IB Physics SL and HL · Theme A.1 · by Ahmed Gomaa

Motion Field Guide

Kinematics in pictures. Every figure moves: press play, drag the sliders and watch how position, velocity and acceleration are linked.

Position and paths Velocity Acceleration and answers Displacement and areas

Part 1Describing motionSections 01–04

01Distance and displacement

← Back to the notes: Position, displacement and distance

Choose an origin and a positive direction. Displacement is the change in position (with a sign). Distance is the total length of the path. They are equal only if the object never turns back.

Δs = s_final − s_initial · distance = total path length ≥ |Δs|

t = 0.0 s

Watch the two counters. The hiker walks out 12 m at 2.0 m s⁻¹, then back 8 m. At the end: distance = 20 m, displacement = +4 m.

Exam trapA runner who finishes a lap where she started has displacement 0 and average velocity 0, even though her average speed is large.

02Speed and velocity

← Back to the notes: Speed and velocity

Average speed = distance ÷ time. Average velocity = displacement ÷ time. On a position–time graph, the gradient is the velocity: steep means fast, a falling line means moving in the negative direction, flat means at rest.

Three constant velocities. The gradient of each line is its velocity: +6 m s⁻¹, +2 m s⁻¹ and −3 m s⁻¹.

Uniform motion

For constant velocity, s = s₀ + vt. Two objects meet when their positions are equal: write s for each one with the same origin and clock, then solve.

03Average and instantaneous velocity

← Back to the notes: Instantaneous velocity

The gradient of a chord gives the average velocity over an interval. Make the interval shorter and shorter: the chord turns into the tangent, and its gradient is the instantaneous velocity.

time t interval Δt

Average velocity (chord)—

Instantaneous velocity (tangent)—

Difference—

Key ideaInstantaneous velocity = gradient of the tangent to the s–t graph. Instantaneous acceleration = gradient of the tangent to the v–t graph.

04Acceleration and signs

← Back to the notes: Acceleration

Acceleration a = Δv/Δt is a vector. Its sign tells you the direction of the change in velocity, not whether the object speeds up. Compare the signs of v and a:

Velocity	Acceleration	What happens
+	+	speeds up, moving forwards
+	−	slows down, moving forwards
−	−	speeds up, moving backwards
−	+	slows down, moving backwards

Exam trapAt the top of a throw, v = 0 but a = g downwards. Zero velocity does not mean zero acceleration.

Part 2Constant accelerationSections 05–08

05Three linked graphs

← Back to the notes: Motion graphs

Set the initial velocity u and the acceleration a, then press play. The car moves along the track while the three graphs draw themselves. Try a negative a with a positive u: the car slows, stops and comes back.

u a

t0.0 s

s0.0 m

v—

a—

Key ideaConstant a gives a horizontal a–t line, a straight v–t line (gradient a) and a curved s–t parabola (gradient v).

06Area under a v–t graph = displacement

← Back to the notes: The equations of motion

In a very short time the velocity hardly changes, so the displacement is a thin strip v × δt. Add more and thinner strips: the staircase area becomes the exact area under the line, ½(u + v)t.

number of strips

Staircase area—

Exact area ½(u + v)t—

v = u + at · s = ½(u + v)t · s = ut + ½at² · v² = u² + 2as

07Choosing an equation

← Back to the notes: Using the equations

Each equation leaves out one of s, u, v, a, t. Find the quantity that is not given and not asked for, then use the equation without it. Tap a quantity:

Tap the quantity that is missing from the problem.

Try it

"A car at 8.0 m s⁻¹ accelerates at 2.0 m s⁻² over 20 m. Find its final speed." No time → v² = u² + 2as → v = 12 m s⁻¹.
"From rest, 3.0 m s⁻² for 5.0 s. How far?" No v → s = ut + ½at² → s = 37.5 m.
"Slows from 18 to 6.0 m s⁻¹ in 4.0 s. Distance?" No a → s = ½(u + v)t → s = 48 m.

08Free fall

← Back to the notes: Free fall and vertical throws

Without air resistance, every object has acceleration g = 9.8 m s⁻² downwards, all the time. Take up as positive: a = −9.8 m s⁻² for the whole flight. Watch the arrows: the red velocity arrow shrinks, flips and grows; the green acceleration arrow never changes.

throw speed u height of balcony

t0.00 s

height above balcony0.0 m

v (up +)—

a−9.8 m s⁻²

Exam trapThe v–t graph is one straight line with gradient −g, through zero at the top. It does not bend or break at the top.

