Kinematics from Scratch
A flying arrow never moves. A Greek thinker named Zeno argued for this strange claim about 2,500 years ago.
Look at a flying arrow at any single instant. At that instant the arrow fills one spot, exactly its own length, just like an arrow lying still. Time is made of instants. So when does the arrow ever move?
Zeno's arrow paradox, retold in our words (Greece, 5th century BC; Zeno's argument reaches us through Aristotle)
Zeno's argument is hard to beat. Every frame of a video shows a speeding car standing still. Yet the car's speedometer shows a speed at every instant. If no instant holds any motion, where does the speedometer's number come from?
Guess first. A cart rolls up a smooth ramp, stops for an instant at the top, and rolls back down. At the top, the cart's velocity is zero. What is the cart's acceleration at that instant?
- Zero, because the cart is not moving
- Not zero: the same as everywhere else on the ramp
- Not zero, and bigger than anywhere else on the ramp
- The answer depends on how fast the cart started
Show answer
B. Not zero: the same as everywhere else on the ramp
Acceleration measures how fast the velocity changes, not how fast the cart goes. At the top, the velocity is changing from uphill to downhill, so the acceleration is not zero. On a smooth ramp the acceleration is the same all the way up, at the top, and all the way down. Match 3 in the toy below is this motion, laid flat.
Drive the car
Try it: drag the blue car and watch the graphs of position \(x\), velocity \(v\) and acceleration \(a\) draw; then pick a Match and steer so the blue \(v\) line follows the dashed line.
In Match 3, watch the bottom graph as the velocity crosses zero. Here is the strange part: at each instant the car sat in one spot, yet the speedometer showed a speed. Where did the speedometer's number come from?
Where is the car?
Start with what one instant does tell us: where the car is. The car's distance from a starting mark is called the car's position, written \(x\). On the track above, a bigger \(x\) means farther right.
A single snapshot shows \(x\) and nothing else. Each dot below, on its own, looks like a parked car.
Which car is speeding up?
- Car A
- Car B
- Neither: snapshots cannot show speed
- Both cars
Show answer
B. Car B
Car B. The gaps between Car B's dots grow, so Car B covers more ground each half second. Car A's gaps are all equal: a steady speed. No single dot shows either fact. Speed shows up only in the gap between two dots.
Divide the distance covered by the time taken, and you get the average velocity over the gap:
Here \(\Delta x\) (say "delta x") is the change in position, and \(\Delta t\) is the time between the two snapshots. But a speedometer shows the speed right now, not an average over half a second. What happens as the gap between the snapshots shrinks toward zero?
Shrink the gap
Take a car that starts from rest and speeds up, with position \(x = t^2\): 1 m after 1 s, 4 m after 2 s, 9 m after 3 s. Point P is the moment \(t = 2\) s. Point Q comes a gap \(\Delta t\) later.
Guess first. As the gap between P and Q shrinks toward zero, what happens to \(\Delta x / \Delta t\)?
- The ratio drops to zero: no time, no distance
- The ratio blows up: dividing by almost nothing
- The ratio settles on one number
- The ratio jumps around and never settles
Show answer
C. The ratio settles on one number
\(\Delta x\) and \(\Delta t\) shrink together, so the ratio closes in on one number: 4 m/s, the car's speed at the instant P.
Try it: drag the \(\Delta t\) slider all the way to the right and watch where the big number settles.
The ratio goes 5, 4.5, 4.1, 4.01, 4.001, closing in on exactly 4 m/s. Zoomed in far enough, the curve looks like a straight line through P, called the tangent line. The tangent line's slope is the velocity at the instant P.
Velocity is how fast the position changes. But in the toy the velocity changed too: the example car sped up, cruised, then braked. How fast does the velocity change?
When velocity changes
Use the same trick on the velocity graph. How fast the velocity changes is called acceleration, written \(a\): the slope of the velocity graph.
An acceleration of \(2\ \text{m/s}^2\) means the velocity grows by 2 m/s every second. A negative acceleration means the velocity shrinks, or grows in the negative direction.
Back to the warm-up guess, the cart at the top of the ramp. Try it: find where the blue velocity line crosses zero, then read the grey acceleration line at that moment.
| t (s) | velocity v (m/s) | acceleration a (m/s²) |
|---|---|---|
| 0 | 2.4 | -0.8 |
| 1 | 1.6 | -0.8 |
| 2 | 0.8 | -0.8 |
| 3 | 0 | -0.8 |
| 4 | -0.8 | -0.8 |
| 5 | -1.6 | -0.8 |
| 6 | -2.4 | -0.8 |
The velocity crosses zero at 3 s, when the cart stops at the top. The acceleration never flickers: \(-0.8\ \text{m/s}^2\) the whole time, because the velocity keeps falling at the same rate, from uphill, through zero, to downhill. Zero velocity for one instant does not mean the velocity has stopped changing.
Slopes take us down a chain: position gives velocity, and velocity gives acceleration. Can we climb back up? If a graph shows only the velocity, can we work out where the car went?
Area climbs back up
Yes, with the reverse of a slope: an area. Slice the time into thin strips of width \(\Delta t\). During one thin strip the velocity hardly changes, so the car moves about \(v\,\Delta t\): the strip's height times the strip's width.
