P1 Describing motion

In this chapter

• Choose a reference frame, and tell position, displacement and distance apart.
• Tell average speed, average velocity and instantaneous velocity apart, and read velocity as the slope of a position–time graph.
• Recognise all three kinds of acceleration (speeding up, slowing down, turning), and read acceleration as the slope of a velocity–time graph and displacement as its area.
• Derive the four constant-acceleration equations from the velocity–time graph, and choose the right one for a problem.
• Use free fall, \(g \approx 9.8\ \text{m/s}^2\), for dropped and thrown objects, and say when air resistance can be ignored.
• Follow changing acceleration in small time steps.

Where, and relative to what?

Checkpoint: Where, and relative to what? (1 of 2)

Checkpoint: Where, and relative to what? (2 of 2)

Position, displacement and distance

Position, displacement and distance

Checkpoint: Position, displacement and distance (1 of 2)

Checkpoint: Position, displacement and distance (2 of 2)

Speed and velocity

Speed and velocity

Speed and velocity

Speed and velocity

On a position–time graph, velocity is the slope: rising means moving forwards, flat means stopped, falling means coming back. Instantaneous velocity is what the average velocity approaches as the time interval shrinks: the slope of the tangent at a point, which is what a speedometer shows. (This answers Zeno's arrow: at an instant the arrow moves no distance, but it still has a velocity.)

Speed and velocity

Speed and velocity

Speed and velocity

Checkpoint: Speed and velocity (1 of 3)

Checkpoint: Speed and velocity (2 of 3)

Checkpoint: Speed and velocity (3 of 3)

Acceleration

Acceleration

• 0 to 100 km/h (27.8 m/s) in 10 s: about \(2.8\ \text{m/s}^2\).
• Slowing down is acceleration too. 20 m/s to rest in 4 s: \(a = -5\ \text{m/s}^2\). If \(v\) and \(a\) point the same way, the object speeds up; opposite ways, it slows down.
• Turning is acceleration too. A car going round a roundabout at steady speed has a changing velocity, because its direction changes.
• John Paul Stapp (1954) stopped from 632 mph (283 m/s) in about 1.4 s: an average of about \(200\ \text{m/s}^2\), around 20 g (peak 46.2 g).

On a velocity–time graph, the slope is the acceleration and the area is the displacement. Going down from position to velocity to acceleration, take slopes; going up, take areas (and add the initial conditions: where it started and how fast).

Acceleration

Checkpoint: Acceleration (1 of 2)

Checkpoint: Acceleration (2 of 2)

The equations of uniform acceleration

The equations of uniform acceleration

The equations of uniform acceleration

The equations of uniform acceleration

Recipe: choose a positive direction; list which of \(s, u, v, a, t\) you know; pick the equation that leaves out the one you neither know nor need; rearrange; check units and size.

Braking to rest (\(v = 0\), deceleration of size \(a\)) gives \(s = u^2/2a\): double the speed, four times the braking distance.

The equations of uniform acceleration

The equations of uniform acceleration

Checkpoint: The equations of uniform acceleration (1 of 3)

Checkpoint: The equations of uniform acceleration (2 of 3)

Checkpoint: The equations of uniform acceleration (3 of 3)

Falling

Falling

• Galileo's ramps (published 1638): a ball rolling down a gentle slope covers distances in the ratio 1 : 4 : 9 : 16 after 1, 2, 3, 4 equal times: the steps go 1, 3, 5, 7, just as \(s = \tfrac12 a t^2\) predicts.
• Heavy and light: Aristotle taught that heavier bodies fall faster. Stevin and de Groot (Delft, 1586) dropped two lead balls, one ten times heavier, and heard a single bang. Galileo argued it from logic (tie a light body to a heavy one). The Leaning Tower of Pisa drop is probably a legend. On the Moon (Apollo 15, 1971), a hammer and a feather landed together.

Falling

Falling

Falling

Air resistance (drag) grows with speed until it balances the weight: terminal velocity, about 55 m/s (200 km/h) for a skydiver lying flat, reached after roughly 12 s. Ignore air for dense, compact objects over short falls; don't for light, spread-out objects or long, fast falls.

Checkpoint: Falling (1 of 3)

Checkpoint: Falling (2 of 3)

Checkpoint: Falling (3 of 3)

Step by step

Step by step

Step by step

Step by step

Checkpoint: Step by step (1 of 2)

Checkpoint: Step by step (2 of 2)

Key terms

Kinematics

The description of motion, without asking what causes it.

