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A.1 Kinematics

Syllabus
First assessment 2025
Topic
Level
HL

Describe Motion with Position, Velocity and Acceleration

Choose a reference

Position r\vec r specifies where an object is relative to a chosen origin. A position is not meaningful without a reference frame and coordinate direction.

Track change in position

Velocity v\vec v describes how position changes with time. It is a vector: its direction is the direction of motion at that instant, and on a curved path the velocity arrow is tangent to the path.

Track change in velocity

Acceleration a\vec a describes how velocity changes with time. A change in speed, direction, or both is acceleration; an object can accelerate even while its instantaneous speed is zero.

Common trap

Do not use “velocity” as a synonym for speed. Speed gives only magnitude; velocity also requires direction relative to the chosen coordinate system.

A.1.1 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

The evidence uses a motion-path diagram and asks for the velocity vector at a point. The mark scheme rewards a tangent arrow with the correct direction and origin at the specified point.

Command terms

Label / Identify / Draw

What earns marks

When a diagram asks for velocity at a point on a path, draw the arrow tangent to the path, beginning at the stated point, and orient it in the direction of motion. Name the quantity and include direction whenever the question requires a vector.

Watch for

Drawing the velocity arrow radially or along the wrong chord instead of tangent to the path at the specified point.

Representative question

Question 1

[Maximum number: 1]

the velocity of the ball at P . Label this arrow v.

Relate Velocity and Acceleration to Rates of Change

Velocity is a rate

Velocity is the rate of change of position:

v=drdt\vec v=\frac{d\vec r}{dt}

Over a finite interval, average velocity is displacement divided by elapsed time.

Acceleration changes velocity

Acceleration is the rate of change of velocity:

a=dvdt\vec a=\frac{d\vec v}{dt}

A constant acceleration gives equal changes in velocity during equal time intervals.

Read the gradient

On a position–time graph, the gradient represents velocity. On a velocity–time graph, the gradient represents acceleration. The graph’s slope, not its height alone, carries the rate-of-change meaning.

Common trap

Distance divided by time gives average speed, not instantaneous velocity. Likewise, a large velocity does not imply a large acceleration unless the velocity is changing rapidly.

A.1.2 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

The evidence includes a definition multiple-choice question and a falling-stone question about the change in velocity during consecutive equal time intervals.

Command terms

State / Identify

What earns marks

State the definition exactly: instantaneous velocity is the rate of change of position. For constant acceleration, connect equal time intervals with equal changes in velocity; do not substitute distance or speed when the question asks for displacement or velocity.

Watch for

Using distance divided by time as the definition of instantaneous velocity.

Representative question

Question 1

[Maximum number: 1]

Instantaneous velocity is defined as...

A

 displacement  time taken \frac{\text { displacement }}{\text { time taken }}.

B

rate of change of position.

C

 distance moved  time taken \frac{\text { distance moved }}{\text { time taken }}.

D

rate of change of distance.

Define Displacement as Change in Position

Displacement is a vector

Displacement is the change in position:

Δr=rfinalrinitial\Delta\vec r=\vec r_{final}-\vec r_{initial}

It has a magnitude and a direction from the initial position to the final position.

Ignore the route for displacement

The path taken between the two positions does not determine displacement. A curved or complicated journey can still have a straight-line displacement between its endpoints.

Use components when needed

For perpendicular changes, resolve displacement into components and combine them vectorially. A signed one-dimensional displacement is positive or negative according to the chosen axis.

Common trap

A return to the starting point gives zero displacement even though the distance travelled is non-zero.

A.1.3 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

The evidence uses projectile and circular-track contexts to test whether the learner chooses the endpoint-to-endpoint displacement rather than the distance along the path.

Command terms

Calculate / Identify

What earns marks

Find the vector from the initial position to the final position. In two dimensions, use the component changes and combine them; in a circular path, use the chord between endpoints rather than the arc length. State the magnitude and unit.

Watch for

Using the distance travelled along the trajectory or circular track as the displacement.

Representative question

Question 1

[Maximum number: 1]

A stone is kicked horizontally at a speed of 1.5 ms11.5 \mathrm{~ms}^{-1} from the edge of a cliff on one of Jupiter's moons. It hits the ground 2.0 s later. The height of the cliff is 4.0 m .
Air resistance is negligible.
What is the magnitude of the displacement of the stone?

A

7.0 m7.0 \mathrm{~m}

B

5.0 m

C

4.0 m

D

3.0 m

Distinguish Distance and Displacement

Distance

Distance is the total path length travelled. It is a scalar, so it has magnitude only and cannot be negative.

Displacement

Displacement is the vector change in position from start to finish. Its magnitude is the shortest endpoint-to-endpoint separation, not generally the length of the route.

