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Study Notes

Unit 1 — Kinematics

Describing motion without asking what causes it: scalars and vectors, displacement, velocity, acceleration, graphs, relative motion, and projectiles. Each section ends with a short practice set — click "Show answer" to check yourself.

1.1

Scalars and Vectors in One Dimension

A scalar has magnitude (size) only. A vector has both magnitude and direction. In one dimension, direction is shown with a sign: choose a positive direction (e.g. right or up) and treat the opposite as negative.

ScalarVector
Distance (d)Displacement (Δx)
SpeedVelocity (v)
Time, mass, energyAcceleration (a), force
  • Distance is the total length of the path travelled; it is never negative and never decreases.
  • Displacement is the straight-line change in position, from start to finish, with direction.
  • The magnitude of a vector is always positive; the sign only encodes direction.
Why it matters: a runner who completes one full lap has travelled a large distance but has a displacement of zero.
Practice

1. A student walks 5 m east, then 3 m west. Find the distance travelled and the displacement (take east as positive).

Show answer
Distance = 5 + 3 = 8 m. Displacement = +5 + (−3) = +2 m, i.e. 2 m east.

2. Classify each as scalar or vector: 30 m/s, 30 m/s north, 12 kg, −4 m/s².

Show answer
30 m/s → scalar (speed). 30 m/s north → vector (velocity). 12 kg → scalar. −4 m/s² → vector (acceleration; the sign gives direction).
1.2

Displacement, Velocity, and Acceleration

These three quantities describe how position changes with time.

QuantityDefinitionUnits
DisplacementΔx = xf − xim
Average velocityvavg = Δx / Δtm/s
Average speeddistance / Δtm/s
Average accelerationaavg = Δv / Δtm/s²

For motion with constant acceleration in one dimension, use the kinematic equations:

v  = v₀ + a·t
Δx = v₀·t + ½·a·t²
v² = v₀² + 2·a·Δx
  • If v and a have the same sign, the object speeds up; if opposite signs, it slows down.
  • An object can have zero velocity and non-zero acceleration (a ball at the top of its flight).
  • Free fall near Earth's surface (ignoring air resistance): a = −g ≈ −9.8 m/s² (down), whether the object is rising or falling.
Common mix-up: negative acceleration does not always mean "slowing down." It means acceleration points in the negative direction.
Practice

1. A car speeds up from 10 m/s to 30 m/s in 5.0 s. What is its average acceleration?

Show answer
a = Δv/Δt = (30 − 10)/5.0 = 4.0 m/s².

2. A ball is thrown straight up. At its highest point, what are its velocity and acceleration?

Show answer
Velocity is 0 for an instant, but acceleration is still −9.8 m/s² (downward). Gravity acts the whole time.
1.3

Representing Motion

Motion can be shown with words, equations, motion diagrams (dots at equal time intervals), and graphs. Graphs are the most tested form.

GraphSlope givesArea under curve gives
Position vs. time (x–t)Velocity—
Velocity vs. time (v–t)AccelerationDisplacement
Acceleration vs. time (a–t)—Change in velocity
  • x–t: a straight line means constant velocity; a curve means changing velocity; a flat line means at rest.
  • v–t: a horizontal line means constant velocity (a = 0); a straight sloped line means constant acceleration.
  • On a v–t graph, area above the axis is positive displacement and area below is negative.
  • Motion diagrams: evenly spaced dots mean constant velocity; dots spreading apart mean speeding up; dots bunching together mean slowing down.
Practice

1. An object's v–t graph rises in a straight line from 0 to 6 m/s over 3 s, then stays at 6 m/s for 4 s. Find its total displacement.

Show answer
Triangle: ½(3)(6) = 9 m. Rectangle: (6)(4) = 24 m. Total = 33 m.

2. On an x–t graph, the curve gets steeper as time passes. What does this say about the object's motion?

Show answer
The slope (velocity) is increasing in magnitude, so the object is accelerating (speeding up).
1.4

Reference Frames and Relative Motion

Velocity is always measured relative to a reference frame, the observer or object you treat as at rest. A passenger sitting on a train is at rest relative to the train but moving relative to the ground.

v(A rel. C) = v(A rel. B) + v(B rel. C)
  • In one dimension, add velocities with signs; the "B" terms cancel like fractions.
  • Displacement and velocity depend on the frame; acceleration is the same in all frames moving at constant velocity relative to each other.
  • Always state the frame explicitly: "relative to the ground," "relative to the water."
Tip: draw a small sketch and label each velocity with its frame before adding anything.
Practice

1. A train moves at 20 m/s east relative to the ground. A passenger walks at 2 m/s toward the rear (west) relative to the train. What is the passenger's velocity relative to the ground?

Show answer
Taking east as positive: +20 + (−2) = +18 m/s, i.e. 18 m/s east.

2. A boat moves at 5 m/s relative to the water. The river flows at 3 m/s downstream. Find the boat's speed relative to the shore going (a) downstream and (b) upstream.

Show answer
(a) 5 + 3 = 8 m/s. (b) 5 − 3 = 2 m/s (heading upstream).
1.5

Vectors and Motion in Two Dimensions

In two dimensions, break each vector into perpendicular components and treat x and y separately.

Aₓ = A·cos θ        A_y = A·sin θ
A  = √(Aₓ² + A_y²)   θ = tan⁻¹(A_y / Aₓ)

Projectile motion (only gravity acts) is 2D motion where the two directions are independent:

Horizontal (x)Vertical (y)
Acceleration0−g (≈ −9.8 m/s²)
VelocityConstantChanges uniformly
Equationx = vₓ·ty = vy0·t − ½gt²
  • Time is the only link between the x and y equations.
  • At the peak of the trajectory, vy = 0 but vₓ is unchanged.
  • A ball dropped and a ball launched horizontally from the same height hit the ground at the same time.
  • For launch and landing at equal heights, the range is greatest at 45° (and equal for complementary angles).
Key idea: gravity only changes the vertical velocity. Horizontal velocity stays constant when air resistance is ignored.
Practice

1. A ball is kicked at 20 m/s at 30° above the horizontal. Find its initial horizontal and vertical velocity components.

Show answer
vₓ = 20 cos 30° ≈ 17.3 m/s; vy = 20 sin 30° = 10 m/s.

2. A rock is thrown horizontally at 8.0 m/s from a 45 m high cliff. How long does it take to land, and how far from the base does it land? (Use g = 9.8 m/s².)

Show answer
Vertical: 45 = ½(9.8)t² → t ≈ 3.0 s. Horizontal: x = vₓt = 8.0 × 3.0 ≈ 24 m.