Reference frame

The point of view you measure from: an origin, a positive direction and a clock.

Position / displacement / distance

Where something is / its change in position, with a direction / the length of the path.

Average speed / velocity

Distance ÷ time / displacement ÷ time.

Key terms

Instantaneous velocity

The velocity at one instant: the slope of the position–time graph there.

Acceleration

The rate of change of velocity (m/s²).

Uniform acceleration

Constant acceleration.

Initial conditions

Where an object starts and how fast.

Key terms

Free fall

Motion under gravity alone.

Air resistance (drag)

A push against the motion through air.

Terminal velocity

The steady top speed when drag balances weight.

Time step

A small slice of time, \(\Delta t\), in a step-by-step calculation.

Galileo's ship (1632)

Walter Arnold, the first speeding conviction (Kent, 1896)

Zeno's arrow (about 450 BC)

Usain Bolt's 9.58 s (Berlin, 16 August 2009)

John Paul Stapp (New Mexico, 10 December 1954)

Galileo's inclined planes (1638)

Simon Stevin and Jan Cornets de Groot (Delft, 1586)

The Leaning Tower of Pisa

The hammer and the feather (Apollo 15, 2 August 1971)

Numbers used:

Watch the worked example

The problem

Problem. The UK Highway Code's typical stopping distances (dry road) are thinking + braking: 20 mph 6 + 6 m; 30 mph 9 + 14 m; 40 mph 12 + 24 m; 50 mph 15 + 38 m; 60 mph 18 + 55 m; 70 mph 21 + 75 m. Why does doubling the speed double the thinking distance but roughly quadruple the braking distance?

Solution

1. Convert: 1 mph ≈ 0.447 m/s, so 30 mph ≈ 13.4 m/s.
2. Thinking distance: the car keeps its speed for the reaction time (about 0.67 s): \(s = ut = 13.4 \times 0.67 \approx 9\) m. So thinking distance \(\propto u\): 18 m at 60 mph. ✓
3. Braking distance: with constant deceleration \(a\) to rest, \(v^2 = u^2 + 2as\) with \(v = 0\) gives \(s = \dfrac{u^2}{2a}\), so braking distance \(\propto u^2\).

Solution

4. The deceleration assumed: \(a = \dfrac{u^2}{2s} = \dfrac{13.4^2}{2 \times 14} \approx 6.4\ \text{m/s}^2\), about two-thirds of \(g\).
5. Check the whole table: with that one value, \(u^2/2a\) gives 6, 14, 25, 39, 56 and 76 m: within about a metre of every figure.
6. Pictures: on a velocity–time graph, braking from \(2u\) with the same slope makes a triangle twice as tall and twice as wide: four times the area. And kinetic energy \(\propto v^2\): four times the energy to remove.

Solution

7. Checks: units \((\text{m/s})^2/(\text{m/s}^2) = \text{m}\); \(u \to 0\) gives \(s \to 0\). On wet roads (Highway Code rule 227) stopping distances are at least double.

★ Warm-up

1. A bus travels 600 m along a straight road in 50 s, then returns 200 m in 20 s. Find its average velocity for the whole trip, in metres per second. Then find its average speed and say why the two differ.

   Hint

   Average velocity uses the net displacement; average speed uses the total distance walked. Both divide by the total time.

   Your answer m/sCheck

   Solution

   Displacement +400 m in 70 s: \(400/70 \approx +5.7\ \text{m/s}\). Distance 800 m in 70 s: average speed \(800/70 \approx 11.4\ \text{m/s}\). They differ because the bus turned round, so the distance is twice the displacement.

Problem 2 (★ Warm-up)

2. A train speeds up uniformly from rest to 25 m/s in 50 s. Find its acceleration, and then how far it travels, in metres.

   Hint

   Find the acceleration from the change in velocity over the time, then the distance from the area under the velocity–time graph.

   Your answer mCheck

   Solution

   \(a = (25 - 0)/50 = 0.5\ \text{m/s}^2\). \(s = \tfrac12(0 + 25)\times50 = 625\ \text{m}\).

Problem 3 (★ Warm-up)

3. A stone is dropped from rest from a 20 m cliff. Ignoring air resistance, how long does it take to fall, in seconds? Then find how fast it is going at the bottom.

