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A.1 Kinematics

Distance and displacement, speed, velocity and acceleration, motion graphs, the suvat equations, free fall, projectiles launched at any angle, the qualitative effect of air resistance, and the Paper 1B graph skills behind measuring g.

Displacement & velocityMotion graphssuvat equationsFree fallProjectile motionFluid resistanceMeasuring g & uncertainties
RSby Riyaz Sir
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No tag = SL and HLHL ONLY HL students onlyDATA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Kinematics describes motion without asking what causes it. Position, velocity and acceleration are vectors: velocity is the rate of change of displacement, acceleration is the rate of change of velocity.
  • Graphs carry the physics: the gradient of s–t is velocity, the gradient of v–t is acceleration, and the area under v–t is displacement.
  • For constant acceleration use the four suvat equations from the data booklet. Choose a positive direction and stick to it; in free fall a = −9.8 m s−2 if up is positive.
  • A projectile is two motions at once: constant velocity horizontally, constant acceleration g vertically, linked only by time. Air resistance lowers the height and the range and makes the path lopsided.
  • A.1 is the same at SL and HL. There is no additional higher level content in this topic.

By the end you can

  • Distinguish distance from displacement and speed from velocity, and find average and instantaneous values.
  • Draw and interpret displacement–time, velocity–time and acceleration–time graphs, using gradients and areas.
  • Choose and apply the equations of motion for uniform acceleration, including free fall.
  • Solve projectile problems for launches that are horizontal, above the horizontal and below it.
  • Describe qualitatively how fluid resistance changes a projectile's path.
  • Linearise s = ½gt2, find g from a gradient and work out its uncertainty.

1Describing motion

Distance, displacement, speed and velocity

A (start)Bpath travelled = distance (scalar)displacement (vector)straight line A → B, with a directionA full lap of a 400 m track: distance = 400 m, displacement = 0.
Distance follows the path. Displacement only cares where you started and where you ended.

Distance — the total length of the path travelled. A scalar, in m.

Displacement — the change in position: the straight line from start to finish, with a direction. A vector, in m.

Speed — distance ÷ time. A scalar, in m s−1.

Velocity — rate of change of displacement, v = Δs/Δt. A vector, in m s−1.

Acceleration — rate of change of velocity, a = Δv/Δt. A vector, in m s−2. Slowing down is an acceleration opposite to the velocity.

Worked example

A runner goes 100 m east, then 40 m west, in 20 s. Find the distance, the displacement, the average speed and the average velocity.

  1. distance = 100 + 40 = 140 m
  2. displacement = 100 − 40 = 60 m east
  3. average speed = 140/20 = 7.0 m s−1
  4. average velocity = 60/20 = 3.0 m s−1 east
Common mistakeGiving a velocity or displacement without a direction (or a sign). "3.0 m s−1" is a speed; "3.0 m s−1 east" or "+3.0 m s−1" is a velocity.

Average and instantaneous values

t / ss / m123451020300tangent at t = 2 schordaverage velocity,1 s to 4 s = 21/3 = 7 m s⁻¹instantaneous velocityat 2 s = tangent gradient= 6 m s⁻¹
The chord gives the average velocity over an interval. The tangent gives the velocity at one instant.
  • Average velocity = total displacement ÷ total time = gradient of the chord on an s–t graph.
  • Instantaneous velocity = gradient of the tangent at that instant: v = ds/dt. Likewise a = dv/dt. LEARN IT
  • Drawing a tangent: use a long ruler, touch the curve at one point and make the line long. A longer triangle gives a smaller percentage uncertainty in the gradient.
Linking questionThe tangent gradient is exactly the derivative from maths. See IB Maths AA Differentiation for v = ds/dt and a = dv/dt done with calculus.

