Real-life linear motion problems can be modelled and analysed using the calculus relationships between displacement, velocity and acceleration. A structured problem-solving framework helps us determine key kinematic quantities such as instantaneous velocity, maximum or minimum values, instantaneous stops and total distance travelled.
See how differentiation and integration connect displacement, velocity and acceleration, and follow the four-step problem-solving cycle.
Velocity is the rate of change of displacement: \(v = \dfrac{ds}{dt}\).Acceleration is the rate of change of velocity: \(a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}\).
Velocity is obtained from acceleration: \(v = \int a \, dt\).Displacement is obtained from velocity: \(s = \int v \, dt\).
Understand the problem: identify given equations, domain of \(t\), initial conditions and what needs to be calculated.
Plan the strategy: choose differentiation or integration and identify conditions such as \(a = 0\) or \(v = 0\).
Implement the strategy: carry out the calculus steps, find constants of integration and evaluate functions at specific values of \(t\).
Check and reflect: verify by reverse substitution or by sketching a velocity-time graph and computing definite integrals.
Velocity from displacement
Differentiate displacement with respect to time.
Displacement from velocity
Integrate velocity with respect to time.
Choose differentiation or integration based on the given and required quantities.
If displacement \(s\) is given, differentiate with respect to time \(t\) to find velocity \(v = \dfrac{ds}{dt}\), then differentiate again to find acceleration \(a = \dfrac{dv}{dt}\).
If acceleration \(a\) is given, integrate with respect to \(t\) to find velocity \(v = \int a \, dt\), remembering to find the constant of integration \(c\) using initial conditions. Then integrate velocity to find displacement \(s = \int v \, dt\).
"Stops momentarily" or "at rest" means \(v = 0\)."Maximum velocity" or "minimum velocity" means \(a = \dfrac{dv}{dt} = 0\)."Maximum height" or "maximum displacement" means \(v = \dfrac{ds}{dt} = 0\)."Initial" means \(t = 0\).
Displacement is given directly by \(s(t)\) or \(\int_{t_1}^{t_2} v \, dt\). Total distance accounts for changes in direction where \(v = 0\), and corresponds to the total area bounded between the \(v-t\) curve and the time axis: \(\int |v| \, dt\).
Velocity: \(v = 0.9t - 0.03t^2\) for \(0 \leqslant t \leqslant 30\), with \(s = 0\) when \(t = 0\).
Acceleration: \(a = \dfrac{dv}{dt} = 0.9 - 0.06t\).
Acceleration is zero when \(0.9 - 0.06t = 0 \implies t = 15\text{ s}\).
Displacement: \(s = \int (0.9t - 0.03t^2) \, dt = 0.45t^2 - 0.01t^3 + c\). Since \(s = 0\) at \(t = 0\), \(c = 0\).
Distance travelled in 30 s
Checking: \(\int_0^{30} (0.9t - 0.03t^2) \, dt = 135\text{ m}\).
Water rocket maximum height
The rocket stops momentarily at \(t = 2\text{ s}\), which is its maximum height.
Answer in your mind, then press "Check Answer".
A water rocket is launched vertically upward such that its velocity after \(t\) seconds is \(v = 20 - 10t\text{ ms}^{-1}\). Find the time, in seconds, when the rocket stops momentarily.
The velocity of a particle is given by \(v = 0.9t - 0.03t^2\text{ ms}^{-1}\). Find the acceleration of the particle when \(t = 5\text{ s}\).
Given the acceleration of an object is \(a = 12 - 6t\text{ ms}^{-2}\) and its initial velocity at \(t = 0\) is \(44\text{ ms}^{-1}\). Find the time \(t\) when the velocity is at its maximum.
Zero velocity means acceleration must also be zero.
When an object stops momentarily, \(v = 0\), but acceleration is not necessarily zero. For example, at the highest point of vertical motion, acceleration due to gravity is still acting.
Maximum velocity occurs when \(v = 0\).
Maximum velocity is a turning point of the velocity function, which occurs when its rate of change is zero, meaning acceleration \(a = \dfrac{dv}{dt} = 0\).
Distance travelled is always equal to displacement.
Distance is a scalar representing the total length of path travelled, whereas displacement is a vector representing the net change in position from the starting point.
Objective: Identify kinematic conditions, calculus operations and key terms in applications of kinematics of linear motion.
1 Fariza's velocity is given by \(v = 0.9t - 0.03t^2\text{ ms}^{-1}\). At what time, in seconds, is her acceleration equal to zero?
2 An object moves along a straight line with velocity \(v\text{ ms}^{-1}\). Which mathematical condition represents the instant when the object attains its maximum velocity?
3 If a particle stops momentarily at time \(t\), its instantaneous velocity \(v\) is equal to ______ \(\text{ms}^{-1}\).
4 To find the displacement function \(s\) from a given velocity function \(v\), we use the calculus operation of ______.
Drag each kinematic condition or calculus operation to the correct mathematical expression. On a phone, tap a card first, then tap the matching box.
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