Match the average quantity

Average speed uses total distance divided by total time. Average velocity uses displacement divided by total time. A route with turns can therefore have average speed greater than the magnitude of average velocity.

Common trap

For a complete oscillation, the displacement is zero but the distance is four times the amplitude. Choose the quantity named in the question before substituting.

A.1.4 Exam Analysis

Assessment in practice

1–3 marks
How it is assessed

The evidence compares average speed and average velocity for a person taking two legs and asks for distance travelled during one complete oscillation.

Command terms

Calculate / Distinguish

What earns marks

Use total path length for average speed and endpoint displacement for average velocity. In a complete oscillation, calculate distance from the repeated path segments; for a route with perpendicular legs, use the resultant displacement and total distance separately.

Watch for

Using displacement in the average-speed calculation or using total distance in the average-velocity calculation.

Representative question

Question 1

[Maximum number: 1]

A person walks 40 m due west and then 30 m due north. The total walking time is 100 s . What are the average speed and the magnitude of the average velocity of the person?

Average speed/m s 1{ }^{-1}

Magnitude of average
velocity /ms1/ \mathrm{m} \mathrm{s}^{-1}

0.5

0.5

0.5

0.7

0.7

0.5

0.7

0.7

Distinguish Instantaneous and Average Motion

Average values use an interval

Average speed is total distance divided by total time. Average velocity is displacement divided by elapsed time. Average acceleration is change in velocity divided by elapsed time.

Instantaneous values use one moment

An instantaneous value describes the motion at a particular instant. On a position–time graph, instantaneous velocity is the tangent gradient at that instant; on a velocity–time graph, instantaneous acceleration is the tangent gradient.

Connect graph quantities

The gradient of a velocity–time graph is acceleration, and the area under it is displacement. A constant acceleration therefore produces a straight-line velocity–time graph.

Common trap

Do not use the average gradient when a question asks for an instantaneous value. Use the tangent at the specified time.

A.1.5 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

The evidence tests graph transformation from acceleration–time to velocity–time, requiring the correct gradient/integration relationship and recognition of the initial condition.

Command terms

Identify / Determine

What earns marks

For an average value, use the whole interval and the appropriate total quantity. For an instantaneous value, read the tangent gradient at the stated time. In a velocity–time graph, integrate acceleration to obtain the change in velocity before applying the initial condition.

Watch for

Treating the acceleration value as the velocity value, or using the graph height instead of the gradient/area relationship.

Representative question

Question 1

[Maximum number: 1]

The graph shows the variation of the acceleration a with time t of an object moving in a straight line.

Which graph shows the variation of the velocity v of the object with time t ?

A
B
C
D

Apply SUVAT Equations to Uniform Acceleration

Uniform-acceleration model

SUVAT equations apply when acceleration is constant along the chosen one-dimensional axis:

v=u+at,s=ut+12at2,v2=u2+2as,s=u+v2tv=u+at,\quad s=ut+\frac12at^2,\quad v^2=u^2+2as,\quad s=\frac{u+v}{2}t

Choose an equation

List the known and unknown quantities s,u,v,a,ts,u,v,a,t. Select an equation containing the required unknown and only known quantities; keep signs consistent with the positive direction.

Check the model

A constant acceleration means equal changes in velocity in equal time intervals. Use separate horizontal and vertical equations only when the motion has been resolved into independent components.

Common trap

Do not use SUVAT when acceleration varies significantly with time or position. A formula can produce a neat number while still violating the model’s constant-acceleration assumption.

Explore Displacement as Area Under a Velocity-Time Graph

A.1.6 Exam Analysis

Assessment in practice

2–4 marks
How it is assessed

The evidence uses constant-acceleration motion from rest and measurement of displacement over time, rewarding the correct equation, data selection and calculation.

Command terms

Determine / Calculate

What earns marks

Check that acceleration is constant, choose a SUVAT equation containing the known quantities, and define the positive direction before substituting. For a photograph or position-time data, use consistent intervals and show how the measured displacement enters the equation.

Watch for

Applying a SUVAT equation without checking that acceleration is constant or mixing signed and unsigned distances.

Representative question

Question 1

[Maximum number: 3]

Determine g using the photograph.

Recognize Uniform and Non-Uniform Acceleration

Uniform acceleration

Acceleration is uniform when the velocity changes by equal amounts in equal time intervals. The velocity–time graph is a straight line with constant gradient.

Non-uniform acceleration

Acceleration is non-uniform when its magnitude or direction changes. The velocity–time graph then has a changing gradient, and a single SUVAT value cannot describe the entire interval.

Model versus reality

A constant-acceleration model can be useful over a limited interval even when real forces vary. State the approximation and identify the neglected force or changing condition.

Common trap

A curved trajectory does not by itself prove that acceleration is non-uniform: projectile motion without drag has constant downward acceleration while its velocity direction changes.