2Motion graphs

tss–t: a curvegradient = velocitytvv–t: a straight linegradient = accelerationarea = displacementtaa–t: flat linearea = change in velocityUniform acceleration from rest: the same motion, three graphs
One motion, three graphs. Each graph is the gradient of the one before it.
GraphGradient givesArea under it gives
displacement–time (s–t)velocitynothing useful
velocity–time (v–t)accelerationdisplacement
acceleration–time (a–t)not neededchange in velocity
  • Uniform acceleration: a straight v–t line. Non-uniform acceleration: a curved v–t line. Find the acceleration at a point from a tangent and the displacement by counting squares under the curve.
  • Area below the time axis counts as negative displacement (moving backwards). Distance adds all the areas as positive; displacement adds them with signs.
Worked example

A car accelerates uniformly from rest to 20 m s−1 in 8.0 s, travels at 20 m s−1 for 12 s, then brakes uniformly to rest in 5.0 s. Find the total distance and the average speed.

  1. area 1 (triangle) = ½ × 8.0 × 20 = 80 m M1
  2. area 2 (rectangle) = 12 × 20 = 240 m; area 3 (triangle) = ½ × 5.0 × 20 = 50 m
  3. total distance = 370 m A1
  4. average speed = 370/25 = 14.8 m s−1 A1
Common mistakeReading the displacement off the height of a v–t graph. The height is the velocity. Displacement is the area.

3Equations of motion

Where the suvat equations come from

timevelocityuvtrectangle = uttriangle = ½(v − u)t= ½at²gradient:a = (v − u)/tso v = u + atarea:s = ut + ½at²or s = ½(u + v)tBoth suvat results come straight from the v–t graph. Only for constant acceleration.
Gradient and area of a straight v–t line give two equations. The other two follow by algebra.
v = u + at DATA BOOKLET
s = ut + ½at2 DATA BOOKLET
v2 = u2 + 2as DATA BOOKLET
s = (u + v)t2 DATA BOOKLET
s displacement, u initial velocity, v final velocity, a acceleration, t time
  • Only for constant acceleration in a straight line. If a changes, use the graph (gradient and area) instead.
  • Choose the equation that leaves out the quantity you neither know nor need:
Equationdoes not contain
v = u + ats
s = ut + ½at2v
v2 = u2 + 2ast
s = ½(u + v)ta
Worked example

A car travelling at 25 m s−1 brakes to rest in 40 m. Find its deceleration and the time taken.

  1. u = 25, v = 0, s = 40; t is not needed, so use v2 = u2 + 2as M1
  2. 0 = 625 + 80a ⇒ a = −7.8 m s−2 (deceleration 7.8 m s−2) A1
  3. v = u + at ⇒ t = 25/7.81 = 3.2 s A1

Free fall and the sign convention

t / sv / m s⁻¹123-15-10-5510150top: v = 0, a = −9.8going upcoming downgradient = −9.8 m s⁻² everywhere, even at the top+ upthrownup at15 m s⁻¹green area = red area:back to the start
The v–t line keeps the same gradient the whole time, through the top and on the way down.
g = 9.8 m s−2 DATA BOOKLET
acceleration of free fall near Earth's surface, always directed downwards
  • Pick a positive direction at the start and write it down ("up is +"). Then a = −9.8 m s−2, velocities going down are negative, and points below the launch have negative s.
Worked example

A ball is thrown vertically upwards at 15 m s−1 from 1.5 m above the ground. Find its maximum height above the ground and the time it takes to hit the ground.

  1. up is +, a = −9.8 m s−2
  2. at the top v = 0: 0 = 152 − 2(9.8)s ⇒ s = 11.5 m, so 13.0 m above the ground M1A1
  3. it hits the ground at s = −1.5 m: −1.5 = 15t − 4.9t2 M1
  4. 4.9t2 − 15t − 1.5 = 0 ⇒ t = 15 + √(225 + 29.4)9.8 = 3.16 s (reject the negative root) A1
Common mistakeSaying the acceleration is zero at the top. Only the velocity is zero there. The acceleration is still 9.8 m s−2 downwards, which is why the ball comes back.
Exam tipA height question with no time in it is a v2 = u2 + 2as question. A "when does it land" question usually becomes a quadratic in t: keep the positive root.

