The Particle in the Groove
A particle slides along a straight horizontal groove. Its position from the origin is $x = 2 + t(t - 3)$ metres, where $t$ is the time in seconds. Move the time slider: the particle sometimes travels forwards, sometimes backwards, and at one instant it is momentarily still. Watch the position, the displacement from the start and the velocity.
position $x$ = 0.00 m · displacement from the start = -2.00 m · velocity $v$ = -1.00 m s-1
The particle starts at $x = 2$ m. A negative velocity means it is moving in the negative direction, but its speed is never negative.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. A scalar quantity is completely described by its:
R2. Which of these is a vector quantity?
R3. Average speed is calculated as:
Every motion quantity is either a scalar (magnitude only) or a vector (magnitude and direction). The direction is what decides the sign in a calculation.
| Quantity | Type | Unit | Meaning |
|---|---|---|---|
| Distance | scalar | metre (m) | total length of the path travelled |
| Displacement | vector | metre (m) | change in position, from the start, with a direction |
| Speed | scalar | m s-1 | how fast the body moves |
| Velocity | vector | m s-1 | how fast, and in which direction |
| Acceleration | vector | m s-2 | rate of change of velocity |
| Time | scalar | second (s) | when, and for how long |
Distance counts every metre travelled; displacement uses only the starting and finishing positions. Move $5\ \text{m}$ forwards and $3\ \text{m}$ back and you have travelled $8\ \text{m}$ (distance) but your displacement is only $+2\ \text{m}$.
$\text{displacement} = \text{final position} - \text{initial position}$, with a sign that shows the direction.
The sum of the lengths of all the pieces of the journey, counted positive. A body that returns to its starting point has zero displacement but a non-zero distance travelled.
| Quantity | Definition | Formula |
|---|---|---|
| Velocity | rate of change of position | $v = \dfrac{dx}{dt}$ |
| Speed | size of the velocity | $\text{speed} = |v|$ |
| Acceleration | rate of change of velocity | $a = \dfrac{dv}{dt}$ |
| Average velocity | displacement per unit time | $\bar{v} = \dfrac{\text{displacement}}{\text{time}}$ |
| Average speed | distance per unit time | $\bar{v} = \dfrac{\text{distance}}{\text{time}}$ |
Worked mini-example: a train covers $24\ \text{km}$ in $45\ \text{min}$, so its average speed is $24 \div 0.75 = 32\ \text{km h}^{-1}$.
Groups of 4: tap a card to pick it up, then tap a zone to file it — "Scalar: magnitude only" or "Vector: magnitude and direction". Discuss each quantity with your group, then press Check cards. Six motion quantities.
Choose a scenario, then scrub the time slider. Six readouts update together: position, displacement, distance travelled, velocity and acceleration. Watch how displacement and distance differ, and how speed ignores the sign of the velocity. Then answer the three challenges.
The position of a particle moving along a straight horizontal groove is given by $x = 2 + t(t - 3)$ for $0 \leqslant t \leqslant 5$, where $x$ is in metres and $t$ in seconds. Find (i) the position at $t = 0, 1, 1.5, 2, 3, 4, 5$, (ii) the displacement at $t = 5$ and (iii) the total distance travelled during the motion.
Substitute each time into $x = 2 + t(t - 3)$. For the distance, first find when the velocity is zero — that is where the particle turns round.
$t = 0, 1, 1.5, 2, 3, 4, 5$ give $x = 2,\ 0,\ -0.25,\ 0,\ 2,\ 6,\ 12$ m.
$x(5) - x(0) = 12 - 2 = 10\ \text{m}$ in the positive direction.
$v = \dfrac{dx}{dt} = 2t - 3$, which is zero at $t = 1.5$. So the particle moves backwards until $t = 1.5$ and forwards afterwards.
Backwards: $2 - (-0.25) = 2.25\ \text{m}$. Forwards: $12 - (-0.25) = 12.25\ \text{m}$. Distance $= 2.25 + 12.25 = 14.5\ \text{m}$.
