IB DP Physics: A.1 Kinematics, graphs, projectiles and fluid resistance (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions follow the IB Diploma Programme Physics guide, first assessment 2025, section A.1 Kinematics. The content is the same at SL and HL. Use g = 9.81 m s⁻² and ignore fluid resistance unless told otherwise.

Links: revision notes for this set, the A.1 study guide, the earlier A.1 practice questions (a different set), the IB Physics hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. A drone flies 150 m due north and then 90 m due east in a total time of 40 s.

(a) Give the total distance flown and the size of the displacement from the start. [2] (b) Find the average speed and the size of the average velocity. [2]

2. A body moves in a straight line. Its displacement s from a start point is recorded.

t / s	0	1.0	2.0	3.0	4.0	5.0
s / m	0	1.5	6.0	13.5	24.0	37.5

(a) Calculate the average velocity between t = 3.0 s and t = 4.0 s. [2] (b) Use the table to estimate the instantaneous velocity at t = 3.0 s. [2] (c) State how more data could improve the estimate in (b). [1]

3. A driver travelling at 18 m s⁻¹ sees a hazard. The car keeps its velocity for 0.70 s before the brakes act, then slows uniformly at 5.0 m s⁻² to rest.

(a) Find the distance covered during the first 0.70 s. [1] (b) Find the distance covered while braking. [2] (c) Find the total time from seeing the hazard to stopping. [2]

4. A lift moves upward from rest. Its velocity-time graph has three straight sections: velocity rises uniformly to 4.0 m s⁻¹ over 2.0 s, stays at 4.0 m s⁻¹ for 6.0 s, then falls uniformly to zero over 3.0 s.

(a) Calculate the total displacement of the lift. [3] (b) Calculate the acceleration in the first and last sections, with signs. [2] (c) Calculate the average velocity over the whole journey. [1]

5. The velocity of a sledge on a slope is measured at 2.0 s intervals and plotted as a smooth curve that becomes less steep.

t / s	0	2.0	4.0	6.0	8.0
v / m s⁻¹	0	2.4	4.1	5.0	5.4

(a) Explain why the equations for uniformly accelerated motion cannot be used for the whole 8.0 s. [1] (b) Estimate the displacement in the 8.0 s using trapeziums. [3] (c) Estimate the acceleration at t = 4.0 s. [2]

6. A rugby ball is punted from level ground at 21 m s⁻¹, 52° above the horizontal. Ignore fluid resistance.

(a) Resolve the launch velocity into horizontal and vertical components. [2] (b) Find how long the ball is in the air. [2] (c) Find the range. [1] (d) Find the greatest height reached. [1]

7. A ball is thrown from a window 15 m above level ground at 8.0 m s⁻¹, directed 25° below the horizontal. Ignore fluid resistance.

(a) Resolve the launch velocity into horizontal and vertical components. [2] (b) Find the time the ball takes to land. [3] (c) Find the horizontal distance from the window to the landing point. [1] (d) Find the speed and direction of the ball just before it lands. [3]

8. A supply aircraft flies level at 55 m s⁻¹ at a height of 250 m and releases a package. Ignore fluid resistance.

(a) Find the time the package takes to fall. [2] (b) Find how far forward the package travels. [1] (c) State where the aircraft is, relative to the package, when it lands, and give a reason. [1]

9. A ball is launched at an angle above level ground. Compare its motion with fluid resistance acting to the motion with none, in terms of the range, the greatest height, the shape of the trajectory, its speed on landing and its acceleration at the highest point. [6]

10. A skydiver falls from a stationary aircraft with the parachute closed. Explain the shape of the velocity-time graph, using forces and acceleration, until a constant speed is reached. [5]

11. A diver leaves a platform 5.0 m above the water at 3.2 m s⁻¹, directed 30° above the horizontal. Ignore fluid resistance.

