P1, Q1 - Mechanical quantities and modelling

P1, Q1 - Mechanical quantities and modelling

Use SI units consistently and describe a journey using position, displacement, distance, velocity, speed and acceleration. You will also choose modelling assumptions that keep the information needed for a particular question.

SI quantities and consistent units

A car travels along a straight road for 100 km100\,\mathrm{km} at 40 km h−140\,\mathrm{km\,h^{-1}}, then returns along the same route at 60 km h−160\,\mathrm{km\,h^{-1}}. Why is its average speed 48 km h−148\,\mathrm{km\,h^{-1}}, rather than 50 km h−150\,\mathrm{km\,h^{-1}}? First make sure the numbers and units describe the quantities you intend to calculate.

You already know

A rate compares two quantities by division. Multiplying both the numerator and denominator of a fraction by the same non-zero number leaves its value unchanged. Keep the units alongside a conversion so that you can see which units cancel.

Italic ss may denote displacement in an equation; upright s\mathrm{s} denotes the unit second. Read the role as well as the letter.

The SI base quantities needed in mechanics are length, time and mass, measured in metres, seconds and kilograms. A measurement is a number together with its unit: changing the unit changes the number, not the physical quantity. For example, 1 km=1000 m1\,\mathrm{km}=1000\,\mathrm m, 1 h=3600 s1\,\mathrm h=3600\,\mathrm s and 11 tonne is 1000 kg1000\,\mathrm{kg}. These conversions are not in the formula booklet: learn them.

Base and derived units

A base unit measures a fundamental quantity. A derived unit combines base units. Velocity and speed use m s−1\mathrm{m\,s^{-1}}; acceleration uses m s−2\mathrm{m\,s^{-2}}. The notations m/s2\mathrm{m/s^2} and m s−2\mathrm{m\,s^{-2}} mean the same unit.

An acceleration of 2 m s−22\,\mathrm{m\,s^{-2}} means a velocity change of 2 m s−12\,\mathrm{m\,s^{-1}} each second. It does not mean a distance of 2 m2\,\mathrm m travelled each second squared.

QuantitySI unitDistinction to remember
Lengthm\mathrm mMetre
Times\mathrm sSecond
Masskg\mathrm{kg}Retained even in a particle model
Force, including weightN=kg m s−2\mathrm N=\mathrm{kg\,m\,s^{-2}}Newton
Moment of a forceN m\mathrm{N\,m}Force multiplied by perpendicular distance

Mass measures the amount of matter; weight is a gravitational force. Their units are different. For example, a mass of 2 kg2\,\mathrm{kg}, with g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}, gives the unrounded weight calculation W=2(9.8)=19.6 NW=2(9.8)=19.6\,\mathrm N. The relationship W=mgW=mg, the unit facts in the table and the meaning of gg are not in the formula booklet: learn them. Force and moment methods are developed later. Trigonometric ratios and the coefficient of friction are dimensionless: they have no physical unit.

Worked example 1: metres per second

Question. Convert 72 km h−172\,\mathrm{km\,h^{-1}} into m s−1\mathrm{m\,s^{-1}}.

Plan. Replace kilometres by metres and hours by seconds in the same fraction.

72 km h−1=72×10003600 m s−1convert both units=20 m s−1divide\begin{aligned} 72\,\mathrm{km\,h^{-1}}&=\frac{72\times1000}{3600}\,\mathrm{m\,s^{-1}} && \text{convert both units} \\ &=20\,\mathrm{m\,s^{-1}} && \text{divide} \end{aligned}

Why this step? Multiplying by 10001000 alone would give metres per hour, so the time unit must change too.

Answer. The speed is 20 m s−120\,\mathrm{m\,s^{-1}}.

Check. In one hour, 20×3600=72000 m=72 km20\times3600=72000\,\mathrm m=72\,\mathrm{km}.

The next example changes only the direction of conversion. Reverse the unit factors rather than trying to remember whether a decimal multiplier should go above or below the line.

Worked example 2: kilometres per hour

Question. Convert 25 m s−125\,\mathrm{m\,s^{-1}} into km h−1\mathrm{km\,h^{-1}}.

Plan. Find the metres covered in an hour, then convert metres to kilometres.

25 m s−1=25×36001000 km h−1seconds to hours, metres to kilometres=90 km h−1simplify\begin{aligned} 25\,\mathrm{m\,s^{-1}}&=\frac{25\times3600}{1000}\,\mathrm{km\,h^{-1}} && \text{seconds to hours, metres to kilometres} \\ &=90\,\mathrm{km\,h^{-1}} && \text{simplify} \end{aligned}

Why this step? The distance in an hour is 36003600 times the distance in a second; the number of kilometres is then 10001000 times smaller than the number of metres.

Answer. The speed is 90 km h−190\,\mathrm{km\,h^{-1}}.

Check. 90×1000/3600=2590\times1000/3600=25, recovering the original speed.

Both conversions describe exactly the same motion. Converting first also prevents a subtler error: adding quantities expressed in incompatible units.

Watch out

Wrong: 72 km h−1=72000 m s−172\,\mathrm{km\,h^{-1}}=72000\,\mathrm{m\,s^{-1}}. Only the length unit has changed. Right: 72000 m h−1=20 m s−172000\,\mathrm{m\,h^{-1}}=20\,\mathrm{m\,s^{-1}}. Convert the denominator as well as the numerator.

Try the reverse conversion once more, with a new speed.

Position, displacement and rates of motion

Choose an origin and a positive direction before using a sign to describe motion.

On a straight road, take a lamp post as the origin and east as positive. Position x=−8 mx=-8\,\mathrm m means 8 m8\,\mathrm m west of the lamp post. A displacement of +3 m+3\,\mathrm m from there ends at x=−5 mx=-5\,\mathrm m: the displacement is a change, not the final position.

Displacement and distance

Displacement is the change in position, including direction. From x1x_1 to x2x_2, it is x2−x1x_2-x_1. Distance travelled is the total length of the route, regardless of direction. A return to the starting point has zero displacement, but need not have zero distance.

A scalar has magnitude only; a vector has magnitude and direction. Distance, speed, time and mass are scalars. Displacement, velocity, acceleration and force are vectors. Along one axis, a signed number records a vector's direction; reversing the positive direction reverses its sign, without changing the motion.

Velocity is the rate of change of position. Speed is its magnitude, so in straight-line motion speed is ∣v∣|v|. Acceleration is the rate of change of velocity. These definitions and the speed–velocity relationship are not in the formula booklet: learn them. A velocity of −3 m s−1-3\,\mathrm{m\,s^{-1}} means speed 3 m s−13\,\mathrm{m\,s^{-1}} in the negative direction.

Average rates

Over an interval of duration Δt>0\Delta t>0,

average velocity=displacementΔtaverage speed=total distanceΔt\text{average velocity}=\frac{\text{displacement}}{\Delta t} \qquad \text{average speed}=\frac{\text{total distance}}{\Delta t}

Not in the formula booklet: learn it.

An average describes the whole interval. It need not equal the velocity or speed at any particular stated instant.

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