4Projectile motion

Two independent motions

20 m cliffthrown sideways at 8 m s⁻¹ (orange)dropped at the same moment (blue)one dot every 0.5 s:always at the same heightlands 2.02 s later, 16.2 m out
The thrown ball moves sideways at a steady rate, but its fall matches the dropped ball exactly.
  • Horizontal: no force (without air resistance), so constant velocity: x = uxt.
  • Vertical: constant acceleration g downwards, so use suvat with uy.
  • Time links them. Find t from one direction, then use it in the other.
  • Launch at speed u and angle θ above the horizontal: ux = u cos θ and uy = u sin θ.
Worked example

A ball rolls off a 20 m high cliff at 8.0 m s−1. Find how far out it lands and its velocity on impact.

  1. vertical: uy = 0, s = 20 m down: 20 = ½(9.8)t2 ⇒ t = 2.02 s M1A1
  2. horizontal: x = 8.0 × 2.02 = 16.2 m A1
  3. at impact vy = 9.8 × 2.02 = 19.8 m s−1 down and vx = 8.0 m s−1 M1
  4. speed = √(8.02 + 19.82) = 21.4 m s−1, at arctan(19.8/8.0) = 68° below the horizontal A1

Launched at an angle

H = 12.0 mR = 40.2 mtop: v = u cos θ onlyu = 20 m s⁻¹ at 50°horizontal: samevertical: changesactual velocity
The horizontal arrow never changes. The vertical arrow shrinks to zero at the top and grows again downwards.
Worked example

A ball is kicked at 20 m s−1 at 50° above the horizontal on level ground. Find the maximum height, the time of flight and the range.

  1. ux = 20 cos 50° = 12.9 m s−1; uy = 20 sin 50° = 15.3 m s−1 M1
  2. top: 0 = 15.32 − 2(9.8)H ⇒ H = 12.0 m A1
  3. time to top = 15.3/9.8 = 1.56 s, so time of flight = 3.13 s (symmetric path) A1
  4. range = 12.9 × 3.13 = 40.2 m A1
Worked example

A stone is thrown from the top of a 30 m tower at 10 m s−1 at 20° below the horizontal. How far from the tower does it land?

  1. down is +: uy = 10 sin 20° = 3.42 m s−1, a = 9.8 m s−2, s = 30 m M1
  2. 30 = 3.42t + 4.9t2 ⇒ t = 2.15 s A1
  3. ux = 10 cos 20° = 9.40 m s−1
  4. x = 9.40 × 2.15 = 20.2 m A1
  • At the top the velocity is horizontal, v = u cos θ, not zero.
  • On level ground without air resistance, the path is symmetric and the range is greatest at 45°. Angles that add to 90° (30° and 60°) give equal ranges.
  • The trajectory is a parabola. You do not need its equation, and the range formula R = u2 sin 2θ / g is not in the data booklet: derive the range from ux × time of flight.
Common mistakeUsing u instead of u cos θ for the horizontal motion, or forgetting that uy is negative (or positive, if down is +) for a launch below the horizontal.

Fluid resistance (qualitative)

no air resistance: symmetric parabolawith air resistance:lower maximum heightshorter rangesteeper fall after the topway down takes longer than way upThe red path is not a parabola. You only need to describe it, not calculate it.
Air resistance grows with speed and always acts against the velocity, so it slows the horizontal motion as well as the vertical one.
  • What changes with air resistance: lower maximum height; shorter range; the horizontal velocity decreases instead of staying constant; the path is no longer symmetric and comes down more steeply; the way down takes longer than the way up; the best launch angle for range is less than 45°.
  • Falling objects reach a terminal speed when the drag force equals the weight, so the acceleration falls to zero. The forces are covered in A.2.
Common mistakeSaying air resistance acts only horizontally, or only upwards. It acts opposite to the velocity, so its direction changes along the path.
Exam tipExam questions on this point ask you to sketch the new path on the same axes and describe the changes. No calculation is expected. Draw it lower, shorter and steeper at the end.

