Measurements and motion

Watch a small cart move along a straight track. Mark its position with a ruler and record the time with a clock. From those readings, can we work out how fast it is moving—and predict where it will be a moment later?

We will use this experiment to connect calculus with motion. Then we will add the ideas of force, energy, and flow that we need for gravity. If you have studied mechanics before, you can start with the five checks in §0.9.

0.1 A derivative is a local prediction#

Choose a mark on the track as the starting position and call it $x=0$. The number $x$ tells us how far the cart is from that mark, with positive values to the right. Start the clock at $t=0$. Suppose the measurements follow this pattern:

During successive seconds the cart travels 3, then 9, then 15 metres. It is speeding up. One formula that fits these measurements is

Four readings do not prove that this formula works at every time. We will use it as a model and see what it predicts.

The coefficient $A$ has units too: multiplying metres per second squared by seconds squared gives metres. A unit specifies what we count, such as metres or seconds. The dimension describes the kind of quantity, such as length or time. Metres and centimetres are different units of the same dimension.

The cart’s velocity is its rate of change of position, including direction. Its acceleration is the rate of change of velocity:

At $t=2\,\mathrm s$, these give $x=12\,\mathrm m$, $v=12\,\mathrm{m/s}$ and $a=6\,\mathrm{m/s^2}$. The velocity predicts the next small change. Over $\Delta t=0.1\,\mathrm s$, it predicts an extra $v\Delta t=1.2\,\mathrm m$. The exact change is $A[(2.1\,\mathrm s)^2-(2\,\mathrm s)^2]=1.23\,\mathrm m$. The extra $0.03\,\mathrm m$ comes from the increase in velocity during that tenth of a second.

Taylor’s formula keeps track of that correction:

The symbol $\Delta$ means a change; the dots stand for higher powers of that change. For this quadratic motion the displayed expression is exact. For a general smooth motion, keeping only the velocity term gives a local approximation. Shorten the time step, and the acceleration correction shrinks faster than the velocity contribution. A derivative predicts a small change, not a whole future at one fixed rate.

WORKED EXAMPLE

One distance, two ways to count it

A table is 2 metres long, or 200 centimetres long. What changed when the number changed?

See the idea

The table stayed the same. A centimetre is a smaller measuring step, so it takes more of them to span the table. A measurement needs both a number and a unit. Keeping those two parts together will help us read every physical equation in this book.

Work it out

1. 01
   Convert the unit and keep the length

   One metre contains 100 centimetres. Multiply by a ratio equal to one, written with the new unit on top and the old unit below. The old unit cancels.

   $$2\,\mathrm m\;\frac{100\,\mathrm{cm}}{1\,\mathrm m}=200\,\mathrm{cm}.$$

   Why this step works Changing the unit changes the number used to describe a fixed length.

2. 02
   Give a rate its two units

   If a cart travels 2 metres in 4 seconds, its average speed is distance divided by elapsed time. Using centimetres gives a different number for the same average speed.

   $$\frac{2\,\mathrm m}{4\,\mathrm s}=0.5\,\mathrm{m/s}=50\,\mathrm{cm/s}.$$

   Why this step works Metres per second and centimetres per second describe the same kind of quantity: length divided by time.

3. 03
   Count area in two directions

   A square 2 metres on each side has area 4 square metres. Each side is also 200 centimetres, giving 40,000 square centimetres. Converting a squared length squares the conversion factor.

   $$1\,\mathrm{m^2}=(100\,\mathrm{cm})^2=10{,}000\,\mathrm{cm^2}.$$

   Why this step works An area contains one length factor from each direction, so both factors must be converted.

Go deeper

A dimension records the kind of quantity independently of the chosen unit. Write $L$ for the dimension of length and $T$ for time. Then $[x]=L$, $[v]=L/T$, and $[a]=L/T^2$; square brackets here mean “the dimensions of.” If $x=At^2$, matching dimensions gives $[A]=L/T^2$. Dimensional agreement is necessary but does not establish an equation: both $x=At^2$ and $x=2At^2$ have consistent dimensions, but they predict different distances for the same $A$ and $t$. Measurements or other physical reasoning must determine the numerical factor.

