Linking displacement, velocity and acceleration

In Kinematics of Linear Motion, a particle moves along a straight line and its displacement $s$ from a fixed point $O$ is given as a function of time $t$. This method connects the three quantities that describe the motion: displacement $s$, velocity $v$ and acceleration $a$.

The link is calculus. Velocity measures how fast displacement changes, so $v=\frac{ds}{dt}$.

Acceleration measures how fast velocity changes, so $a=\frac{dv}{dt}=\frac{d^{2}s}{dt^{2}}$. To go the other way, from acceleration back to velocity, or velocity back to displacement, you integrate and use given conditions to find the constant.

With this one framework you can answer typical questions: the velocity at a given instant, the times when the particle is momentarily at rest, the acceleration at a moment, and the direction of motion from the sign of $v$.

Reach for this method whenever a question describes a particle moving in a straight line and gives one of $s$, $v$ or $a$ as a function of $t$. Phrases like 'displacement from $O$', 'velocity after $t$ seconds', 'momentarily at rest', 'instantaneously stationary' or 'maximum velocity' are all signals.

Decide the direction first. If you are given displacement and asked for velocity or acceleration, you differentiate (once for $v$, twice for $a$).

If you are given acceleration or velocity and asked for velocity or displacement, you integrate and use the stated initial values to fix the constant. Key phrases map to equations: 'at rest' means $v=0$; 'maximum or minimum velocity' means $a=0$; 'returns to $O$' means $s=0$.

1. 1

   Read the function

   Write down what is given, usually $s$ as a function of $t$, and what is asked.

2. 2

   Differentiate for velocity

   Find $v=\frac{ds}{dt}$.

3. 3

   Differentiate again for acceleration

   Find $a=\frac{dv}{dt}$.

4. 4

   Translate the condition

   'At rest' → set $v=0$; 'maximum/minimum velocity' → set $a=0$; 'at $O$' → set $s=0$.

5. 5

   Solve or substitute

   Solve the resulting equation for $t$, or substitute a given $t$ to get a value.

6. 6

   Interpret signs and units

   A negative velocity means motion in the negative direction; state units such as m/s and m/s².

• Mixing up direction: to go from $s$ to $v$ to $a$ you differentiate; to go back you integrate. Doing the wrong one is a common slip.
• Setting $s=0$ instead of $v=0$ for 'at rest'. 'At rest' always means the velocity is zero.
• Forgetting that speed is the magnitude of velocity, so a velocity of $-3$ m/s is a speed of $3$ m/s.
• When integrating, forgetting the constant of integration and the given initial condition needed to find it.
• Dropping units, or giving acceleration in m/s instead of m/s².

The step students most often get wrong is translating the words into an equation, reading 'at rest' and setting $s=0$ instead of $v=0$, or differentiating when they should integrate. In a one-to-one lesson our teachers drill the small dictionary of phrases with you until 'at rest', 'maximum velocity' and 'returns to $O$' each trigger the right equation instantly.

We also keep the direction, differentiate down, integrate up, visible on every question so it becomes second nature. Because your working is shown line by line, each step earns its method mark.

Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English. To see how we teach kinematics, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.