typical timing for this distance. To study how velocity changes during one run, use positions recorded throughout the motion.
Estimate velocities and plot them. Use this illustrative video dataset with 0.50 s windows. From 0.25 s to 0.75 s the cart moves from +0.010 m to +0.065 m. The average velocity is +0.055 m / 0.50 s = +0.11 m/s. Plot this estimate at the midpoint, 0.50 s, and repeat for each window.
Add a best-fit line. The points lie close to a straight line. Fit one line through the trend. Some readings lie above it and some lie below it.
Estimate the gradient. Read two well-separated points on the fitted line. At 0.50 s it gives about +0.12 m/s and at 2.00 s about +0.50 m/s. The gradient is approximately (0.50 − 0.12)/(2.00 − 0.50) = +0.25 m/s².
Relate scatter to measurement limits. Small differences from the line are consistent with an approximately constant acceleration. Video frame spacing limits the timing resolution, and the cart’s position may fall between scale marks. A hand-timed run also depends on the observer’s reaction time.
Improve one measurement. Use automatic timing gates at the marked start and finish when repeating the timed run. This reduces variation from pressing a stopwatch by hand and makes the elapsed times easier to compare.

A mean summarizes repeated measurements. A best-fit line summarizes the trend in estimated velocities. Small scatter around a straight line supports an approximately constant acceleration model over the sampled interval. Measurement limits affect how precisely the gradient can be estimated.

Worked example

Find a mean from repeated times

A cart travels the same 0.60 m in three runs. The measured times are 2.18 s, 2.20 s and 2.22 s. Find the mean and state what you would need to assume to use it in a constant-acceleration prediction.

Show solution

Mark the same start and the same 0.60 m finish on the ramp for every run. Each stopwatch reading measures the duration of that interval.
For three repeats under the same conditions, mean time = (2.18 s + 2.20 s + 2.22 s)/3 = 6.60 s/3 = 2.20 s.
Check that the mean is between the shortest and longest readings. The 0.04 s spread shows the times varied slightly.
Use the mean to find average velocity for the measured interval. A from-rest prediction also assumes the cart starts at u = 0 and has approximately constant acceleration under the same ramp conditions.

Reference

Look up terms, symbols, equations and the conditions for using them.

Key terms

Origin: The point you choose to label x = 0. Measure positions from this point. Construct an axis
Positive direction: The direction in which the coordinate increases. In this chapter right is positive unless a scene says otherwise. Construct an axis
Scale: The distance represented by one interval between equally spaced ticks, for example 1 m per interval. Construct an axis
Position: The signed coordinate of an object on a chosen axis. Place an object
Separation: The non-negative length between two positions, |x_B − x_A|. Compare two positions
Relative position: The signed difference x_B − x_A. It tells you where B is when you start your comparison at A. Compare two positions
Distance travelled: The total length of the path travelled, including any parts covered more than once. Turn around
Displacement: The signed change in position, Δx = x_f − x₀. Add a traveller and clock
Duration: The time between two events, Δt = t_f − t_i. Add a traveller and clock
Speed: The rate at which distance is travelled. Speed is zero or positive. Constant speed and velocity
Velocity: The rate of change of position. Its sign gives the direction along the chosen axis. Constant speed and velocity
Uniform motion: Motion with equal signed displacements in equal time intervals. Velocity is constant and the position–time graph is straight. Constant speed and velocity
Average speed: Total distance travelled divided by total elapsed time. Average speed and average velocity
Average velocity: Total displacement divided by total elapsed time. Average speed and average velocity
Instantaneous velocity: Velocity at a particular moment. Read it from the height of the velocity–time graph at that moment. What an average speed tells you
Acceleration: The rate of change of velocity with time. Acceleration has a sign on a one-dimensional axis. Acceleration from v(t)
Constant acceleration: Equal changes in velocity in equal time intervals. The velocity–time graph is straight. Acceleration from v(t)
Model: A simplified representation used to describe or predict motion. Its assumptions tell you when it applies. Choose the right model
Gradient: Change in the vertical quantity divided by change in the horizontal quantity. Use two points and keep the sign. Interpreting the x(t) gradient
Signed area: Area above the time axis counts positive and area below counts negative. On a velocity–time graph the sum gives displacement. A multi-stage velocity graph

Notation

Symbols and SI units
Symbol	Meaning	Unit
x	Position	m
x₀	Starting position	m
x_f	Final position	m
Δx	Displacement	m
d	Total distance travelled	m
t	Elapsed time or the time coordinate	s
Δt	Duration of an interval	s
u	Initial velocity in an accelerated interval	m/s
v	Velocity, or final velocity when paired with u	m/s
Δv	Change in velocity	m/s
v_avg	Average velocity	m/s
a	Acceleration	m/s²