Further readingBIPM · The International System of Units

Test the idea

FIRST, PREDICT

A cart’s speed is reported as 0.5 metres per second and 50 centimetres per second. Which statement is justified?

These can be two descriptions of the same speed.The second speed must be 100 times faster.A speed needs only a number, because the units cancel.

Compare the reasoning

These can be two descriptions of the same speed.

There are 100 centimetres in one metre; both the number and the unit changed together.

The second speed must be 100 times faster.

Compare quantities in the same unit before comparing their numbers.

A speed needs only a number, because the units cancel.

Dividing a length by a time leaves length-per-time units.

A hint

Express both distances in centimetres while keeping the second unchanged.

NOW CHANGE THE EXAMPLE

A rectangle is 2 metres long and 30 centimetres wide. What is its area in square metres?

A hint

Convert the width to metres before multiplying.

Work through the solution

$30\,\mathrm{cm}=0.30\,\mathrm m$, so the area is $(2\,\mathrm m)(0.30\,\mathrm m)=0.60\,\mathrm{m^2}$.

A physical measurement is a number with a unit. Convert both together before comparing quantities.

0.2 Partial derivatives: change one input at a time#

Suppose a calculator takes two numbers and returns $f(x,y)=x^2+3y$. Here $x$, $y$, and $f$ are dimensionless numbers. There are two different ways to change its output: change $x$, or change $y$.

The partial derivative $\partial f/\partial x=2x$ holds $y$ fixed and differentiates with respect to $x$. The other partial derivative is $\partial f/\partial y=3$. We also write them as $\partial_x f$ and $\partial_y f$. The symbol $\partial$ is a derivative with an instruction about what to hold fixed.

If a path through the inputs supplies $x=x(s)$ and $y=y(s)$, both can change. The chain rule adds their contributions:

Try $x=s$, $y=s^2$, with dimensionless $s$. Substitution gives $f=4s^2$, so $df/ds=8s$. The chain rule gives $(2s)(1)+(3)(2s)=8s$ too. Later we will use the same operation when a measurement depends on where and when we make it.

0.3 An integral adds local measurements#

Water flows into an initially empty tank at a rate $q(t)$. The rate is measured in litres per second. If it is nearly constant during a short time $\Delta t$, the added volume is approximately $q(t)\Delta t$. Add the contributions from many short time steps. An integral is the limit as those steps become arbitrarily fine:

For $q(t)=bt$ with $b=2\,\mathrm{litres/s^2}$, the flow increases steadily. After three seconds,

The brackets mean evaluate at the upper limit and subtract the value at the lower limit. The $dt$ contributes a time unit: flow rate multiplied by time gives volume. This is why checking units is part of understanding an integral.

The same addition works across a region of space. Mass density, written $\rho$, is mass per volume. A small piece of volume $dV$ contains approximately $\rho\,dV$ of mass. Adding all the pieces gives $M=\int\rho\,dV$. For a uniform density $2\,\mathrm{kg/m^3}$ in a rectangular box of volume $3\,\mathrm{m^3}$, the integral is simply $M=6\,\mathrm{kg}$.

0.4 A differential equation needs starting measurements#

Near Earth’s surface, an ideal falling ball speeds up downward at nearly $g=9.8\,\mathrm{m/s^2}$. Ignore air resistance and restrict attention to heights small compared with Earth’s radius. Choose upward as the positive $x$ direction. Its acceleration is then negative:

This is a differential equation: it specifies a derivative of the unknown function $x(t)$. Integrate once and then again:

The two constants have physical meanings: starting position $x_0$ and starting velocity $v_0$. For a ball released from rest 20 metres above the ground, set $x_0=20\,\mathrm m$ and $v_0=0$. After one second, the formulas give $x=15.1\,\mathrm m$ and $v=-9.8\,\mathrm{m/s}$. The negative velocity means downward motion. The model describes the fall until the ball hits the ground.