(a) Resolve the launch velocity into horizontal and vertical components. [2] (b) Find the time until the diver reaches the water. [3] (c) Find the horizontal distance from the platform edge to the entry point. [1] (d) Find the vertical velocity, the speed and the angle below the horizontal at entry. [3] (e) Suggest one way the real motion would differ because of fluid resistance. [1]

Answers

1. (a) Distance = 150 + 90 = 240 m [1]. Displacement = √(150² + 90²) = 175 m [1]. (b) Average speed = 240 ÷ 40 = 6.0 m s⁻¹ [1]. Average velocity = 175 ÷ 40 = 4.4 m s⁻¹ [1]. Examiner insight: Displacement is a straight-line vector, so adding the two legs for it is a common slip.

2. (a) v = (24.0 − 13.5) ÷ (4.0 − 3.0) [1] = 10.5 m s⁻¹ [1]. (b) Use a chord centred on t = 3.0 s: (24.0 − 6.0) ÷ (4.0 − 2.0) [1] = 9.0 m s⁻¹ [1]. (c) Use a shorter interval around t = 3.0 s, or a tangent on a graph [1]. Examiner insight: An average over a whole second starting at 3.0 s is not the instantaneous value at 3.0 s, so quoting it in (b) is a typical way to lose credit.

3. (a) s = 18 × 0.70 = 12.6 m [1]. (b) v² = u² + 2as with v = 0: 0 = 18² + 2(−5.0)s [1], so s = 324 ÷ 10 = 32.4 m [1]. (c) Braking time = 18 ÷ 5.0 = 3.6 s [1]. Total time = 0.70 + 3.6 = 4.3 s [1]. Examiner insight: Treat the two stages separately, because one equation across both would assume a single constant acceleration.

4. (a) Area of first section = ½ × 2.0 × 4.0 = 4.0 m [1]. Middle section = 6.0 × 4.0 = 24 m [1]. Last section = ½ × 3.0 × 4.0 = 6.0 m, so total = 34 m [1]. (b) First section a = 4.0 ÷ 2.0 = +2.0 m s⁻² [1]. Last section a = −4.0 ÷ 3.0 = −1.3 m s⁻² [1]. (c) Total time = 11 s, so average velocity = 34 ÷ 11 = 3.1 m s⁻¹ [1]. Examiner insight: Show each area separately with its shape, and give the sign of the deceleration.

5. (a) The v-t graph is curved, so the acceleration is not constant and the equations need constant acceleration [1]. (b) Each 2.0 s strip: ½(v₁ + v₂) × 2.0 [1]. Strips give 2.4, 6.5, 9.1 and 10.4 m [1]. Total ≈ 28 m (28.4 m) [1]. (c) Use the chord from t = 2.0 s to t = 6.0 s, centred on 4.0 s [1]: a = (5.0 − 2.4) ÷ 4.0 ≈ 0.65 m s⁻² [1]. Examiner insight: Say the result is an estimate.

6. (a) uₓ = 21 cos 52° = 12.9 m s⁻¹ [1]. uᵧ = 21 sin 52° = 16.5 m s⁻¹ [1]. (b) Vertical motion returns to launch height: 0 = 16.5t − ½(9.81)t² [1], so t = 2 × 16.55 ÷ 9.81 = 3.37 s [1]. (c) Range = 12.93 × 3.374 = 43.6 m [1]. (d) H = 16.55² ÷ (2 × 9.81) = 14.0 m [1]. Examiner insight: Keep extra figures from the components until the end, since rounded components can change the third figure.

7. (a) uₓ = 8.0 cos 25° = 7.25 m s⁻¹ [1]. Vertical component downward = 8.0 sin 25° = 3.38 m s⁻¹ [1]. (b) With down positive: 15 = 3.38t + ½(9.81)t² [1]. Rearrange: 4.905t² + 3.38t − 15 = 0 [1]. Positive root: t = 1.44 s [1]. (c) x = 7.25 × 1.438 = 10.4 m [1]. (d) vᵧ = 3.38 + 9.81 × 1.438 = 17.5 m s⁻¹ downward [1]. Speed = √(7.25² + 17.5²) = 18.9 m s⁻¹ [1]. Angle = tan⁻¹(17.5 ÷ 7.25) = 67.5° below the horizontal [1]. Examiner insight: Say which root you reject, and give a direction as well as a speed.

