Vectors and motion in one dimension

Quantities that need orientation

A mass of 2 kg or a duration of 6 s is determined by value and unit: these are scalars. A displacement of 6 m needs orientation: which way did it occur? Displacement, velocity, acceleration and force are vectors. A vector has a magnitude and a direction. Here, we distinguish the line along which it points from the way it points along that line. The line may be horizontal, with the vector pointing right or left.

We represent vectors with arrows. The arrow’s length, on a stated scale, represents magnitude; its tip represents the sense. Along a straight line, it is enough to choose an oriented axis and work with signed components. If positive is to the right, v = −4 m/s means a velocity of magnitude 4 m/s to the left.

A vector’s magnitude is never negative. Its component can be negative because it depends on the chosen orientation. Reversing the axis changes the signs of the components without changing the physical experiment. That is why you should state the axis before using equations, rather than adding signs intuitively at the end.

Adding vectors along the same line

Vector addition can be drawn by placing the tail of one arrow at the tip of the previous one. The resultant connects the initial tail to the final tip. Moving an arrow parallel to itself without changing length or orientation allows this construction for the free vectors used in this description of motion.

Along a straight line, the construction is equivalent to the algebraic sum of the components. A displacement of +9 m followed by −4 m gives +5 m. The distance travelled is 13 m and is not a vector sum: it adds the magnitudes of the sections. Subtraction A − B is addition A + (−B); vector −B has the same magnitude as B and the opposite sense.

When vectors are no longer collinear, we cannot simply add their magnitudes. A displacement of 3 m east followed by 4 m north has a resultant magnitude of 5 m by Pythagoras’ theorem, not 7 m. In this lesson the dynamics stays in one dimension, but the distinction prepares for motion in a plane.

Position, velocity and acceleration

Choose x(t) as position in metres and t as time in seconds. Displacement Δx uses final x minus initial x. Average velocity over the interval is Δx/Δt. Instantaneous velocity is the rate of change of position at that instant, represented by the slope of the tangent to the x(t) graph. Instantaneous speed is |v|.

Average acceleration is Δv/Δt. Instantaneous acceleration is the local rate of change of velocity and is measured in m/s². On a v(t) graph, the local slope represents acceleration. On an a(t) graph, the signed area gives the change in velocity. Notice the chain: each quantity describes the change in the previous one.

Being at x = 0 does not require v = 0. Having v = 0 at an instant does not require a = 0: an object thrown vertically upwards stops instantaneously at the highest point, but remains accelerated by gravity. A condition at an instant is not equivalent to a condition throughout an interval.

Building the constant-acceleration model

If acceleration a is constant, every equal time interval produces the same change in velocity. The definition a = (v − v₀)/Δt can be rearranged: v = v₀ + aΔt. v₀ is the velocity at the initial instant t₀; Δt = t − t₀. Only if t₀ = 0 can we write Δt simply as t without losing information.

The v(t) graph is a straight line. Its average velocity over that interval is (v₀ + v)/2, because the trapezium’s area is average height times base. Then Δx = [(v₀ + v)/2]Δt. Substituting v = v₀ + aΔt gives x = x₀ + v₀Δt + a(Δt)²/2.

We can also eliminate time. Multiplying v − v₀ = aΔt by v + v₀ and using 2Δx = (v + v₀)Δt gives v² − v₀² = 2aΔx. This expression is useful when duration is unknown. All three relationships require constant acceleration; they are not universal definitions for any motion.

The geometry of graphs checks the algebra

With non-zero constant acceleration, x(t) is a parabola, v(t) is a sloping straight line and a(t) is a horizontal line. The sign of a determines the concavity of the x(t) graph. However, a rising parabola can have negative a: it starts rising more and more slowly until it reaches zero slope.

The signed area under v(t) is displacement. If v crosses zero, split the interval at the crossing instant. For distance, add the magnitudes of the two areas. Do not simply take the magnitude of the total area: that would erase the path travelled before reversal.

A table calculated at successive instants helps identify patterns. With constant a and steps of 1 s, the differences in v are equal. The differences in x are not equal, because position varies quadratically. The graph is not an afterthought: it can show a reversal that would remain hidden in an uninterpreted calculation.

Signs, stops and validity

A body slows down when velocity and acceleration point in opposite directions; it speeds up when they point in the same direction. “Negative acceleration” only tells us the direction along the chosen axis. A vehicle travelling left can gain speed with both v and a negative.

To find a stop, solve v = 0. Then check whether the calculated instant belongs to the physical interval. An equation with negative acceleration may predict that the vehicle moves in the opposite direction after stopping; during real braking, the driver may change the force and acceleration no longer remains the same. The model must stop where the conditions change.

In free fall near Earth, neglecting air, vertical acceleration is approximately constant. With the positive axis upwards, a = −g and g = 9,8 m/s² in this lesson. Mass does not enter this model’s equations. For objects with significant air resistance, such as an unfolded sheet of paper, the assumption does not adequately describe the experiment.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

A negative vector has negative magnitude.

The component can be negative; magnitude is always non-negative.

v = v₀ + at works for any acceleration.

The form with the product at requires constant a over the interval.

Distance is always the magnitude of displacement.

If there is a reversal, add the section lengths separately.