The acceleration law alone does not say whether the ball was dropped, thrown up, or thrown down. Those choices give different motions that obey the same law. A law plus starting measurements gives a prediction.

WORKED EXAMPLE

A law needs a starting state

What is missing when someone hands you an equation of motion?

Builds onDerivatives

See the idea

A rule for change is not one particular history. Two identical swings obey the same equation but start at different positions and speeds. A solution must pass two tests: it must obey the differential equation everywhere in its domain, and it must match the stated initial measurements.

Work it out

1. 01
   Solve proportional growth

   For $dy/dt=ky$, the exponential $y(t)=Ae^{kt}$ works because differentiating it multiplies it by $k$. Initial data $y(0)=y_0$ fix $A=y_0$. Positive $k$ means growth, negative $k$ decay. The product $kt$ must be dimensionless.

   $$y(t)=y_0e^{kt}.$$

   Why this step works Differentiate the proposed solution and then check the starting value.

2. 02
   Build an oscillator from two independent solutions

   A dot means a time derivative: $\dot x=dx/dt$ and $\ddot x=d^2x/dt^2$. Let $\omega>0$ be a constant with units of inverse time. The oscillator equation $\ddot x+\omega^2x=0$ says that acceleration points toward $x=0$ and grows with displacement. Both $\cos(\omega t)$ and $\sin(\omega t)$ return minus $\omega^2$ times themselves after two derivatives. The constants multiplying them are fixed by $x(0)=x_0$ and $\dot x(0)=v_0$.

   $$x(t)=x_0\cos(\omega t)+\frac{v_0}{\omega}\sin(\omega t),\qquad \omega>0.$$

   Why this step works The position and velocity supply the two constants in a second-order equation.

3. 03
   Rewrite second order as a system

   Introduce a second variable $v=\dot x$. The pair $\dot x=v$, $\dot v=-\omega^2x$ advances the state $(x,v)$. A numerical method approximates this advance in small time steps; it must be checked against an exact solution and an error estimate.

   $$\frac{d}{dt}\begin{pmatrix}x\\v\end{pmatrix}=\begin{pmatrix}0&1\\-\omega^2&0\end{pmatrix}\begin{pmatrix}x\\v\end{pmatrix}.$$

   Why this step works Adding a state variable changes the packaging, not the equation.

4. 04
   Recognize rotating components

   The transport system $\dot A=-\Omega B$, $\dot B=\Omega A$ has solution $A=A_0\cos\Omega t-B_0\sin\Omega t$, $B=A_0\sin\Omega t+B_0\cos\Omega t$. Differentiate and verify both equations. Also differentiate $A^2+B^2$: the cross terms cancel.

   $$\frac{d}{dt}(A^2+B^2)=0.$$

   Why this step works Components can change while the length of the represented arrow stays fixed.

Go deeper

Initial and boundary data answer different questions. For $\ddot x+\omega^2x=0$, prescribing $x(0)$ and $\dot x(0)$ determines one solution. Prescribing $x(0)=x(\pi/\omega)=0$ leaves any $x=B\sin\omega t$, so this boundary-value problem is not unique. Existence and uniqueness of a first-order system are local under suitable smoothness conditions on its right-hand side; a solution can still become unbounded in finite time. For example, $\dot y=y^2$ with dimensionless $t$ and $y(0)=1$ gives $y=1/(1-t)$, valid until $t=1$. Always state the interval on which you use it.

Further readingMIT · Differential equations and initial data

Test the idea

FIRST, PREDICT

A particle obeys $\ddot x=-4x$. You know only $x(0)=0$. Is its history determined?

Yes: it stays at the origin.No solution exists at the origin.No: its initial velocity is still free.