Add up all the strips and you get the total change in position, called the displacement. As the strips get thinner, the sum becomes exactly the area under the velocity graph. Area below the time axis counts as negative, because below the axis the car moves left.
Keep this cheat sheet handy for the game below.
| The line is | Position graph | Velocity graph |
|---|---|---|
| Flat | Standing still | Steady velocity |
| Sloped | Steady velocity | Steady acceleration |
| Steeper | Faster | Bigger acceleration |
| Crossing zero | Passing \(x = 0\) | Turning around |
| Area under the line | Nothing useful | Displacement |
Try it: for each of the three velocity graphs, pick what the car does, then watch the car act the graph out.
Slopes and areas work for any motion, even a wobbly hand-drawn one. Many real motions are simpler: the acceleration stays the same, like a car braking steadily. With a steady acceleration, can we predict exactly where a car will stop?
Predict the stop
The car below has steady brakes, the same in every round. Try it: watch round 1, then tap the road where you think the car will stop and press Brake.
Did round 2 surprise you? The common guess is 20 m: twice the speed, twice the distance. The car needed 40 m. Why does twice the speed take four times the distance?
Why four times?
Draw the velocity graph of each stop. With steady brakes the velocity falls in a straight line to zero, so each graph is a triangle. Each stopping distance is the area of its triangle.
Twice the speed makes the triangle twice as tall. With the same brakes, shedding twice the speed takes twice as long, so the triangle is also twice as wide. Twice as tall and twice as wide makes four times the area.
The same picture gives a formula. The triangle's height is the starting speed \(v_0\). The triangle's width is the stopping time \(v_0 / a\), where \(a\) is the size of the braking acceleration (5 m/s² in the game). The area is half of height times width:
Check round 2: \(v_0 = 20\ \text{m/s}\) and \(a = 5\ \text{m/s}^2\) give \(d = 400 / 10 = 40\) m. On a real road a driver also needs time to react, still at full speed. That reaction distance only doubles when the speed doubles; the braking distance quadruples.
The triangle worked because the acceleration never changed. What else can a steady acceleration predict?
Three equations
With a steady acceleration, every velocity graph is a straight line, and the area under a straight line is a trapezoid. One straight line and one trapezoid give the three equations on the AP equation sheet. Each equation leaves out one quantity, so pick the equation that leaves out the quantity you don't know and don't need.
| Equation | Leaves out |
|---|---|
| \(v = v_0 + at\) | position \(x\) |
| \(x = x_0 + v_0 t + \tfrac12 a t^2\) | final velocity \(v\) |
| \(v^2 = v_0^2 + 2a(x - x_0)\) | time \(t\) |
A car brakes steadily from 30 m/s to a stop, with \(a = -6\ \text{m/s}^2\). You want the stopping distance, and nobody tells you the time. Which equation goes straight to the answer?
- \(v = v_0 + at\)
- \(x = x_0 + v_0 t + \tfrac12 a t^2\)
- \(v^2 = v_0^2 + 2a(x - x_0)\)
Show answer
C. \(v^2 = v_0^2 + 2a(x - x_0)\)
The third equation leaves out time. Put in \(v = 0\), \(v_0 = 30\ \text{m/s}\) and \(a = -6\ \text{m/s}^2\): \(0 = 900 - 12\,(x - x_0)\), so \(x - x_0 = 75\) m. The stop game's formula \(v_0^2 / 2a\) is the same equation with \(v = 0\).
The derivation: where the three equations come from
Take a steady acceleration \(a\). The velocity graph is then a straight line that starts at \(v_0\) and has slope \(a\). After a time \(t\), the velocity has grown by \(at\). The straight line gives the first equation: \(v = v_0 + at\).
The displacement is the area under the velocity line from \(0\) to \(t\). The area is a trapezoid with parallel sides \(v_0\) and \(v\), and width \(t\). So \(x - x_0 = \tfrac12 (v_0 + v)\,t\).
In the trapezoid, replace \(v\) with \(v_0 + at\) from the first equation: \(x - x_0 = \tfrac12 (2v_0 + at)\,t = v_0 t + \tfrac12 a t^2\). The result is the second equation.
For the third equation, remove \(t\). The first equation gives \(t = (v - v_0)/a\). Put this expression for \(t\) into the trapezoid: \(x - x_0 = \tfrac12 (v + v_0)\cdot\dfrac{v - v_0}{a} = \dfrac{v^2 - v_0^2}{2a}\). Multiply both sides by \(2a\) and rearrange: \(v^2 = v_0^2 + 2a(x - x_0)\).
All three equations need a steady acceleration. The runs you drew by hand in the first toy had a changing acceleration, so for those runs you read slopes and areas off the graphs instead.
Back to the arrow
Look back at the trail. Driving the car and freezing time at P gave one idea: the slope. The slope answered Zeno and explained the cart at the top of the ramp.
Reading graphs and predicting stops gave the reverse idea: the area. The area showed why twice the speed takes four times the distance, and built the three equations.
So Zeno's arrow does move. At every instant the arrow has a position, and the slope of those positions is the arrow's speed.
Ask Phy
- Why is the area under a velocity graph the displacement and not the total distance traveled?
- What do the position, velocity and acceleration graphs look like for a ball thrown straight up?
- With a 1.5 s reaction time, how far does a car travel before stopping from 10 m/s and from 20 m/s?