Motion in One Dimension

A car is doing 25 m/s and the driver stands on the brakes. Before the car stops it travels a certain distance. Double the speed to 50 m/s, brake just as hard, and the car does not travel twice as far. It travels four times as far. Nothing in that sentence is about forces or engines or tyres. It follows from the bookkeeping of position, velocity and time alone.

That bookkeeping is what this lesson is. In the previous lesson you were reading slopes and areas off graphs and worrying about how much you could trust a number. Now you get the small set of relations that govern any object moving along a line, and the habits that turn a wordy problem into two equations you can actually solve.

Position, velocity, acceleration

Pick a line to move along and an origin on it. The position $x$ is how far the object is from the origin, with a sign: negative means the other side. It is measured in metres.

Velocity is the rate at which position changes. If the object moves from $x_1$ at time $t_1$ to $x_2$ at time $t_2$, the average velocity over that stretch is

$\bar v = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\Delta x}{\Delta t}.$

Shrink the interval until it is vanishingly small and you get the velocity at one instant, which is written $v = dx/dt$. That is the derivative of $x$ with respect to $t$, and if you have not met derivatives, here is all you need for now: the derivative of $t^n$ is $nt^{n-1}$, and the derivative of a sum is the sum of the derivatives. So if $x(t) = 3 + 4t - t^2$ in metres, then $v(t) = 4 - 2t$ in m/s, and at $t = 3.0$ s that is $4 - 6 = -2.0$ m/s. Negative: the cart is heading back toward the origin.

Acceleration is the same operation applied once more. It is the rate at which velocity changes, $a = dv/dt$, in metres per second per second, m/s². Note what it is not. Acceleration is not "how fast you are going" and it is not "speeding up". A car braking from 30 m/s has a negative acceleration while its velocity is still large and positive. A car reversing and speeding up has negative velocity and negative acceleration. The two signs are independent, and the only rule connecting them is this: when $v$ and $a$ have the same sign the speed grows, and when they have opposite signs the speed shrinks.

The constant-acceleration equations, derived

Most problems hold $a$ fixed. That single assumption gives you everything.

Constant $a$ means velocity changes by the same amount every second, so after a time $t$ it has changed by $at$:

$v = v_0 + at.$

Here $v_0$ is the velocity at the moment you start your clock. On a velocity-time graph this is a straight line with slope $a$ and intercept $v_0$.

Now for position. The area under a velocity-time graph is the displacement, because a thin strip of width $dt$ and height $v$ has area $v\,dt$, which is the small distance covered in that sliver of time. Add up the strips and you have $\Delta x$. Under a straight line the region is a trapezoid: two parallel sides of length $v_0$ and $v$, separated by a base of length $t$. Its area is the average of the parallel sides times the base, so

$\Delta x = \frac{v_0 + v}{2}\,t.$

That is worth a name. The bracket is the average velocity, and for constant acceleration (and only for constant acceleration) it equals the plain average of the starting and ending velocities.

Substitute $v = v_0 + at$ into it and the trapezoid splits into a rectangle plus a triangle:

$\Delta x = \frac{v_0 + v_0 + at}{2}\,t = v_0 t + \tfrac{1}{2}a t^2.$

Finally, take $t = (v - v_0)/a$ from the first equation and put it into the trapezoid form:

$\Delta x = \frac{v_0 + v}{2}\cdot\frac{v - v_0}{a} = \frac{v^2 - v_0^2}{2a},$

which rearranges to

$v^2 = v_0^2 + 2a\,\Delta x.$

That last one has no $t$ in it. Whenever a problem hands you two speeds and a distance and never mentions seconds, it is asking for that equation. It is also the answer to the puzzle at the top: stopping from $v_0$ to rest takes $\Delta x = v_0^2/(2|a|)$, which goes as the square of the speed.

Worked example 1. A cyclist is moving at 4.0 m/s and accelerates uniformly at 1.2 m/s² over the next 60 m. What is her final speed?

