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IB Maths AA SL · Unit 5: Calculus

IB Maths AA SL Kinematics Questions

Exam-style IB Maths AA SL kinematics questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.

Practise Kinematics questions → AA SL formula booklet

What you need to know

s(t), v(t) = s'(t), a(t) = v'(t). SL AA Paper 2 asks 'when does the particle come to rest?' and expects you to spot v = 0. Kinematics — displacement, velocity, acceleration overview →

What's examined in AA SL kinematics

The question bank covers these kinematics question types (number of questions in brackets):

Kinematics worked examples

Worked example 1: Displacement versus total distance · easy

A particle moves along a straight horizontal line. Its velocity, $v\text{ ms}^{-1}$, is given by $v(t) = 2t - 2$ for $0 \le t \le 4$. State the time when the particle comes instantaneously to rest, and calculate the total distance travelled by the particle in the first 4 seconds.

Solution

1. Find when $v = 0$: $2t - 2 = 0 \implies \mathbf{t = 1\text{ s}}$.

2. Recognise the difference: displacement is $\int v \,dt$; total distance is $\int |v| \,dt$. Since the particle stops at $t = 1$ it changes direction.

3. Split: $\text{Distance} = \left|\int_0^1 (2t - 2)\,dt\right| + \int_1^4 (2t - 2)\,dt$.

4. Integrate: $\int (2t - 2)\,dt = t^2 - 2t$.

5. First segment ($0$ to $1$): $(1 - 2) - 0 = -1$, so distance $= |-1| = 1\text{ m}$.

6. Second segment ($1$ to $4$): $(16 - 8) - (1 - 2) = 8 - (-1) = 9\text{ m}$. Total $= 1 + 9 = \mathbf{10\text{ m}}$.

Examiner tip: For total distance, find every root of $v(t)$ in the domain and split the integral there, or use $\int |v(t)| \,dt$ on the GDC directly.

Worked example 2: Recovering velocity from acceleration · medium

A particle has acceleration $a\text{ ms}^{-2}$ defined by $a(t) = \frac{1}{t^2} + \sin t$ for $t \ge 1$. Given that $v(1) = 1\text{ ms}^{-1}$, find a specific expression for $v(t)$.

Solution

1. Recall: $v(t) = \int a(t) \,dt$.

2. Rewrite with negative indices: $\int (t^{-2} + \sin t) \,dt$.

3. Integrate: $\frac{t^{-1}}{-1} - \cos t + C = -\frac{1}{t} - \cos t + C$.

4. State the general form: $v(t) = -\frac{1}{t} - \cos t + C$.

5. Apply the boundary condition $v(1) = 1$: $1 = -1 - \cos 1 + C$.

6. Solve: $C = 2 + \cos 1$. Specific expression: $\mathbf{v(t) = -\frac{1}{t} - \cos t + 2 + \cos 1}$.

Examiner tip: Do NOT evaluate trigonometric constants like $\cos 1$ into decimals unless instructed — leave them exact. Remember the argument is $1$ radian, not $1$ degree.

Worked example 3: Kinematics via graphing technology · hard

A particle moves so that its displacement from a fixed point $P$, after $t$ seconds, is $s(t) = 15 - 6e^{0.8t - 0.25t^2}$. Use your GDC to find the value of $t$ when the particle first returns to $P$, and find its initial velocity at $t = 0$.

Solution

1. Interpret: "returns to $P$" means displacement is zero, so solve $s(t) = 0$.

2. Set up: $15 - 6e^{0.8t - 0.25t^2} = 0 \implies e^{0.8t - 0.25t^2} = 2.5$.

3. Solve on the GDC (Equation Solver or G-Solv $\to$ ROOT). Discard the negative root. Valid root: $\mathbf{t = 4.12\text{ s}}$ (3 s.f.).

4. Interpret initial velocity: $v(0) = s'(0)$.

5. Use the numerical derivative on the GDC (OPTN $\to$ CALC $\to$ d/dx) — no need to hand-differentiate.

6. Evaluate $\frac{d}{dt}(15 - 6e^{0.8t - 0.25t^2}) \big|_{t = 0} = \mathbf{-4.80\text{ ms}^{-1}}$.

Examiner tip: In Paper 2, whenever the derivative is needed at a specific number, use the $\frac{d}{dx}$ button. Hand-differentiating chained exponentials wastes time and is error-prone.

Try these IB Maths AA SL kinematics questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.

Question 1 · easy · 4 marks · Paper 2

A cricket ball is projected directly upwards from ground level. The vertical height of the cricket ball, \(h\) metres above the ground, after \(t\) seconds is modelled by the function: \[h(t) = 13t - 4.9t^2, \quad t \ge 0\]

  1. Use your graphic display calculator to find the times at which the cricket ball is exactly \(3\text{ m}\) above the ground.

  2. Calculate the maximum height reached by the cricket ball.

Attempt it and see the mark scheme →

Question 2 · medium · 5 marks · Paper 1

A particle is found to have a velocity, \(v\text{ ms}^{-1}\), that can be expressed by the function: \[v(t) = t^3 \cos t, \quad t \ge 0\]

  1. Find an expression for the acceleration, \(a(t)\), of the particle.

  2. Hence, evaluate the exact acceleration of the particle at time \(t = \pi\).

Attempt it and see the mark scheme →

Question 3 · hard · 5 marks · Paper 1

A particle, moving in a straight line, is found to have a velocity \(v(t) = \sin t + \cos(2t)\) where \(v\) is measured in \(\text{ms}^{-1}\) and \(0 \le t \le 2\pi\). By using a suitable double angle trigonometric identity, analytically find the exact times \(t\) when the particle is instantaneously at rest.

Attempt it and see the mark scheme →

All 34 kinematics questions with mark schemes →

FAQ

How many IB Maths AA SL kinematics questions are there?

There are 34 exam-style kinematics questions in the AA SL question bank (Paper 1: 19 · Paper 2: 15), graded 4 easy, 10 medium, 15 hard, 3 very hard, 2 starter. Every question has a full IB-style mark scheme (M, A and R marks).

Is kinematics on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 19 · Paper 2: 15. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Where can I get the mark schemes?

Open the AA SL Unit 5 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

More AA SL Unit 5 topics

Kinematics in other IB Maths courses

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