   Hint

   It starts from rest, so use \(s = \tfrac12 g t^2\) for the time, then \(v = gt\) for the speed.

   Your answer sCheck

   Solution

   \(s = \tfrac12 g t^2 \Rightarrow t = \sqrt{2s/g} = \sqrt{40/9.8} \approx 2.0\ \text{s}\). \(v^2 = 2gs \Rightarrow v = \sqrt{2\times9.8\times20} \approx 19.8\ \text{m/s}\).

Problem 4 (★ Warm-up)

4. A car moving at +15 m/s brakes with an acceleration of −3 m/s². Is it speeding up or slowing down? Then find how long, in seconds, until it stops.

   Your answer sCheck

   Solution

   Velocity and acceleration point opposite ways: slowing down. \(t = (0 - 15)/(-3) = 5\ \text{s}\).

★★ Standard

5. Two-stage journey. A cyclist speeds up from rest to 12 m/s in 4 s, rides at 12 m/s for 10 s, then brakes uniformly to rest in 3 s. Sketch the velocity–time graph, find the braking acceleration, and find the total distance, in metres.

   Hint

   Split the graph into a rising line, a flat line and a falling line. The braking acceleration is the slope of the last one; the distance is the total area.

   Your answer mCheck

   Solution

   Braking: \(a = (0 - 12)/3 = -4\ \text{m/s}^2\). Distance = area = \(\tfrac12\times4\times12 + 12\times10 + \tfrac12\times3\times12 = 24 + 120 + 18 = 162\ \text{m}\).

Problem 6 (★★ Standard)

6. A ball is thrown straight up at 20 m/s. (a) How high does it rise, in metres? (b) How long until it returns to the thrower's hand? (c) What is its acceleration at the top?

   Hint

   Take up as positive, so \(a = -9.8\ \text{m/s}^2\). At the top the velocity is zero. The return trip takes as long as the climb.

   Your answer mCheck

   Solution

   Up positive, \(a = -9.8\ \text{m/s}^2\). (a) \(0 = 20^2 - 2\times9.8\times s \Rightarrow s \approx 20.4\ \text{m}\). (b) Up: \(t = 20/9.8 \approx 2.04\ \text{s}\); the fall back takes the same time, so about 4.1 s in all (or \(s = 0\) in \(s = ut + \tfrac12 at^2\): \(t = 2u/g\)). (c) 9.8 m/s² downwards; only the velocity is zero there.

Problem 7 (★★ Standard)

7. From Bolt's splits, he passed 40 m at 4.64 s and 60 m at 6.29 s. What was his average speed over that 20 m, in m/s?

   Your answer m/sCheck

   Solution

   \(20/(6.29 - 4.64) = 20/1.65 \approx 12.1\ \text{m/s}\), times 3.6 ≈ 44 km/h.

Problem 8 (★★ Standard)

8. A car brakes at 6.4 m/s² after a reaction time of 0.67 s. Find the total stopping distance from 100 km/h, in metres. Then find it from 50 km/h, and the factor by which it grows.

   Hint

   Thinking distance is speed times reaction time. Braking distance is \(u^2/2a\). Convert km/h to m/s first (divide by 3.6).

   Your answer mCheck

   Solution

   (a) 100 km/h ≈ 27.8 m/s: thinking ≈ 18.6 m; braking \(27.8^2/(2\times6.4) \approx 60.3\ \text{m}\); total ≈ 79 m. (b) 50 km/h ≈ 13.9 m/s: thinking \(13.9\times0.67 \approx 9.3\ \text{m}\); braking ≈ 15.1 m; total ≈ 24 m. Factor ≈ 3.2: between 2 (thinking part) and 4 (braking part).

Problem 9 (★★ Standard)

9. Dropped from 2.0 m, how long does an object take to land on the Moon (\(g = 1.62\ \text{m/s}^2\)), in seconds? How long on Earth, without air?

   Hint

   Dropped from rest means \(s = \tfrac12 g t^2\), so \(t = \sqrt{2s/g}\). Only \(g\) changes between the two worlds.

   Your answer sCheck

   Solution

   \(t = \sqrt{2s/g}\): Moon \(\sqrt{4/1.62} \approx 1.57\ \text{s}\); Earth \(\sqrt{4/9.8} \approx 0.64\ \text{s}\). Check: the ratio is \(\sqrt{9.8/1.62} \approx 2.5\).