Write down what you have: $v_0 = 4.0$ m/s, $a = 1.2$ m/s², $\Delta x = 60$ m, and $v$ is wanted. No time appears anywhere, so reach for the timeless equation.

$v^2 = v_0^2 + 2a\Delta x = (4.0)^2 + 2(1.2)(60) = 16 + 144 = 160\ \mathrm{m^2/s^2}.$

So $v = \sqrt{160} = 12.6$ m/s. Taking the positive root is a choice: she was moving forward and speeding up, so the negative root is not physical here.

Averages, and the trap in them

Average velocity is always $\Delta x/\Delta t$: total displacement over total time. Average speed is total distance over total time. Neither is the average of the speeds you held, unless you held each one for the same time.

Say you drive the first half of a trip's distance at 40 mph and the second half at 60 mph. Call the total distance $2d$. The first leg takes $d/40$ hours, the second $d/60$. The total time is

$\frac{d}{40} + \frac{d}{60} = d\left(\frac{3}{120} + \frac{2}{120}\right) = \frac{5d}{120} = \frac{d}{24}.$

Average speed is total distance over total time: $2d \div (d/24) = 48$ mph. Not 50. You spend longer on the slow half, so the slow half gets more weight. Same arithmetic, different numbers: 40 km at 80 km/h then 40 km at 40 km/h takes $0.5 + 1.0 = 1.5$ h for 80 km, giving 53.3 km/h.

Free fall and the moving platform

Near the ground, ignoring air resistance, everything falls with the same downward acceleration $g$. Use $g = 9.8$ m/s², or $10$ m/s² when a problem says so. Mass does not appear. A bowling ball and a marble released together land together.

The acceleration is almost never the hard part. The initial condition is. An object released from a moving platform leaves with the platform's velocity: it does not start at rest just because you let go of it.

Worked example 2. A balloon rises at a steady 5 m/s. At a height of 60 m a sandbag is released. With $g = 10$ m/s², how long until it hits the ground?

Set up axes first, before touching a number: let $y$ be measured upward from the ground, so $y_0 = +60$ m, the bag's initial velocity is $v_0 = +5$ m/s (it was moving up with the balloon), and $a = -10$ m/s².

$y(t) = 60 + 5t - 5t^2.$

The bag lands when $y = 0$:

$5t^2 - 5t - 60 = 0 \quad\Longrightarrow\quad t^2 - t - 12 = 0 \quad\Longrightarrow\quad (t-4)(t+3) = 0.$

So $t = 4.0$ s, discarding the negative root, which describes a time before the bag existed. Notice that the bag first goes up: its velocity is zero at $t = v_0/g = 0.5$ s, at a height of $60 + 5(0.5) - 5(0.25) = 61.25$ m. If you had set $v_0 = 0$ you would have got $t = \sqrt{12} = 3.46$ s, too small.

Motion in stages

Real problems rarely hold one acceleration for the whole trip. The method is always the same: break the motion at every instant where $a$ changes, solve each stage separately, and pass the final velocity and position of one stage in as the initial conditions of the next. Draw the velocity-time graph while you do it; the shape usually tells you the answer.

A rocket sled starting from rest at 8.0 m/s² for 10 s reaches $v = 80$ m/s having covered $\tfrac12(8)(10)^2 = 400$ m. Stage two starts at 80 m/s and brakes at 4.0 m/s², needing $80/4 = 20$ s and covering $80^2/(2\cdot 4) = 800$ m. Total: 1200 m. Each stage is a fresh set of initial conditions.

Sometimes the graph shape alone is enough. A tram accelerates uniformly from rest, then decelerates uniformly to rest, 200 m in 20 s. Its velocity-time graph is a triangle of base 20 s and peak $v_{\max}$, so the area is $\tfrac12(20)v_{\max} = 200$, giving $v_{\max} = 20$ m/s. You never need to know where the switch happened.

Worked example 3. A shuttle must cover 1000 m, starting and ending at rest. It can accelerate at most 2.0 m/s² and brake at most 4.0 m/s², with no speed limit. What is the minimum travel time?