8. (a) 250 = ½(9.81)t² [1], so t = √(500 ÷ 9.81) = 7.14 s [1]. (b) x = 55 × 7.14 = 393 m [1]. (c) The aircraft is directly above the package [1] because both keep the same horizontal velocity when there is no fluid resistance (one reason is enough). Examiner insight: Name the constant horizontal velocity and the absence of a horizontal force.

9. Any six of the following, each worth one mark [6]:

• The range is smaller with resistance.
• The greatest height is smaller.
• The trajectory is no longer a parabola, and it is steeper on the way down than on the way up.
• The horizontal velocity decreases, since drag opposes the motion.
• The landing speed is less than the launch speed (on level ground).
• At the highest point the acceleration is no longer g: there is a horizontal drag component as well as weight, so it is not purely downward.
• The fall takes longer than the rise. Examiner insight: Pair each effect with a short cause such as “drag opposes the velocity”.

10. At release the speed is small, so drag is small and the acceleration is close to g, giving a steep gradient [1]. As speed rises, drag rises [1]. Weight minus drag, the net force, shrinks, so the acceleration shrinks and the graph curves over and flattens [1]. Once drag matches weight the net force vanishes [1], giving zero acceleration and a constant velocity, the terminal speed [1]. Examiner insight: Link the gradient to the acceleration at each stage, and say the forces are balanced.

11. (a) uₓ = 3.2 cos 30° = 2.77 m s⁻¹ [1]. uᵧ = 3.2 sin 30° = 1.60 m s⁻¹ upward [1]. (b) With up positive, final displacement −5.0 m: −5.0 = 1.60t − ½(9.81)t² [1]. Rearrange: 4.905t² − 1.60t − 5.0 = 0 [1]. Positive root: t = 1.19 s [1]. (c) x = 2.771 × 1.186 = 3.29 m [1]. (d) vᵧ = 1.60 − 9.81 × 1.186 = −10.0 m s⁻¹, so 10.0 m s⁻¹ downward [1]. Speed = √(2.771² + 10.03²) = 10.4 m s⁻¹ [1]. Angle = tan⁻¹(10.03 ÷ 2.771) = 74.6° below the horizontal [1]. (e) The entry point would be nearer, since horizontal velocity would decrease [1]. Examiner insight: Declare the sign convention before the quadratic, and keep it for every term.

Where marks are usually lost

• Giving distance when displacement is asked for, or a speed with no direction when a velocity is needed.
• Quoting an average over an interval as an instantaneous value.
• Applying one equation across two stages with different accelerations.
• Dropping the sign on a deceleration or a downward component.
• Mixing horizontal and vertical quantities in one equation, or choosing the wrong root of a quadratic.
• Listing fluid-resistance effects with no cause.
• Saying that terminal speed means there is no force, not balanced forces.

Next steps

• Recap with the revision notes and the study guide.
• Move on to A.2 Forces and momentum.
• Go to the IB Physics hub and the printable checklist.
• Try the free 10-minute diagnostics for all subjects.
• Book a free trial class.

Frequently asked questions

Are these IB DP Physics kinematics questions from past papers? No. They are original questions written for revision and practice, and they do not replicate the exam’s structure, question count or mark tariffs. Use them alongside the official IB past papers from your school or the IB store.

Are these A.1 kinematics questions for SL or HL? Both. The content of A.1 Kinematics is the same at SL and HL.

What value of g and assumptions do these questions use? Use g = 9.81 m s⁻² and ignore fluid resistance unless a question tells you otherwise.

Is this the only set of A.1 practice questions? No. This is a second set; the earlier A.1 practice questions are a different set, linked at the top of this page.

Official syllabus

International Baccalaureate Organization, Physics guide (Diploma Programme), first assessment 2025, section A.1 Kinematics.