A.1 Kinematics
- Syllabus
- First assessment 2025
- Topic
- —
- Level
- HL
Choose a reference
Position 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 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 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.
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.
Label / Identify / Draw
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.
Drawing the velocity arrow radially or along the wrong chord instead of tangent to the path at the specified point.
Representative question
the velocity of the ball at P . Label this arrow v.
arrow tangent to the path in the correct direction
If the line when produced backwards goes below the curve - no mark.
Arrows not beginning at P score [0]
Velocity is a rate
Velocity is the rate of change of position:
v=dtdr
Over a finite interval, average velocity is displacement divided by elapsed time.
Acceleration changes velocity
Acceleration is the rate of change of velocity:
a=dtdv
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.
The evidence includes a definition multiple-choice question and a falling-stone question about the change in velocity during consecutive equal time intervals.
State / Identify
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.
Using distance divided by time as the definition of instantaneous velocity.
Representative question
Instantaneous velocity is defined as...
time taken displacement .
rate of change of position.
time taken distance moved .
rate of change of distance.
B
Displacement is a vector
Displacement is the change in position:
Δr=rfinal−rinitial
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.
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.
Calculate / Identify
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.
Using the distance travelled along the trajectory or circular track as the displacement.
Representative question
A stone is kicked horizontally at a speed of 1.5 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?
7.0 m
5.0 m
4.0 m
3.0 m
B
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.
The evidence compares average speed and average velocity for a person taking two legs and asks for distance travelled during one complete oscillation.
Calculate / Distinguish
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.
Using displacement in the average-speed calculation or using total distance in the average-velocity calculation.
Representative question
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
Magnitude of average
velocity /ms−1
0.5
0.5
0.5
0.7
0.7
0.5
0.7
0.7
C
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.
The evidence tests graph transformation from acceleration–time to velocity–time, requiring the correct gradient/integration relationship and recognition of the initial condition.
Identify / Determine
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.
Treating the acceleration value as the velocity value, or using the graph height instead of the gradient/area relationship.
Representative question
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
Uniform-acceleration model
SUVAT equations apply when acceleration is constant along the chosen one-dimensional axis:
v=u+at,s=ut+21at2,v2=u2+2as,s=2u+vt
Choose an equation
List the known and unknown quantities s,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.
Loading interaction assets…
The evidence uses constant-acceleration motion from rest and measurement of displacement over time, rewarding the correct equation, data selection and calculation.
Determine / Calculate
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.
Applying a SUVAT equation without checking that acceleration is constant or mixing signed and unsigned distances.
Representative question
Determine g using the photograph.
Read at least two points correctly and consistently
Use 21⋅g⋅t2 with a length interval from two non-consecutive points OR for two (or more) length intervals, using consistent time intervals
Correct calculation of g
Marking guidance:
Award [2] max if they use one single length interval of consecutive points.
Award [1] max if they miss to subtract the initial point in their length interval or if they use inconsistent time intervals.
Do not penalize significant figures in the final answer.
2 3
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.
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.
State / Outline
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.
Giving a vague statement such as “the model is not realistic” without naming a force, changing condition, or neglected parameter.
Representative question
State one reason why the acceleration of the spacecraft will not be constant.
The intensity of light on the sail will not remain constant/the force due to the Sun/Earth/Jupiter/other planets is ignored/sail may not be flat/light may be incident at an angle/part of the radiation may be absorbed/any other reasonable
statement
[1]
Separate the axes
With negligible fluid resistance, projectile motion is independent horizontal and vertical motion. Resolve the launch velocity:
ux=ucosθ,uy=usinθ
Horizontal motion
There is no horizontal acceleration in the ideal model, so vx=ux and x=uxt. Use the horizontal displacement to find time or horizontal speed.
Vertical motion
Use one-dimensional constant-acceleration equations vertically, usually with ay=−g if upward is positive. The horizontal and vertical equations share the same time t.
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.
The evidence uses launch angle and horizontal distance to determine time or initial speed, rewarding the correct trigonometric component and the shared-time model.
Calculate / Show
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.
Using u sin θ for horizontal motion or forgetting that the vertical and horizontal calculations refer to the same elapsed time.
Representative question
The ball leaves the ground at an angle of 22∘. 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.
horizontal speed =19×cos22 « =17.6 m s−1 »
time =≪ speed distance =19cos2211=>0.62<s≫
Marking guidance:
Allow ECF for MP2
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.
The evidence compares actual motion with an ideal no-drag path and asks where acceleration has greatest magnitude during a drag-affected vertical throw.
Describe / Identify / Compare
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.
Assuming acceleration is always g when drag is present, or assuming the trajectory remains a symmetric parabola.
Representative question
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 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 increases.
Which statements are correct?
I and II only
I and III only
II and III only
I, II and III
D
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.