Compare the reasoning

Yes: it stays at the origin.

That solution requires zero initial velocity too.

No solution exists at the origin.

Both the stationary solution and moving oscillator solutions can pass through the origin.

No: its initial velocity is still free.

The coefficient of sin(2t) is not fixed by the initial position.

A hint

Try a sine wave with an arbitrary amplitude.

NOW CHANGE THE EXAMPLE

Use seconds and metres. For $\ddot x+4x=0$, $x(0)=0$, $\dot x(0)=6$, find $x(\pi/4)$.

A hint

Here $\omega=2\,\mathrm{s^{-1}}$.

Work through the solution

$x=3\sin(2t)$, so $x(\pi/4)=3\,\mathrm m$.

Check a proposed history by substitution and initial data; neither check replaces the other.

0.5 The mechanics we will use#

A force is a push or pull, such as a floor pushing on your shoes. To calculate what it does, introduce momentum $\mathbf p=m\mathbf v$: mass times velocity. Bold symbols here are arrows with a magnitude and a direction. In Newton’s mechanics the total force changes momentum:

The force unit is the newton: $1\,\mathrm N=1\,\mathrm{kg\,m/s^2}$. These equations describe ordinary speeds well; Chapter 3 develops their relativistic replacements.

Push a cart with a constant force of 6 newtons while it moves 2 metres in the direction of the push. The force transfers $6\times2=12$ joules of energy to the cart. This transfer is called work. For a force along a straight track,

A force in the direction of motion does positive work; a force opposing the motion does negative work. One joule is one newton metre, so $1\,\mathrm J=1\,\mathrm{kg\,m^2/s^2}$.

Where does this energy appear? For a cart of constant mass whose wheels have negligible mass, use $F=ma$ and $v=dx/dt$:

Thus the work changes the quantity $K=mv^2/2$, called kinetic energy. If our cart starts at rest and no other force does work, it gains 12 joules of kinetic energy. For a mass of 6 kilograms, $mv^2/2=12\,\mathrm J$ gives $v=2\,\mathrm{m/s}$.

A thrown ball gives us another way to store energy. As it rises, gravity does negative work and the ball loses kinetic energy. Near Earth’s surface, we assign it gravitational potential energy $U=mgh$, where $h$ is its height above a chosen zero. A rise of $\Delta h$ increases $U$ by $mg\Delta h$, exactly the kinetic energy lost to gravity. Ignoring air resistance, $K+U$ stays constant. On the way down, the exchange reverses.

A spring lets us watch this exchange repeatedly. Attach a cart to a spring on a level, frictionless track. Call its displacement from the unstretched position $q$, positive to the right. A stretched spring pulls the cart left; a compressed spring pushes it right. Its stiffness $k$ is the restoring force per metre of displacement. The experiment starts with a 1 kg cart, a stiffness of 1 N/m, and a stretch of 1 metre.

Release the cart and follow the two energy amounts. At the middle, the spring is unstretched but the cart is moving. At either end of the motion, the cart is momentarily at rest but the spring stores energy. The cart’s momentum changes sign when it reverses; its kinetic energy remains nonnegative.

MECHANICS LAB / 01

Watch the energy change places.

Release a stretched spring. Follow the cart through the middle, then watch it turn. The two energy amounts change; their sum stays fixed.

The cart on its track

Momentarily at restMomentum 0.00 kg m/s

Total energy0.50 J

Motion · kinetic energy0.00 J

Spring · potential energy0.50 J

At a turning point the cart is momentarily at rest. The spring force is still acting, so the motion reverses.

What is being calculated?

The default cart has mass 1 kg and spring stiffness 1 N/m. It starts 1 m to the right of the unstretched position, at rest. The track has no friction; the spring has negligible mass. Motion is shown at one simulated second per second.