Any journey

$Separation$ $= | x_{B} - x_{A} |$
Use the magnitude to find the gap.
$B relative to A$ $= x_{B} - x_{A}$
Start the comparison at A and keep the sign.
$Δx$ $= x_{f} - x_{0}$
Use the start and finish positions on the same axis.
$d_{total}$ $= d_{1} + d_{2} + \dots$
Add non-negative leg lengths.
$Δt$ $= t_{f} - t_{i}$
Subtract the starting clock reading.
$Average speed$ $= \frac{d_{total}}{{Δt}_{total}}$
Use the whole path and all elapsed time.
$v_{avg}$ $= \frac{{Δx}_{total}}{{Δt}_{total}}$
Use the signed net change and all elapsed time.
$t_{mean}$ $= \frac{(t_{1} + t_{2} + \dots + t_{n})}{n}$
Use repeated times for the same event under comparable conditions.

Constant velocity

$v$ $= \frac{Δx}{Δt}$
Use the signed gradient of a straight x(t) segment.
$Δx$ $= vΔt$
The signed area of a rectangular v(t) segment gives the same result.
$x$ $= x_{0} + vt$
Measure t from the starting event and include x₀.

Acceleration

$Δv$ $= v - u$
Subtract initial velocity from final velocity, keeping both signs.
$a$ $= \frac{Δv}{Δt}$ $= \frac{(v - u)}{Δt}$
This gives the average acceleration over the interval. A straight v(t) segment has this acceleration throughout.
$v$ $= u + at$
Use constant acceleration and elapsed time t.
$v_{avg}$ $= \frac{(u + v)}{2}$
Use only for constant acceleration. Keep the sign.
$Δx$ $= [\frac{(u + v)}{2}] t$
Use constant acceleration. This is the signed trapezoid area on v(t).
$Δx$ $= ut + \frac{1}{2} a t^{2}$
Use constant acceleration and include the initial velocity term.
$x$ $= x_{0} + ut + \frac{1}{2} a t^{2}$
Add the starting position to the predicted displacement.
$v$ $= at$ $v_{avg}$ $= \frac{v}{2}$ $Δx$ $= \frac{1}{2} vt$ $= \frac{1}{2} a t^{2}$
Use only for constant acceleration from rest (u = 0).
$a$ $= \frac{2 Δx}{t^{2}}$ $t$ $= \sqrt{\frac{2 Δx}{a}}$
For constant acceleration from rest. The time formula uses nonzero acceleration and a positive interval.

Read a graph

$Gradient of x (t)$ $\to velocity$
Calculate Δx/Δt on a straight segment. A horizontal segment means position stays fixed.
$Gradient of v (t)$ $\to acceleration$
Calculate Δv/Δt on a straight segment. A horizontal segment means velocity stays fixed.
$Signed area on v (t)$ $\to displacement$
Split at v = 0 for total distance and add the magnitudes of the separate areas.
$Signed area on a (t)$ $\to velocity change$
For constant a, multiply the signed height a by the interval duration.
$Intersection of two x (t) graphs$ $\to meeting$
Both objects have the same position at the same time. Their gradients can differ.

Common mistakes

For a coordinate of −3 m, the minus sign locates the point to the left. Its distance from zero is 3 m.
Distance follows the whole path. Displacement uses only the start and finish.
Average speed weights each leg by time. Divide the total distance by the total duration.
A position–time graph shows position versus time along the route. Read velocity from gradient.
A horizontal position–time line shows that position stays fixed.
A horizontal velocity–time line shows constant velocity. The object rests only at v = 0.
A velocity–time area below the time axis contributes a negative displacement.
Compare velocity and acceleration signs to decide whether speed increases or decreases. Matching nonzero signs increase speed.
The mean of endpoint velocities gives average velocity when acceleration is constant.
The shortened formula Δx = ½at² needs both constant acceleration and a start from rest.
An intersection on x(t) identifies a shared time and position. Compare the slopes to find each velocity.

Start with one dimension

Choose an origin

Set the scale

Place two objects

Read signed positions

Find the separation

Move the origin

Worked example

Distance and displacement after a turn

Worked example

Two averages for a two-leg journey

Worked example

What two cameras can measure

Worked example

Predict a position

Rightward

Leftward

Stationary

Faster rightward

Worked example

Find a meeting time and place

Worked example

Read displacement and distance

Worked example

Find a negative acceleration

Rightward and speeding up

Rightward and slowing down

Leftward and speeding up

Reversing direction

Worked example

Predict motion with non-zero u

Observed interval

Predicted interval

Worked example

Estimate acceleration from a timed run

Worked example

Predict the motion over 1.20 m

Constant velocity

Piecewise velocity

Constant acceleration

Worked example

Find a mean from repeated times

Reference

Key terms

Notation

Any journey

Constant velocity

Acceleration

Read a graph

Common mistakes