5Data skills for Paper 1B

Measuring g with a straight-line graph

t² / s²s / m0.050.100.150.200.20.40.60.81.00s = ½ g t² so s ∝ t²gradient = g/2best fit: 4.88 → g = 9.8steepest 5.1, shallowest 4.7so g = 9.8 ± 0.4 m s⁻²Error bars show the uncertainty in t². Draw the best line and the steepest and shallowest lines through all bars.
Plotting s against t2 turns the curve s = ½gt2 into a straight line through the origin with gradient g/2.
  • Linearise: compare s = ½g t2 with y = mx. Plot y = s and x = t2, so the gradient m = g/2 and g = 2 × gradient.
  • Error bars: if t has an uncertainty, t2 has twice the percentage uncertainty. Draw the best-fit line, then the steepest and the shallowest lines that pass through every error bar.
  • Uncertainty in g: Δg ≈ ½(gmax − gmin), using the gradients of the steepest and shallowest lines.
Worked example

A ball falls 1.20 ± 0.01 m in 0.50 ± 0.01 s from rest. Find g and its uncertainty.

  1. g = 2s/t2 = 2(1.20)/0.502 = 9.6 m s−2 A1
  2. % uncertainty: s gives 0.83 %, t2 gives 2 × 2.0 % = 4.0 % M1
  3. total = 4.8 %, and 4.8 % of 9.6 is 0.46
  4. g = 9.6 ± 0.5 m s−2 A1
  • Systematic error: a line that misses the origin points to one. A positive intercept on the t2 axis (every time too long) often comes from reaction time, or a delay in the release mechanism. Random errors scatter the points; systematic errors shift them all.
  • Nature of science: Galileo slowed free fall down on a gentle ramp so that he could time it, and found that distance grows with time squared. Linearising data the same way is still how physicists test a model.
Idea for your IA / explorationMeasure how the range of a small ball depends on launch angle with a simple launcher and phone video (Tracker is free). Compare your best angle with the 45° prediction and use air resistance to explain the difference.
Linking questionFor a falling object, v2 = u2 + 2as and conservation of energy (A.3) give the same speed: ½mv2 = ½mu2 + mgh. Which approach still works if the path is curved?

Test yourself — cover the answers first

Quick check

  1. A runner completes one lap of a 400 m track. Give the distance and the displacement.
  2. What does the gradient of a velocity–time graph represent?
  3. A car accelerates uniformly from 4.0 m s−1 to 16 m s−1 in 6.0 s. Find a and the distance travelled.
  4. What are the velocity and the acceleration of a ball at the top of a vertical throw?
  5. Without air resistance, what happens to a projectile's horizontal velocity?
  6. One ball is dropped and another is thrown horizontally from the same height at the same moment. Which lands first?
  7. Which launch angle gives the greatest range on level ground without air resistance?
  8. How does air resistance change the maximum height and the range?

Exam-style

  1. A stone is thrown at 12 m s−1 at 35° above the horizontal from the edge of a 25 m cliff. Find the time of flight and how far from the foot of the cliff it lands.
  2. A lift starts from rest, accelerates at 1.2 m s−2 for 2.5 s, moves at constant velocity for 6.0 s, then decelerates uniformly to rest in 2.0 s. Find the total distance.
  3. A student plots s against t2 for a falling ball and gets a best-fit gradient of 4.85 m s−2, with steepest and shallowest gradients of 5.05 and 4.65 m s−2. Find g with its uncertainty.
Show answers
  1. 400 m and 0
  2. acceleration
  3. a = 2.0 m s−2; s = ½(4.0 + 16)(6.0) = 60 m
  4. v = 0; a = 9.8 m s−2 downwards
  5. it stays constant
  6. both together: their vertical motions are identical
  7. 45°
  8. both decrease
  9. uy = 12 sin 35° = 6.88 m s−1 up; ux = 12 cos 35° = 9.83 m s−1
    up is +: −25 = 6.88t − 4.9t2
    t = 6.88 + √(6.882 + 490)9.8 = 3.07 s
    x = 9.83 × 3.07 = 30.2 m
  10. top speed = 1.2 × 2.5 = 3.0 m s−1
    areas: ½(2.5)(3.0) + 6.0(3.0) + ½(2.0)(3.0) = 3.75 + 18 + 3.0
    total = 24.8 m (25 m)
  11. g = 2 × 4.85 = 9.7 m s−2
    Δg = ½(10.1 − 9.3) = 0.4
    g = 9.7 ± 0.4 m s−2
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.