Force and momentum carry a direction. Energy is a number. For this experiment, $p=m\frac{dq}{dt},\quad F=-kq$, while $K=\frac{p^2}{2m},\quad U=\frac12kq^2$. The bar is divided in the ratio of these two energies. Its total is $E=K+U$.

The motion obeys $m\frac{d^2q}{dt^2}=-kq$. Write $\omega=\sqrt{k/m}$; its units are inverse seconds. With starting displacement $q_0$ and momentum $p_0$, the solution is

$$q(t)=q_0\cos(\omega t)+\frac{p_0}{m\omega}\sin(\omega t)$$

Differentiating twice gives the required acceleration. At time zero it reproduces the starting position and momentum. One complete cycle takes $T=2\pi\sqrt{m/k}$. The model has no damping, impacts, external driving, or relativistic corrections.

For this ideal spring the force is $F=-kq$, called Hooke’s law. To stretch it slowly, we apply the opposite force, $kq$. The work stored in the spring is

This also gives positive stored energy for compression, where $q<0$: our applied force and displacement both point left. The default stretch stores $0.5\,\mathrm J$. At the middle, all of that energy is kinetic, so $mv^2/2=0.5\,\mathrm J$ gives a speed of $1\,\mathrm{m/s}$. On successive passes the momentum is $+1$ or $-1\,\mathrm{kg\,m/s}$, although the kinetic energy is the same. A real spring can dissipate energy into heating; this experiment omits that effect so the exchange is visible on its own.

More generally write $U=m\Phi$, where $\Phi$ is potential energy per unit mass. Outside a spherical body of mass $M$,

Here $r$ is distance from the center and $G_N$ is Newton’s gravitational constant. This choice makes $\Phi$ approach zero far away. This is Newton’s model; the near-Earth experiments here lie in its useful range. Later chapters quantify its limits at high speeds and near very compact massive objects. The gradient $\boldsymbol\nabla\Phi$ is the arrow whose Cartesian components are the partial derivatives of $\Phi$. It points toward fastest increase. For this spherical example, differentiating with respect to $r$ gives $d\Phi/dr=G_NM/r^2$. The minus sign in the acceleration law therefore gives inward acceleration of magnitude $G_NM/r^2$.

Pressure is force per area. A gas exerts pressure on a wall because collisions transfer momentum to it. The pressure unit is the pascal, $1\,\mathrm{Pa}=1\,\mathrm{N/m^2}$. Pressure will matter when we ask how fluids move and how they affect gravity. The worked example below first shows how pressure can transfer energy.

WORKED EXAMPLE

Motion, stored energy, and pressure work

How can a force exchange energy without making energy disappear?

Builds onChain rule

See the idea

Force measures how momentum changes. Work measures how much energy that force transfers while its point of application moves. A book held motionless can experience a force without receiving mechanical work. These statements concern an ideal point particle or a specified mechanical system; your muscles can consume chemical energy while holding the book.

Work it out

1. 01
   Differentiate kinetic energy

   For constant mass $m$, momentum is $\mathbf p=m\mathbf v$ and Newton’s law is $\dot{\mathbf p}=\mathbf F$. Dot with velocity and use the product rule. Integrating relates work to kinetic-energy change.

   $$\frac{d}{dt}\left(\frac12m v^2\right)=\mathbf F\cdot\mathbf v,\qquad \Delta K=\int\mathbf F\cdot d\mathbf x.$$

   Why this step works Only the force component along the instantaneous displacement contributes to work.

2. 02
   Introduce potential energy with its sign

   For a time-independent potential $U(\mathbf x)$, define $\mathbf F=-\nabla U$. Along a trajectory the chain rule gives $\dot U=\nabla U\cdot\mathbf v$. Therefore $K+U$ is constant. Near Earth $U=mgz$ predicts downward force $F_z=-mg$.

   $$\frac{d}{dt}(K+U)=0.$$

   Why this step works A loss of potential energy balances a gain of kinetic energy under the stated conservative force.

