UNSW Sydney · FACULTY OF PHYSICS

DPST1021 Chap.1 Kinematics, Vectors and Motion Graphs

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Chapter 1 of 15 · DPST1021

Kinematics, Vectors and Motion Graphs

The first two lectures of Physics 1A set up the language that the whole mechanics block depends on. You learn to tell a vector from a scalar, to write vectors with the unit vectors i, j and k, and to keep displacement, distance, velocity and speed as four separate ideas.

Much of the one-dimensional material is treated as assumed knowledge from Year 12, with revision resources supplied, so the lectures move quickly to the parts that use calculus.

That calculus link is the centre of the chapter. Velocity is the time derivative of position and acceleration is the derivative of velocity, so you can move between x(t), v(t) and a(t) by differentiating or by integrating with an initial value.

On a graph the same operations become the gradient and the area: the slope of a position-time graph is the velocity, while the area under a velocity-time graph is the displacement.

When acceleration is constant, as in free fall near the Earth's surface, the four kinematic equations on the course formula sheet do the integration for you.

Questions in this block usually combine a sketch, a sign convention and one of these tools.

The skill that earns marks is choosing the right tool for the information given and then keeping every sign consistent through to the answer, quoted with sensible significant figures.

In this chapter

What this chapter covers

  • 01

    Scalars and vectors: sort quantities by whether direction matters, and list familiar examples of each such as mass and time against displacement and force.

  • 02

    Unit-vector notation: write a vector as components along i, j and k, add vectors component by component and find a magnitude with Pythagoras.

  • 03

    Direction angles: convert components to an angle and correct the calculator result using the signs of the components to find the right quadrant.

  • 04

    Average versus instantaneous: divide displacement by elapsed time for an average velocity, and differentiate the position function for the value at one instant.

  • 05

    Motion graphs: read velocity as the gradient of a position-time graph and displacement as the signed area under a velocity-time graph.

  • 06

    Constant acceleration and free fall: apply the four kinematic equations with g = 9.80 m/s squared and a fixed sign convention, keeping the physically meaningful root.

Worked example · free

A cyclist braking to rest

Q [5 marks]. Original AskSia exercise: a cyclist travelling at 8.0 m/s brakes with constant deceleration and stops in 20 m. Find the acceleration, the stopping time, and how far the bike travels during the first second of braking. AskSia practice weighting: 5 marks; this is not a university marking scheme.
Slope gives acceleration, area gives displacement.Slope gives acceleration, area gives displacement. The three segments have slopes of +2, 0 and −2 m/s², and the shaded area totals 9 + 24 + 9 = 42 m.time t (s)velocity v (m/s)0371069 m24 m9 ma = +2a = 0a = −2
Slope gives acceleration, area gives displacement. The three segments have slopes of +2, 0 and −2 m/s², and the shaded area totals 9 + 24 + 9 = 42 m.
  • 1Take the direction of travel as positive. Known: v0 = 8.0 m/s, v = 0, x − x0 = 20 m. Acceleration is constant, so the formula-sheet equations apply.
  • 1Use v² = v0² + 2a(x − x0): 0 = 64 + 2a(20), giving a = −1.6 m/s². The negative sign says the acceleration opposes the motion.
  • 1Use v = v0 + at: 0 = 8.0 − 1.6t, so the bike stops after t = 5.0 s.
  • 1First second: x = v0t + ½at² = 8.0(1.0) + ½(−1.6)(1.0)² = 8.0 − 0.8 = 7.2 m.
  • 1Check: the average velocity over the stop is (8.0 + 0)/2 = 4.0 m/s, and 4.0 m/s × 5.0 s = 20 m, matching the given distance.
The acceleration is −1.6 m/s², the bike stops in 5.0 s, and it covers 7.2 m in the first second, which is more than a third of the total stopping distance because it is moving fastest at the start.
Sia tip — When a problem says decelerates, keep your positive direction and let the acceleration take a negative value; do not also put a minus sign into the equation, or the braking will be counted twice.
Glossary

Key terms

Scalar
A quantity described completely by a size and a unit, such as mass, time, temperature or speed.
Unit vector
A vector of length one that marks a direction; i, j and k point along the positive x, y and z axes.
Resultant
The single vector equal to the sum of two or more vectors, found by adding their components.
Instantaneous velocity
The derivative dx/dt of position with respect to time, equal to the gradient of the position-time graph at that moment.
Average speed
Total path length divided by elapsed time; it is never negative and exceeds the size of the average velocity whenever the motion reverses.
Free fall
Motion under gravity alone, with a constant downward acceleration of g when air resistance is ignored.
FAQ

Kinematics, Vectors and Motion Graphs FAQ

Can an object have zero velocity but a non-zero acceleration?

Yes. A ball thrown straight up is momentarily at rest at the top of its flight, yet gravity is still changing its velocity at 9.80 m/s squared downward. Velocity says how fast position is changing; acceleration says how fast velocity is changing, and the two are independent at any single instant.

How do I get displacement from a velocity-time graph?

Find the signed area between the graph and the time axis over the interval. Regions above the axis add and regions below subtract. For the total distance travelled, add the sizes of the regions instead, which gives a larger number whenever the velocity changes sign.

When can I use the constant-acceleration equations?

Only when the acceleration stays the same throughout the stage you are analysing. Free fall without air resistance qualifies. If the acceleration depends on time, as in a(t) = 6t, integrate instead, and if the motion has stages with different accelerations, treat each stage separately and match velocities at the join.

Why does my calculator give the wrong angle for a vector?

The inverse tangent only returns angles between minus 90 and plus 90 degrees, so it cannot distinguish opposite quadrants. Look at the signs of the components: a negative x component means adding 180 degrees to the calculator value to point the vector the right way.

Study strategy

Exam move

Start by building fluency with components, because every later topic, from projectiles to torque, depends on them. Practise turning a word description into i and j components, adding several vectors, and recovering the magnitude and the correct quadrant.

Then drill the calculus link in both directions: take a polynomial x(t), differentiate twice, and also start from a given a(t) with initial values and integrate back to position. Sketch the matching graphs each time so you can see gradients and areas. Finally, work free-fall problems with a fixed upward-positive convention and write the sign of g before you substitute.

The first homework quiz covers these lectures and can be repeated, so use it until each question type feels routine, and pay attention to the significant-figure rule when entering answers.

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