Think about what the fastest possible trip looks like. Any moment spent not accelerating as hard as allowed is a moment you could have been going faster, so the optimal run is: full acceleration up to some peak speed $v$, then immediately full braking down to rest. The velocity-time graph is a triangle with peak $v$.

The accelerating leg covers $v^2/(2\cdot 2.0) = v^2/4$ and takes $v/2.0$. The braking leg covers $v^2/(2\cdot 4.0) = v^2/8$ and takes $v/4.0$. The distances must add to 1000 m:

$\frac{v^2}{4} + \frac{v^2}{8} = \frac{3v^2}{8} = 1000 \quad\Longrightarrow\quad v^2 = \frac{8000}{3} = 2666.7,$

so $v = 51.64$ m/s. The total time is

$t = \frac{v}{2.0} + \frac{v}{4.0} = 0.75v = 0.75(51.64) = 38.7\ \mathrm{s}.$

If the problem adds a speed limit, the triangle gets its top sliced off into a trapezoid: accelerate to the limit, cruise, then brake, and the cruise distance is whatever is left over after the two ramps.

When the acceleration is not constant

The four boxed equations are worthless the moment $a$ changes with time. The definitions still hold, though, and they are all you need.

If you are given $v(t)$ directly, displacement is the integral (the signed area under the curve) and distance is the area with every piece counted positive. Take a cart with $v(t) = 12 - 3t$ m/s over the first 8.0 s. It reverses at $t = 4.0$ s. Forward leg: a triangle of base 4 and height 12, area $\tfrac12(4)(12) = 24$ m. Backward leg: from 4 s to 8 s the velocity runs from 0 to $-12$ m/s, another triangle of area 24 m, but travelled backwards. The displacement is $24 - 24 = 0$. The distance is $24 + 24 = 48$ m. Whenever $v$ changes sign inside the interval, split the integral there or you will report displacement when the problem asked for distance.

A harder case is acceleration given as a function of velocity. A boat's engine cuts out at 20 m/s and afterwards $a = -kv$ with $k = 0.50$ s⁻¹. How far does it coast? Solving $dv/dt = -kv$ needs an exponential, but there is a change of variable that avoids it. Write acceleration using the chain rule:

$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\frac{dv}{dx}.$

That identity is worth memorising; it converts any velocity-dependent acceleration into a relation between $v$ and $x$. Here it gives $v\,dv/dx = -kv$, and cancelling the $v$ leaves $dv/dx = -k$: velocity falls off linearly with distance, even though it decays exponentially with time. So $\Delta v = -k\,\Delta x$, and going from 20 m/s to 0:

$\Delta x = \frac{20}{0.50} = 40\ \mathrm{m}.$

The boat takes forever to stop but only ever travels 40 m.

Recognising what a problem wants

Before you write any equation, list the five constant-acceleration symbols and mark which you know: $v_0$, $v$, $a$, $t$, $\Delta x$. Three knowns pick the equation for you.

• No time mentioned, and none wanted (use $v^2 = v_0^2 + 2a\Delta x$. "Stopping distance", "final speed over this stretch", "how high does it go".
• Time is given or asked for, and one velocity is missing) use $\Delta x = v_0t + \tfrac12 at^2$ or, more cheaply, the trapezoid form.
• Total distance with no way to find the individual stages: look for the area of the velocity-time graph as a whole shape. Triangle and trapezoid areas solve a surprising number of problems in one line.
• The word "maximum" or "minimum" with two acceleration limits (the answer is the bang-bang trip: full acceleration, then full braking, possibly with a cruise if there is a speed limit.
• Two moving objects, or a head start, or a delay) one origin, one clock, two position functions, set equal.
• Acceleration given as a formula in $t$ or $v$: the equations do not apply. Integrate, or use $a = v\,dv/dx$ if $a$ depends on $v$.
• "Distance" rather than "displacement", or anything that turns around: find where $v = 0$ and split the calculation there.

Almost every one of these is easier if you sketch the velocity-time graph first. That is the subject of the next lesson, Motion Graphs, where slopes and areas stop being a computational trick and become the fastest way to read a problem.