Part 3Projectiles and airSections 09–12

09Two motions at once

← Back to the notes: Projectiles: two motions in one

One ball is dropped. At the same moment another is thrown sideways. Both fall the same height in the same time, because gravity acts only vertically and the vertical motions start the same.

sideways speed

horizontal: aₓ = 0, vₓ = u cos θ (constant) · vertical: aᵧ = −g, vᵧ = u sin θ − gt

10Projectile lab

← Back to the notes: Projectiles: two motions in one

Choose the launch speed, angle and height. The path is drawn with the velocity arrow and its two components. Turn on air resistance to compare the real path with the ideal one.

speed angle height air resistance

time of flight—

range—

max height (above ground)—

impact speed and angle below horizontal—

Numbers are for no air resistance. With air resistance on, the red path is a model to show the effect; the IB only asks about it in words.

11Angle and range

← Back to the notes: Launch at an angle (level ground)

On level ground, R = u² sin 2θ / g. The range is largest at 45°, and two angles that add up to 90° give the same range.

angle θ

θ and partner—

range (u = 20 m s⁻¹)—

max height: θ / partner—

12Terminal speed

← Back to the notes: Fluid resistance and terminal speed

Drag acts against the motion and grows with speed. As an object falls, drag grows until the resultant force is zero: then the acceleration is zero and the speed is constant (terminal speed). In air, buoyancy is negligible, so this happens when drag = weight. In a liquid, buoyancy also counts: drag + buoyancy = weight. A bigger drag (larger area, like an open parachute) gives a lower terminal speed.

drag size

speed0.0 m s⁻¹

acceleration9.8 m s⁻²

drag ÷ weight0%

Effects of air resistance on a projectile

Lower maximum height · shorter range · steeper fall at the end (not symmetric) · horizontal velocity decreases · landing speed smaller than launch speed · acceleration not constant (its vertical part is more than g going up, exactly g at the top, less than g coming down) · time of flight longer for a horizontal launch, often a little shorter for a high upward launch.

13Practice and self-check

← Back to the notes: Extra practice

Answers checked by calculation. Use g = 9.8 m s⁻². Try each one before you open it.

1. A cyclist rides 6.0 km east, then 2.0 km west, in 0.50 h. Find her average speed and average velocity.

Distance = 8.0 km, so average speed = 8.0/0.50 = 16 km h⁻¹. Displacement = 4.0 km east, so average velocity = 8.0 km h⁻¹ east.

2. A train speeds up uniformly from 5.0 m s⁻¹ to 25 m s⁻¹ in 40 s. Find its acceleration and the distance travelled.

a = (25 − 5.0)/40 = 0.50 m s⁻². s = ½(5.0 + 25) × 40 = 600 m.

3. A car at 24 m s⁻¹ brakes at 6.0 m s⁻². The driver's reaction time is 0.50 s. Find the stopping distance.

Thinking: 24 × 0.50 = 12 m. Braking: 24²/(2 × 6.0) = 48 m. Total = 60 m.

4. A stone is thrown up at 9.8 m s⁻¹. How long until it comes back to the hand?

Time up = u/g = 1.0 s; the motion is symmetric, so total = 2.0 s.

5. A v–t graph is a straight line from 10 m s⁻¹ at t = 0 to −5.0 m s⁻¹ at t = 6.0 s. Find the displacement and the distance.

v = 0 at t = 4.0 s. Area 0–4 s: ½ × 4.0 × 10 = 20 m. Area 4–6 s: ½ × 2.0 × (−5.0) = −5.0 m. Displacement = 15 m, distance = 25 m.

6. A ball is kicked horizontally at 15 m s⁻¹ off a wall 1.8 m high. How far from the wall does it land?

t = √(2 × 1.8/9.8) = 0.606 s. x = 15 × 0.606 = 9.1 m.

7. A ball is launched at 25 m s⁻¹ at 50° from level ground. Find the maximum height and the range.

uᵧ = 19.2 m s⁻¹, uₓ = 16.1 m s⁻¹. H = 19.2²/19.6 = 18.7 m. T = 2 × 19.2/9.8 = 3.91 s, R = 16.1 × 3.91 = 62.8 m.

8. Explain why a skydiver reaches a lower terminal speed after opening the parachute.

The parachute has a much larger area, so the drag at a given speed is much larger. Drag equals the weight (buoyancy in air is negligible) at a lower speed, so the new terminal speed is lower.

Before the exam, can you…

tell distance from displacement and speed from velocity?
find instantaneous velocity with a tangent?
say if an object speeds up or slows down from the signs of v and a?
use gradients and areas of s–t, v–t and a–t graphs?
choose the right equation of motion every time?
solve vertical throws with one sign convention?
split a projectile into horizontal and vertical motion?
solve launches from a height, above and below the horizontal?
explain why 45° gives the longest range on level ground?
describe terminal speed and the effects of air resistance?