Coaching Centre
Kinematics review
Module 1 review: summary of the key ideas and equations, how the topics connect, a problem-solving method, and a mixed HSC-style test on motion in a straight line and motion on a plane.
Every kinematics question in this module comes down to three decisions: which frame of reference, which direction is positive, and which tool to use (an equation, a graph, or a vector diagram). Get those three right and the maths is usually short.
Summary
| Topic | Key relationships | Watch out for |
|---|---|---|
| 1A Scalars and vectors | average speed $= \dfrac{d}{t}$; $v_{\text{avg}} = \dfrac{\Delta r}{\Delta t}$ | distance vs displacement; direction on every vector |
| 1B Instantaneous velocity | chord gradient = average $v$; tangent gradient = instantaneous $v$ | use points on the tangent, far apart |
| 1C Practical | $v = \dfrac{\text{card width}}{\Delta t}$; $a = \dfrac{v_2 - v_1}{t}$ | validity, reliability, accuracy used correctly |
| 1D Equations of motion | $v = u + at$; $s = ut + \tfrac12at^2$; $v^2 = u^2 + 2as$ | constant $a$ only; $a = -9.8$ m s⁻² at the top too |
| 1E Graphs | gradient: $s \to v \to a$; area: $a \to v \to s$ | area below the axis is negative displacement |
| 1F Vectors | $R = \sqrt{x^2 + y^2}$; components $v\cos\theta$, $v\sin\theta$ | bearings are clockwise from north |
| 1G Change in velocity | $\Delta\vec v = \vec v - \vec u$; $\vec a_{\text{avg}} = \dfrac{\Delta\vec v}{\Delta t}$ | constant speed round a bend is still acceleration |
| 1H Relative motion | $\vec v_{A\text{ rel }B} = \vec v_A - \vec v_B$; object/ground = object/medium + medium/ground | crossing time uses the across component only |
Putting it together
The three tools describe the same motion in different ways, and good answers often use two of them as a check:
- Graphs work for any motion, including changing acceleration. Gradients and areas are always valid.
- Equations of motion are faster but need constant acceleration. They are just the gradient and area of a straight $v$–$t$ line written as algebra.
- Vector diagrams handle direction on a plane. Resolving into components turns a 2D problem into two 1D problems, which is exactly how projectile motion is done in Year 12.
A problem-solving method that works for all of them:
- Sketch the situation. Mark the origin and the positive direction (or north on a map).
- List the known quantities with signs and units (convert km/h to m/s if the acceleration is in m s⁻²).
- Choose the tool: equation, graph or vector diagram.
- Solve, keeping one more significant figure in the working than in the data.
- Answer with units, the right number of significant figures, and a direction for every vector. Check it is sensible.
A sprinter accelerates uniformly from rest to 10.0 m s⁻¹ over the first 4.0 s of a 100 m race, then holds that speed. (a) How far has she run after 4.0 s? (b) What is her total race time? (c) What is her average velocity for the race?
Key words and formulas
| Term | Meaning |
|---|---|
| displacement | change in position; a vector from start to finish |
| instantaneous velocity | velocity at an instant; gradient of the tangent to an $s$–$t$ graph |
| uniform motion | constant velocity in a straight line |
| uniformly accelerated motion | constant acceleration; straight-line $v$–$t$ graph |
| resultant | the single vector equal to the sum of two or more vectors |
| component | one of two perpendicular vectors that add to give the original |
| frame of reference | the object or point that motion is measured relative to |
| heading / track | direction a craft points / direction it actually moves over the ground |
Test yourself
A stone is dropped from rest down a well and hits the water after 2.5 s. Ignoring air resistance, the depth to the water is
$s = \tfrac12(9.8)(2.5)^2 = 30.6$ m.
Which graph shows an object moving with constant positive acceleration?
Constant gradient on a $v$–$t$ graph is constant acceleration. Option B is constant velocity.
A ship sails at 12 knots on a bearing of 150°. Its components are
East $= 12\sin150° = 6.0$; north $= 12\cos150° = -10.4$, i.e. 10.4 south.
Two trains approach each other on parallel tracks at 25 m s⁻¹ and 15 m s⁻¹. They are 2.0 km apart. They pass each other after
Relative speed $= 25 + 15 = 40$ m s⁻¹; $2000 \div 40 = 50$ s.
A car's velocity changes from 20 m s⁻¹ north to 20 m s⁻¹ east in 5.0 s. The size of its average acceleration is
$|\Delta\vec v| = \sqrt{20^2 + 20^2} = 28.3$ m s⁻¹; $28.3 \div 5.0 = 5.7$ m s⁻².
In a light-gate experiment, which change would most improve the accuracy of the measured instantaneous velocity of an accelerating trolley?
A narrower card means the velocity changes less while it passes, so the average is closer to the instantaneous value.
Label the river-crossing vectors.
Which tool gives each quantity most directly?
Displacement is a quantity. The equations of motion require the acceleration to be . The area under a velocity–time graph gives . Velocity of A relative to B is $\vec v_A$ $\vec v_B$. A bearing is measured clockwise from .
A hot-air balloon rises vertically at a constant 4.0 m s⁻¹. When it is 60 m above the ground, a passenger drops a sandbag over the side. (a) What is the sandbag's initial velocity relative to the ground? Explain. (b) Calculate the maximum height the sandbag reaches. (c) Calculate the time it takes to reach the ground and its speed on impact. (d) Sketch and describe its velocity–time graph from release to impact.
(a) 4.0 m s⁻¹ upwards. At the moment of release the sandbag is moving with the balloon, so relative to the ground it starts with the balloon's velocity. (Relative to the balloon it starts at rest.)
(b) Up positive, $u = +4.0$, $a = -9.8$. At the top $v = 0$: $0 = 16 - 19.6s$, $s = 0.82$ m. Maximum height $= 60.8$ m above the ground.
(c) The ground is at $s = -60$ m: $-60 = 4.0t - 4.9t^2$, so $4.9t^2 - 4.0t - 60 = 0$ and $t = \dfrac{4.0 + \sqrt{16 + 1176}}{9.8} = 3.93$ s (the negative root is rejected). $v = 4.0 - 9.8(3.93) = -34.5$ m s⁻¹, so it hits the ground at 35 m s⁻¹ downwards.
(d) A straight line starting at +4.0 m s⁻¹, sloping down with a constant gradient of −9.8 m s⁻², crossing the time axis at 0.41 s (the top) and ending at −34.5 m s⁻¹ at 3.93 s. The straight line shows constant acceleration throughout.