IB Physics Kinematics Questions: How to Answer Them

IB Physics kinematics questions become much more manageable when you use the same disciplined method every time: define a positive direction, translate the information into signed variables, identify the physical model, select an equation that contains the required unknown, and check the final answer. Most lost marks come not from difficult algebra but from sign errors, graph misinterpretation, or using constant-acceleration equations when acceleration is changing.

In the current IB Physics course, kinematics is A.1 Kinematics within Theme A, Space, time and motion. It is assessed through multiple-choice, data-based, short-answer, and extended-response formats, so students need conceptual understanding as well as reliable calculation methods.

How kinematics is examined in IB Physics

The current course, first assessed in 2025, has two external examination papers. According to the official IB Physics curriculum update, Paper 1 contains Paper 1A multiple-choice questions and Paper 1B data-based questions, while Paper 2 contains short-answer and extended-response questions.

Kinematics can therefore appear in several predictable forms:

The official IB specimen papers and markschemes are particularly useful because they show how calculations, data analysis, and multi-topic reasoning are presented under the current assessment structure.

The five-step method for IB Physics kinematics questions

1. Sketch the motion and choose a positive direction

Draw a simple diagram, even when the question does not request one. Mark the initial position, direction of motion, displacement, acceleration, and any important final position.

Then state or clearly adopt a positive direction. If upward is positive, gravitational acceleration near Earth is a = -g. If downward is positive, it is a = +g; either convention works if used consistently.

2. Translate the wording into variables

For one-dimensional uniformly accelerated motion, list:

• s: displacement, not necessarily distance
• u: initial velocity
• v: final velocity
• a: constant acceleration
• t: elapsed time

Attach signs and SI units immediately. The phrase “released from rest” means u = 0, while “momentarily at rest” means the velocity at that instant is zero, not that acceleration is zero.

3. Decide which physical model applies

Before selecting an equation, ask:

• Is the acceleration constant?
• Is the motion one-dimensional or two-dimensional?
• Is fluid resistance neglected?
• Is the question asking for displacement or total distance?
• Does a graph provide the information more directly?

The four standard equations apply only to uniformly accelerated motion:

These equations are supplied in the Physics data booklet, but the booklet does not decide which equation or sign convention to use. The RevisionDojo equations of motion notes explain where the relationships come from, which makes selection easier than memorising them as unrelated formulas.

4. Substitute symbols and values clearly

Choose the equation containing the unknown and known variables. Rearranging symbolically before inserting numbers often prevents calculator and unit errors.

In Paper 2, show enough working for the method to be followed. A numerical answer without a valid equation or substitution can lose method marks if the final value is wrong.

5. Check the physics of the result

A strong final check asks:

• Does the sign match the chosen direction?
• Are the units correct?
• Is the magnitude realistic?
• Does the object change direction during the interval?
• Is the requested quantity displacement, distance, speed, or velocity?

Do not automatically remove a negative sign. A negative velocity or displacement often communicates physically important directional information.

Worked example: vertical motion

A ball is thrown vertically upward at 18.0 m s⁻¹. Ignore fluid resistance and take upward as positive. Find its maximum height above the release point using g = 9.81 m s⁻².

At maximum height, the vertical velocity is momentarily zero:

• $u=+18.0\text{ m s}^{-1}$
• $v=0$
• $a=-9.81\text{ m s}^{-2}$

Time is neither known nor required, so use:

$v^2=u^2+2as$

$0=(18.0)^2+2(-9.81)s$

$s=\frac{(18.0)^2}{2(9.81)}=16.5\text{ m}$

The positive displacement is consistent with the ball being above its release point. A common incorrect method uses a = +9.81 m s⁻² despite defining upward as positive, producing an impossible negative height.

How to answer motion graph questions

Graph questions test whether you understand rates of change and accumulation, not merely whether you can identify a line’s shape.

On a velocity-time graph, area below the time axis is negative displacement. To find total distance, calculate the magnitude of each separate area and add them. Adding signed areas gives net displacement instead.

For a curved position-time graph, instantaneous velocity comes from the gradient of a tangent. For a curved velocity-time graph, acceleration is changing, so the four uniform-acceleration equations generally cannot be applied across the whole interval.

