TheIBAxolotl
The language of motion
Kinematics describes motion without asking what causes it. Four quantities do almost all the work:
- Distance is the total path length travelled (scalar).
- Displacement \(s\) is the straight-line change in position, with a direction (vector).
- Velocity \(v\) is the rate of change of displacement; speed is its magnitude.
- Acceleration \(a\) is the rate of change of velocity. A body can accelerate while its speed stays the same, simply by changing direction.
Always pick a positive direction at the start of a problem and stick to it. A ball thrown upward with up taken as positive has \(a = -9.81\ \text{m s}^{-2}\) the whole time it is in the air, including at the very top, where its velocity is momentarily zero but its acceleration is not.
Average velocity is total displacement divided by total time. Instantaneous velocity is the value at one moment, found from the gradient of a tangent on a displacement–time graph.
Reading motion graphs
Graphs turn a story about motion into geometry:
| Graph | Gradient gives | Area under curve gives |
|---|---|---|
| displacement–time | velocity | (not used) |
| velocity–time | acceleration | displacement |
| acceleration–time | (not used) | change in velocity |
A straight, sloping \(v\)–\(t\) line means constant acceleration. A curved \(s\)–\(t\) graph means the velocity is changing. Area below the time axis on a \(v\)–\(t\) graph counts as negative displacement, so the body is moving backwards.
The constant-acceleration (suvat) equations
When \(a\) is constant, five quantities (\(s\), \(u\), \(v\), \(a\), \(t\)) are linked by four equations. Each equation leaves out one quantity, so pick the one that omits the variable you neither know nor want:
- \(v = u + at\) (no \(s\))
- \(s = \frac{(u + v)}{2}t\) (no \(a\))
- \(s = ut + \frac{1}{2}at^2\) (no \(v\))
- \(v^2 = u^2 + 2as\) (no \(t\))
These equations are only valid when the acceleration is constant. They fail for a car whose acceleration fades as it speeds up, or for anything with significant air resistance.
Projectile motion
Without air resistance, a projectile has a constant downward acceleration \(g\) and zero horizontal acceleration. The trick is to treat the two directions completely separately, linked only by the shared time of flight.
- Horizontal: \(v_x = u\cos\theta\) stays constant, so \(x = u\cos\theta \cdot t\).
- Vertical: \(v_y = u\sin\theta - gt\), so the suvat equations apply with \(a = -g\).
The path is a parabola. On level ground, the maximum range occurs at a launch angle of \(45^\circ\), and complementary angles (for example \(30^\circ\) and \(60^\circ\)) give equal ranges.
Fluid resistance (qualitative)
Real projectiles feel drag, which acts opposite to the velocity and grows with speed. Compared with the ideal path, the real trajectory has a lower peak, a shorter range and a steeper descent; it is no longer symmetrical. A falling object accelerates until drag equals its weight, then falls at terminal velocity, where its acceleration is zero.
Worked example 1: uniform acceleration
A car speeds up uniformly from \(5.0\ \text{m s}^{-1}\) to \(21\ \text{m s}^{-1}\) in \(8.0\ \text{s}\). Find its acceleration and the distance covered.
- \(a = \frac{v - u}{t} = \frac{21 - 5.0}{8.0} = 2.0\ \text{m s}^{-2}\)
- \(s = \frac{(u + v)}{2}t = \frac{5.0 + 21}{2} \times 8.0 = 104\ \text{m} \approx 1.0 \times 10^2\ \text{m}\) (2 s.f.)
Worked example 2: horizontal launch
A ball is kicked horizontally at \(12.0\ \text{m s}^{-1}\) from the top of a cliff \(20.0\ \text{m}\) high. Ignore air resistance; \(g = 9.81\ \text{m s}^{-2}\).
- Vertical, starting from \(u_y = 0\): \(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20.0}{9.81}} = 2.02\ \text{s}\)
- Horizontal distance: \(x = 12.0 \times 2.019 = 24.2\ \text{m}\)
- Vertical speed at impact: \(v_y = 9.81 \times 2.019 = 19.8\ \text{m s}^{-1}\)
- Impact speed: \(v = \sqrt{12.0^2 + 19.8^2} = 23.2\ \text{m s}^{-1}\)
Notice that the horizontal speed plays no part in the time of fall.
Common mistakes
- Treating acceleration as zero at the top of a vertical throw.
- Mixing up signs: using \(g = +9.81\) in one line and \(-9.81\) in the next.
- Applying suvat to motion where the acceleration changes.
- Using the full launch speed \(u\) instead of its components \(u\cos\theta\) and \(u\sin\theta\).
- Reading displacement from the gradient of a \(v\)–\(t\) graph instead of the area.
- Rounding intermediate values (such as time of flight) too early, which shifts the final answer.
Exam tips
- Write the five suvat symbols, fill in the three you know, then choose the equation.
- On graph questions, say explicitly whether you are using a gradient or an area.
- For projectiles, draw two columns (horizontal and vertical) and keep them separate.
- Match your final answer to the least precise data value, usually 2 or 3 significant figures.
- "Explain" questions about drag need the chain: speed increases, drag increases, resultant force decreases, acceleration decreases.