3. 03
   Use a piston to define pressure work

   Pressure $p$ is normal force per unit area. A gas pushes a piston of area $A$ outward by $dx$. It does work $pA\,dx=p\,dV$ on the surroundings. In an insulated expansion there is no heat transfer. Assume the piston moves slowly enough for the gas to pass through near-equilibrium states, with negligible friction; this is the quasistatic limit used here. Its internal energy $E$, the energy stored in its microscopic motion and interactions, decreases by the work done. Heat supplied to the gas, called $\delta Q$, adds energy to the gas.

   $$dE=\delta Q-p\,dV.$$

   Why this step works The sign follows which system does the work; pressure has the same dimensions as energy density.

Go deeper

Potential energy has an arbitrary zero. Replacing $U$ by $U+C$ for constant $C$ leaves $\mathbf F=-\nabla U$ unchanged and preserves every energy difference. It changes the energy reference while preserving the motion. Pressure work also provides a useful unit check: $1\,\mathrm{Pa}=1\,\mathrm{N/m^2}=1\,\mathrm{J/m^3}$. Thus pressure times a volume change has energy units. “Insulated” means no heat transfer; it does not prevent work through a moving boundary. The symbol $\delta Q$ denotes energy transferred as heat during a process, while $dE$ is the change in the system’s stored energy. Heat is a transfer, not an extra stored quantity to add to $E$.

Further readingDavid Tong · General relativity David Tong · Statistical physics

Test the idea

FIRST, PREDICT

In an insulated quasistatic expansion against positive pressure, what happens to the gas’s internal energy?

It is constant because insulation prevents all transfers.It decreases because the gas does work.It increases because its volume increases.

Compare the reasoning

It is constant because insulation prevents all transfers.

Insulation blocks heat transfer; mechanical work can still cross the boundary.

It decreases because the gas does work.

With no heat supplied, dE = −p dV and dV is positive.

It increases because its volume increases.

Volume is not energy. The sign follows the direction of energy transfer.

A hint

Choose the gas as the system and track where the piston’s work comes from.

NOW CHANGE THE EXAMPLE

A $2\,\mathrm{kg}$ particle falls from rest through $5\,\mathrm m$ in constant $g=10\,\mathrm{m/s^2}$, without drag. Find its final speed.

A hint

Set the lost $mgh$ equal to the gained $mv^2/2$.

Work through the solution

$v=\sqrt{2gh}=\sqrt{100}=10\,\mathrm{m/s}$. The mass cancels.

Conservation laws become useful when the system, force law, and direction of energy transfer are explicit.

0.6 Measuring with matrices#

A map uses one metre per horizontal square and two metres per vertical square. An instruction to move two squares right and one square up has map components $v=(2,1)$, but its physical displacement is two metres right and two metres up. Pythagoras gives squared length $2^2+2^2=8$ square metres.

We can keep the map components and let a matrix carry the scale factors. With distances expressed in metres, define

The superscript $\mathsf T$ means transpose: turn the column into a row for matrix multiplication. The vertical scale factor is squared because we are computing squared length. If the second instruction is $w=(1,3)$, their physical dot product is

in square metres. The same rule accepts two arrows as inputs. Double either input while holding the other fixed, and the result doubles; adding two arrows in one input adds their results. These two properties are what linear in each input means. A rule with this property is called a bilinear form.

Squared length uses the same arrow in both inputs. Doubling that arrow doubles both inputs, so its squared length becomes four times as large. This is why a bilinear rule can describe lengths even though length squared is not linear in the arrow.

An inverse matrix undoes a linear map. It is not obtained by taking the reciprocal of every entry. For example,

Multiply them to check that the diagonal entries become one and the other entries zero. The result is the identity matrix, which leaves every input unchanged.

0.7 Flow through a box: flux and divergence#

Imagine water flowing through an imaginary box. Let the arrow $\mathbf J$ describe the mass crossing a unit area per unit time, in $\mathrm{kg/(m^2s)}$. This is a flux density. Only the part of the flow perpendicular to a face crosses that face. Multiply that component by the face area to obtain a mass flow rate.