When drawing a best-fit line in a data-based question, use a large gradient triangle based on the line, not two nearby raw data points. Include units in the gradient calculation and avoid forcing the line through the origin unless the physics or evidence justifies it.

How to solve projectile questions

Projectile motion should be divided into two simultaneous motions. With fluid resistance neglected:

• Horizontal acceleration is zero, so horizontal velocity is constant.
• Vertical acceleration is gravitational acceleration.
• The same time applies to both components.

For a launch speed $u$ at angle $\theta$ above the horizontal:

$u_x=u\cos\theta$

$u_y=u\sin\theta$

Use $s_x=u_xt$ horizontally and the constant-acceleration equations vertically. Do not insert the full launch speed into a vertical equation.

Suppose a projectile is launched horizontally at 12 m s⁻¹ from a height of 20 m. Taking downward as positive, its vertical motion has $u_y=0$, $s_y=20\text{ m}$, and :

$20=\frac{1}{2}(9.81)t^2$

$t=2.02\text{ s}$

The horizontal range is then:

$s_x=u_xt=(12)(2.02)=24.2\text{ m}$

The recurring trap is to treat the horizontal and vertical motions as unrelated. They have different accelerations, but they share the same flight time.

Fluid resistance and terminal speed

When resistance is included, do not assume constant acceleration unless the question explicitly provides a suitable interval or model. As a falling object accelerates, drag generally increases, reducing the resultant downward force and therefore the acceleration.

At terminal speed, drag balances weight, so resultant force and acceleration are zero. The object is not stationary; it continues moving at constant velocity. For a parachutist, opening the parachute can produce a large upward resultant force, followed by decreasing speed and eventual motion at a lower terminal speed.

Common mistakes that cost marks

• Treating distance and displacement as interchangeable.
• Treating speed and velocity as interchangeable.
• Substituting $g$ as positive without considering the chosen axis.
• Assuming zero velocity means zero acceleration.
• Using the four motion equations when acceleration varies.
• Calculating displacement when the question asks for total distance.
• Using the full projectile speed in both component equations.
• Assuming launch and landing heights are equal when they are not.
• Giving an answer without units or with excessive significant figures.
• Copying an equation from the data booklet without explaining its relevance.

The RevisionDojo A.1 Kinematics notes can repair conceptual gaps, while the kinematics flashcards are useful for testing definitions and graph relationships.

The fastest way to improve your kinematics method

Re-reading notes can clarify concepts, but it does not fully train equation selection, sign control, or markscheme communication. The fastest practical improvement usually comes from attempting a question independently and then watching the complete method worked through.

Use this cycle:

1. Attempt one question under a short time limit.
2. Commit to a complete answer before checking help.
3. Watch the per-question worked video solution.
4. Compare the setup, signs, equation choice, and final presentation.
5. Record the precise cause of any lost mark.
6. Repeat a similar question without notes within 48 hours.

RevisionDojo’s A.1 Kinematics Questionbank and per-question worked solutions provide targeted practice, while the kinematics video library supports visual review of the method. Once topic questions are reliable, use IB Physics predicted papers to practise recognising kinematics inside mixed and linked questions.

Keep an error log using categories such as sign convention, wrong model, graph interpretation, algebra, units, and question wording. Jojo AI can help explain why a chosen approach fails, but you should still redo the calculation yourself so that the corrected method becomes automatic.

Conclusion

IB Physics kinematics questions follow recurring structures: signed one-dimensional motion, graph analysis, projectiles, data interpretation, and qualitative resistance. Success depends on defining directions, selecting the correct model, separating vector components, and showing a checkable calculation.

Use RevisionDojo Study Notes or Flashcards to establish the concepts, then move quickly to the Questionbank. Attempting questions before watching the per-question video solutions is the most effective way to build a method that remains reliable under exam pressure.

Sources and referenced URLs

• Official IB Physics curriculum page
• Official IB Physics curriculum and assessment update
• Official IB Physics specimen papers and markschemes
• RevisionDojo A.1 Kinematics topic hub
• RevisionDojo A.1 Kinematics Questionbank
• RevisionDojo A.1 Kinematics notes
• RevisionDojo equations of motion notes
• RevisionDojo A.1 Kinematics videos
• RevisionDojo A.1 Kinematics flashcards
• RevisionDojo IB Physics predicted papers