Problem 10 (★★ Standard)

10. Overtaking room. A car speeds up from 10 m/s to 30 m/s while covering 150 m. Find its acceleration, in metres per second squared, assuming it is steady.

    Hint

    You know \(u\), \(v\) and \(s\), and you neither know nor need \(t\). Choose the equation that leaves \(t\) out.

    Your answer m/s^2Check

    Solution

    \(v^2 = u^2 + 2as \Rightarrow a = (v^2 - u^2)/2s = (900 - 100)/300 \approx 2.67\ \text{m/s}^2\). Check: the average speed is 20 m/s, so the time is \(150/20 = 7.5\) s, and \(a = 20/7.5 \approx 2.67\ \text{m/s}^2\) ✓.

Problem 11 (★★ Standard)

11. The rocket sled. John Paul Stapp's sled was stopped from 283 m/s in about 1.4 s. Find the average deceleration, in units of \(g\) (\(g = 9.8\ \text{m/s}^2\)).

    Hint

    Divide the change in velocity by the time to get the acceleration in m/s², then divide by 9.8.

    Your answer Check

    Solution

    \(a = 283/1.4 \approx 202\ \text{m/s}^2\), and \(202/9.8 \approx 20.6\), so about 20.6 g. (His peak was 46.2 g: an average hides the peaks.)

★★★ Stretch

12. Spreadsheet: a falling skydiver. Use the model \(a = 9.8\,(1 - (v/55)^2)\) m/s². Starting from rest, build a table with columns \(t\), \(v\), \(a\), \(x\), using steps \(\Delta t = 1\) s: \(v_\text{new} = v + a\,\Delta t\), \(x_\text{new} = x + v\,\Delta t\). Find \(v\) after 6 s. Then repeat with \(\Delta t = 0.5\) s and 0.1 s. What do you notice?

    Hint

    In a spreadsheet, each row refers to the row above: for example B3 = B2 + C2*$H$1, where H1 holds Δt.

    Your prediction or attempt, before you look

    Show solution

    Solution

    After 6 s: about 45.3 m/s with 1 s steps, 44.3 m/s with 0.5 s steps, 43.6 m/s with 0.1 s steps. The answers settle down as the steps shrink (towards about 43.4 m/s, the exact value for this model). Big steps overshoot because they use the larger acceleration from the start of each step for the whole step. Distance after 12 s with 1 s steps: about 437 m, and about 450 m with fine steps, close to the "about 450 m to reach terminal velocity" figure for skydivers.

Problem 13 (★★★ Stretch)

13. Two cars. Car A passes a junction at a steady 20 m/s. At that moment car B starts from rest at the junction and accelerates at 2.5 m/s² in the same direction. After how many seconds does B catch A? Then find where, and how fast B is going then.

    Hint

    Write each car's position as a function of time and set them equal. Discard the solution at \(t = 0\).

    Your answer sCheck

    Solution

    \(x_A = 20t\), \(x_B = \tfrac12\times2.5\,t^2 = 1.25t^2\). Equal when \(1.25t^2 = 20t \Rightarrow t = 16\ \text{s}\), at \(x = 320\ \text{m}\). B's speed: \(2.5\times16 = 40\ \text{m/s}\), twice A's (as it must be: equal displacements in equal times means equal average velocities, and B's average is half its final speed).

Problem 14 (★★★ Stretch)

14. A well. You drop a stone into a well and hear the splash 3.0 s later. Sound travels at 340 m/s. Estimate the depth, in metres, allowing for the time the sound takes to climb back. By how much does ignoring the sound's travel time overestimate it?

    Hint

    The 3.0 s is the fall time plus the time for sound to come up. Try a fall time a little under 3 s and adjust.

    Your answer mCheck

    Solution

    Ignoring sound: \(\tfrac12\times9.8\times3^2 \approx 44\ \text{m}\). Allowing for it: fall time \(t_f\) plus \(d/340\) equals 3.0 s, with \(d = 4.9t_f^2\). Trying \(t_f = 2.88\ \text{s}\): \(d \approx 40.6\ \text{m}\), and sound takes \(40.6/340 \approx 0.12\ \text{s}\): total 3.00 s ✓. So the well is about 40.7 m deep: the simple estimate is about 8% too deep.

Answer these out loud, from memory. Then try again in a few days.

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Summary