For a numerical example, suppose 10 kilograms per second enter a one-cubic-metre box and 12 kilograms per second leave it. The box loses 2 kilograms each second. Its average mass density therefore decreases at $2\,\mathrm{kg/(m^3s)}$.

To describe a small box at any position, let its side lengths be $\Delta x,\Delta y,\Delta z$. Write $J^x$ for the flux-density component pointing along $x$; the superscript is a direction label, not a power. The left and right faces each have area $\Delta y\Delta z$, so their net outward flow is

The approximation is the local derivative rule from §0.1. Repeat it for the other two pairs of faces, add the results, and divide by the box’s volume. As the box shrinks, the result is

This quantity is divergence, the net outflow per volume. Its units here are $\mathrm{kg/(m^3s)}$. If more mass leaves than enters, the density $\rho$ inside decreases. Local conservation of mass is therefore

Add this accounting over many little boxes. Flow across a shared face is outflow from one box and inflow to its neighbor, so it cancels. Only the outer boundary remains. That cancellation is the idea behind the divergence theorem: total divergence over a volume equals net flux through its boundary.

0.8 Estimating with a small parameter#

Suppose we need $\sqrt{1.04}$. It is close to $\sqrt1=1$. How much should we add? Let $f(q)=\sqrt{1+q}$. Its value at zero is 1 and its derivative there is $f'(0)=1/2$. The local prediction from §0.1 gives $f(q)\simeq1+q/2$, so $\sqrt{1.04}\simeq1.02$.

The symbol $\simeq$ means approximately equal. Here $q=0.04$ is a dimensionless fractional change. The same method gives several useful estimates:

Each approximation discards terms beginning at order $q^2$. For $q=0.01$, those terms are on the scale of $10^{-4}$, though the coefficient depends on the function. “Small” requires a comparison: an extra centimetre compared with one metre is the ratio $0.01$. An extra centimetre compared with a millimetre is not small.

Check dimensions before arithmetic. An acceleration must have length/time squared units on both sides of its equation. A distance and a time cannot be added directly. Different units of the same dimension, such as metres and centimetres, must first be expressed consistently. These checks often catch a mistake before a page of algebra does.

0.9 Check your understanding#

Try these before revealing the answers. They test operations taught above.

1. If $x(t)=Bt^3$ with $B=2\,\mathrm{m/s^3}$, what are its velocity and acceleration?

$v=3Bt^2$ and $a=6Bt$. At $t=1\,\mathrm s$, these are $6\,\mathrm{m/s}$ and $12\,\mathrm{m/s^2}$. The units of $B$ make both dimension checks work.

2. For $f(x,y)=xy^2$, what is $df/ds$ along $x=s$, $y=2s$? All inputs are dimensionless.

Substitution gives $f=4s^3$, so $df/ds=12s^2$. The chain rule gives $y^2(1)+2xy(2)=4s^2+8s^2$, the same result.

3. A tank receives water at $2\,\mathrm{litres/s}$ for three seconds. How much volume is added?

$\Delta V=\int_0^{3\,\mathrm s}(2\,\mathrm{litres/s})dt=6\,\mathrm{litres}$. The time unit cancels the rate’s denominator.

4. Why does $d^2x/dt^2=-g$ fail to specify one particular throw?

It specifies acceleration but leaves starting position and starting velocity free. Integrating twice exposes those two constants.

5. Estimate $\sqrt{1.04}$ to first order. What is the small parameter?

Write $1.04=1+q$ with $q=0.04$. Then $\sqrt{1.04}\simeq1+0.04/2=1.02$. Squaring this estimate gives $1.0404$, close to the original input. The first discarded term in the square-root expansion is $-q^2/8=-0.0002$.

Chapter 1 uses these ideas to ask a new question: what would a scale read if you and the scale were falling together?