A.1.7 Exam Analysis

Assessment in practice

1 marks
How it is assessed

The evidence asks for one reason a spacecraft’s acceleration is not constant and one reason a dancer model is unrealistic, rewarding a specific neglected or changing physical parameter.

Command terms

State / Outline

What earns marks

Give a physical reason why acceleration changes: for example, a changing force, changing radiation intensity, changing force direction, drag, or an omitted interaction. Link the reason to the acceleration rather than merely saying the motion is unrealistic.

Watch for

Giving a vague statement such as “the model is not realistic” without naming a force, changing condition, or neglected parameter.

Representative question

Question 1

[Maximum number: 1]

State one reason why the acceleration of the spacecraft will not be constant.

Resolve Projectile Motion into Components

Separate the axes

With negligible fluid resistance, projectile motion is independent horizontal and vertical motion. Resolve the launch velocity:

ux=ucosθ,uy=usinθu_x=u\cos\theta,\qquad u_y=u\sin\theta

Horizontal motion

There is no horizontal acceleration in the ideal model, so vx=uxv_x=u_x and x=uxtx=u_xt. Use the horizontal displacement to find time or horizontal speed.

Vertical motion

Use one-dimensional constant-acceleration equations vertically, usually with ay=ga_y=-g if upward is positive. The horizontal and vertical equations share the same time tt.

Common trap

Do not use the launch speed as the horizontal speed. Resolve it first, and do not use the horizontal time independently of the vertical motion.

A.1.8 Exam Analysis

Assessment in practice

2–4 marks
How it is assessed

The evidence uses launch angle and horizontal distance to determine time or initial speed, rewarding the correct trigonometric component and the shared-time model.

Command terms

Calculate / Show

What earns marks

Resolve the launch velocity into horizontal and vertical components before using equations. Use the common time for both axes; calculate horizontal time from x=u_xt when horizontal acceleration is zero, then check the vertical condition separately.

Watch for

Using u sin θ for horizontal motion or forgetting that the vertical and horizontal calculations refer to the same elapsed time.

Representative question

Question 1

[Maximum number: 2]

The ball leaves the ground at an angle of 2222^{\circ}. The horizontal distance from the initial position of the edge of the ball to the wall is 11 m . Calculate the time taken for the ball to reach the wall.

Explain How Fluid Resistance Changes Projectile Motion

Drag opposes motion

Fluid resistance acts opposite the instantaneous velocity. Its magnitude depends on speed, shape, fluid and orientation, so the net force and acceleration can change throughout the flight.

Compare with the ideal path

Compared with no-drag motion, drag reduces horizontal speed, shortens range and usually lowers the maximum height. The trajectory is not the symmetric parabola predicted by the ideal model.

Terminal speed

During a vertical fall, drag grows as speed grows. When drag balances weight, the net force and acceleration become zero and the object continues at terminal speed.

Common trap

At the top of a drag-affected trajectory, the velocity may be momentarily horizontal or zero, but the acceleration is determined by the resultant forces, not by the velocity alone.

A.1.9 Exam Analysis

Assessment in practice

1–3 marks
How it is assessed

The evidence compares actual motion with an ideal no-drag path and asks where acceleration has greatest magnitude during a drag-affected vertical throw.

Command terms

Describe / Identify / Compare

What earns marks

State the direction of drag and connect its changing magnitude to the resultant acceleration. For vertical motion, identify the point where drag is greatest or where drag balances weight; for a projectile, compare speed, range, height and symmetry with the no-resistance model.

Watch for

Assuming acceleration is always g when drag is present, or assuming the trajectory remains a symmetric parabola.

Representative question

Question 1

[Maximum number: 1]

The diagram shows the path of a ball in the absence of air resistance. Q is the highest point of the ball's trajectory and a is the vertical acceleration at Q . At impact the velocity makes an angle θ\theta to the horizontal.

Three statements about the actual motion of the ball when there is air resistance are:
I. Q is lower.
II. a remains the same.
III. θ\theta increases.

Which statements are correct?

A

I and II only

B

I and III only

C

II and III only

D

I, II and III

Retrieve the A.1 Kinematics Model

Describe motion

Position locates the object, velocity is the rate of change of position, and acceleration is the rate of change of velocity. Distance and speed are scalar; displacement and velocity are directed quantities.

Use the right model

For uniform acceleration:

v=u+at,\quad s=ut+ rac12at^2,\quad v^2=u^2+2as

For projectiles without drag, solve horizontal and vertical components with a shared time. Do not use these equations when acceleration is non-uniform.

Check the boundary

Ask whether the quantity is average or instantaneous, whether the route or endpoints matter, whether acceleration is constant, and whether a neglected force such as drag changes the model.

ConceptIB Physics HL