Kinematics
🎯What you need to be able to do
- Distinguish distance from displacement, and speed from velocity — and explain why the distinction matters.
- Work with average and instantaneous speed, velocity and acceleration.
- Use the equations of motion for uniformly accelerated motion in a straight line.
- Read motion graphs: what a gradient means, what an area means, and how the three graphs connect.
- Analyse projectile motion by treating horizontal and vertical motion separately.
- Describe qualitatively the effect of fluid resistance on falling objects, including terminal speed.
📚The physics
Distance and displacement. Distance is how far you travelled. Displacement is how far you ended up from where you started, together with the direction. Run one lap of a 400 m track and your distance is 400 m while your displacement is zero. Distance is a scalar; displacement is a vector. Almost every avoidable kinematics error starts with treating one as the other.
Speed is built from distance, so speed is a scalar. Velocity is built from displacement, so velocity is a vector. On that lap of the track your average speed was respectable and your average velocity was exactly zero.
Average and instantaneous. Average velocity is total displacement divided by total time. Instantaneous velocity is what a speedometer reads — the gradient of the displacement–time graph at a single point. A car that averages 60 km/h over a journey may never once have been travelling at 60 km/h.
Acceleration is the rate of change of velocity, \( a = \dfrac{\Delta v}{\Delta t} \), and it is a vector. Two consequences catch people out. An object can accelerate without changing speed — that is exactly what circular motion is. And a negative acceleration does not mean slowing down; it means the acceleration points in the negative direction. Something moving in the negative direction with negative acceleration is speeding up. Drop the word deceleration and let the signs carry the meaning.
The equations of motion hold only when acceleration is constant. Check that condition before writing anything down.
Each one is missing a different quantity: the first has no \(s\), the second no \(v\), the third no \(t\), the fourth no \(a\). List the three quantities you know and the one you want, then choose the equation that omits the one you neither know nor need.
Motion graphs. On a displacement–time graph the gradient is velocity. On a velocity–time graph the gradient is acceleration and the area between the line and the time axis is displacement — signed, so area below the axis subtracts. On an acceleration–time graph the area is the change in velocity. Remember it as gradient going down the list, area coming back up, and you can move freely between all three.
Projectiles. Horizontal and vertical motion are independent, and time is the only quantity they share. Horizontally there is no acceleration, so the horizontal velocity never changes. Vertically the acceleration is \(g\) downwards throughout, including at the highest point, where the vertical velocity is zero but the acceleration certainly is not. Resolve the launch velocity into components first, and never substitute the launch speed itself into an equation of motion.
Fluid resistance and terminal speed. Drag grows with speed. For a falling object the weight stays constant while the drag increases, so the resultant force shrinks and the acceleration falls — the object is still speeding up, just less and less quickly. When drag equals weight the resultant is zero, the acceleration is zero and the speed stops changing: terminal speed. On a velocity–time graph that is a curve flattening towards a horizontal asymptote, never a straight line that suddenly bends over. For a projectile, drag shortens the range and destroys the symmetry, so the descent is steeper than the ascent.
✏️Worked example
Resolve first. \( u_x = 18\cos 35^\circ = 14.7 \) m s\(^{-1}\) and \( u_y = 18\sin 35^\circ = 10.3 \) m s\(^{-1}\). From this point on the figure 18 never appears again, and that discipline is what the whole question rests on.
(a) Time of flight is a vertical question. The ball lands at the height it left, so its vertical displacement over the whole flight is zero. Putting \(s = 0\), \(u = 10.3\) and \(a = -9.81\) into \( s = ut + \tfrac{1}{2}at^{2} \) gives \( 0 = t(10.3 - 4.905t) \), so
(b) Range is a horizontal question, and the horizontal velocity is constant: \( x = 14.7 \times 2.10 = 31 \) m.
(c) At the highest point the vertical velocity is zero, so \( v^{2} = u^{2} + 2as \) vertically with \(v = 0\) gives
🔭See it happen
PhET, Projectile Motion. Two things worth doing rather than just firing the cannon: launch at 30° and then at 60° with the same speed and notice the ranges are identical, because angles adding to 90° always give the same range on level ground. Then switch air resistance on and watch that symmetry break.
📝Practise
Work through these, then reveal the answer. Each question targets a different objective from the list above.
1. A car accelerates uniformly from 12 m s\(^{-1}\) to 30 m s\(^{-1}\) in 6.0 s. Find its acceleration and the distance travelled.
2. A stone is dropped from rest and falls 45 m. Take \( g = 9.81 \) m s\(^{-2}\). Find the time of fall and the speed on impact.
3. A velocity–time graph shows the velocity rising in a straight line from 0 to 20 m s\(^{-1}\) over 5.0 s, then staying constant at 20 m s\(^{-1}\) for a further 3.0 s. Find the total displacement.
4. A ball is thrown horizontally at 15 m s\(^{-1}\) from a cliff 20 m high. Find the time of flight and how far from the base of the cliff it lands.
5. A projectile is launched at 25 m s\(^{-1}\) at 40° above the horizontal. Find its maximum height. Ignore air resistance.
6. Explain, in terms of forces, why the velocity–time graph of a falling object with significant air resistance flattens towards a horizontal asymptote.
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.
- The Physics Classroom — 1D Kinematics tutorial
- The Physics Hypertextbook — kinematics