A plane flies from London to Toronto, a distance of $5700\ \text{km}$, at an average speed of $1280\ \text{km h}^{-1}$. It returns at an average speed of $1200\ \text{km h}^{-1}$. Find the average speed for the round trip.
Average speed is total distance divided by total time, not the mean of the two speeds. Find the two journey times separately first.
Out: $5700 \div 1280 = 4.453\ \text{h}$. Back: $5700 \div 1200 = 4.75\ \text{h}$.
Total distance $= 5700 \times 2 = 11400\ \text{km}$. Total time $= 4.453 + 4.75 = 9.203\ \text{h}$.
$11400 \div 9.203 = 1238.7\ \text{km h}^{-1}$, so the average speed is $1239\ \text{km h}^{-1}$ (3 s.f.). Notice it is not the mean of 1280 and 1200.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. Which pair contains one scalar and one vector?
Q2. A cyclist rides $300\ \text{m}$ east along a straight road and then $100\ \text{m}$ west. Her displacement and distance travelled are:
Q3. A body moves along a straight line at constant speed and then reverses its direction. Which quantity changes at the moment it reverses?
Q4. A car moves along a straight road from the point with position $20\ \text{m}$ to the point with position $-30\ \text{m}$. Its displacement is:
Work each on paper, then reveal the solution to check your working.
A boy throws a ball vertically upwards. The ball rises to a greatest height of $4.8\ \text{m}$ above his hand and is caught again at the same height. (i) State the displacement of the ball at the moment it is caught. (ii) Find the total distance travelled by the ball.
Reveal Solution
(i) It finishes where it started, so the displacement is $0\ \text{m}$. (ii) Distance $= 4.8 + 4.8 = 9.6\ \text{m}$ — up and back down again.
A particle moves along a straight line. It starts at the point with position $-4\ \text{m}$, moves to the point with position $12\ \text{m}$, and then back to the point with position $2\ \text{m}$. Find (i) the total distance travelled and (ii) its final displacement from the starting point.
Reveal Solution
(i) First part $12 - (-4) = 16\ \text{m}$; second part $12 - 2 = 10\ \text{m}$. Distance $= 16 + 10 = 26\ \text{m}$. (ii) Displacement $= 2 - (-4) = +6\ \text{m}$.
A particle moves so that its position $x$ metres at time $t$ seconds is $x = 2t^3 - 18t$. (i) Find the velocity of the particle at $t = 2$. (ii) Find the time when the particle is at the origin for $t > 0$, and state its velocity at that time.
Reveal Solution
(i) $v = \dfrac{dx}{dt} = 6t^2 - 18$, so $v(2) = 24 - 18 = 6\ \text{m s}^{-1}$. (ii) $2t^3 - 18t = 2t(t^2 - 9) = 0$ gives $t = 3$; then $v = 54 - 18 = 36\ \text{m s}^{-1}$.
A train takes $45$ minutes to complete its $24$ kilometre trip, stopping for $1$ minute at each of $7$ stations during the trip. Find the average speed of the train in $\text{km h}^{-1}$.
Reveal Solution
The total time is $45\ \text{min} = 0.75\ \text{h}$ (the stops are already included in the 45 minutes). Average speed $= \dfrac{24}{0.75} = 32\ \text{km h}^{-1}$.
Louise covers distances along main roads at about $100\ \text{km h}^{-1}$ and distances in towns at $30\ \text{km h}^{-1}$. Find her average speed for each journey: (a) $20\ \text{km}$ on main roads and then $10\ \text{km}$ in a town; (b) $20\ \text{km}$ on main roads and then $20\ \text{km}$ in a town.
Reveal Solution
(a) Times $0.2\ \text{h}$ and $10 \div 30 = 0.3333\ \text{h}$; average speed $= \dfrac{30}{0.5333} = 56.25\ \text{km h}^{-1}$. (b) Times $0.2\ \text{h}$ and $0.6667\ \text{h}$; average speed $= \dfrac{40}{0.8667} = 46.2\ \text{km h}^{-1}$ (3 s.f.). The slower miles always dominate.
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