Representing Motion: Study Guide (Physics 1 1.3)

This guide is for the algebra-based Physics 1 course. Everything here uses algebra and graphs. It builds on the definitions in Topic 1.2: displacement, average velocity and average acceleration.

One motion, five representations

Physicists describe the same motion in several ways. Each way makes some features easy to see. Good problem-solvers switch between them and check that they agree.

1. Narrative. Words: “A cart starts from rest and speeds up steadily to the right for 2.0 s.”
2. Motion diagram. Dots showing where the object is at equal time intervals, with velocity arrows and an acceleration arrow.
3. Graphs. Position, velocity and acceleration plotted against time.
4. Equations. Algebra that links x, v_x, a_x and t.
5. Data table. Measured or calculated values, for example from a motion detector.

As always, start by writing your axis and origin, for example “+x to the right, origin at the start”.

Motion diagrams

In a motion diagram, the object (treated as a point) is drawn at equal time intervals.

• Spacing between dots shows speed. Wider gaps mean faster motion.
• Velocity arrows point in the direction of motion. Their length shows speed.
• One acceleration arrow shows the direction in which the velocity is changing.

Notice that both carts move to the right. Only the acceleration arrow tells them apart. When the acceleration arrow points the same way as the velocity arrows, the object speeds up. When it points the opposite way, the object slows down.

Graphs: slopes and areas

Three graphs against time carry the same information in different forms. Two rules link them.

Graph	Its slope gives	The area between it and the time axis gives
position x against t	velocity v_x (tangent slope at an instant)	(no standard meaning in this course)
velocity v_x against t	acceleration a_x (tangent slope at an instant)	displacement Δx
acceleration a_x against t	(not used in this course)	change in velocity Δv_x

Area below the time axis counts as negative. On a curved graph, the slope at an instant is the slope of the tangent line at that point, as in Topic 1.2.

Figure 2 shows one trolley described by all three graphs. Take +x to the right, origin at the start.

• 0 to 2.0 s: it moves at a steady +2.0 m/s.
• 2.0 to 4.0 s: it speeds up steadily, a_x = +1.0 m/s², reaching +4.0 m/s.
• 4.0 to 6.0 s: it brakes steadily, a_x = −2.0 m/s², and stops.

Read the three graphs together:

• 0–2.0 s. x–t is a straight line with slope 4.0 m ÷ 2.0 s = 2.0 m/s. So v_x–t is flat at 2.0 m/s, and a_x = 0.
• 2.0–4.0 s. v_x–t rises with slope (4.0 − 2.0) m/s ÷ 2.0 s = +1.0 m/s². The x–t graph curves upward (gets steeper) because the velocity is increasing.
• 4.0–6.0 s. v_x–t falls with slope −2.0 m/s². The x–t graph still rises, because v_x is still positive, but it gets flatter until it is level at t = 6.0 s, when v_x = 0.

The trolley never moves backwards, so x never decreases. A falling v_x–t line does not mean the object is moving backwards. Only a negative v_x means that.

Where the constant-acceleration equations come from

When a_x is constant, the v_x–t graph is a straight line. That gives the first equation directly, because the slope is a_x:

v_x = v_x0 + a_x t

The displacement is the area under that line from 0 to t. The shape is a trapezium with parallel sides v_x0 and v_x and width t:

Δx = ½(v_x0 + v_x)t = ½(v_x0 + v_x0 + a_x t)t = v_x0 t + ½a_x t²

Writing Δx = x − x₀ gives the second equation. Eliminating t between the first two gives the third.

Equation	Leaves out	Use it when you do not know or need
v_x = v_x0 + a_x t	x	the displacement
x = x₀ + v_x0 t + ½a_x t²	v_x	the final velocity
v_x² = v_x0² + 2a_x(x − x₀)	t	the time

Three points matter:

• The equations are valid only for constant acceleration. If a_x changes during the interval, split the motion into stages where it is constant, or describe it qualitatively.
• They describe instantaneous values at time t, not averages.
• They work in any single dimension. For vertical motion, replace x with y.

How to choose. List the five quantities (Δx, v_x0, v_x, a_x, t). Mark the three you know and the one you want. Use the equation that leaves out the fifth.

Free fall near Earth’s surface

Near Earth’s surface, an object moving only under gravity has a vertical acceleration that is downward, constant, and about 9.8 m/s² in size. We call this size g. With +y upward, a_y = −g. This holds on the way up, at the top and on the way down. It applies whether the object was dropped, thrown up or thrown down, as long as air resistance is small enough to ignore.

Because a_y is constant, the three equations apply with y in place of x and a_y = −g. In exam questions that need a numerical value, g is taken as 10 m/s², and using 9.8 or 9.81 m/s² correctly is not penalised. This guide uses 9.8 m/s².

Worked example 1: choosing an equation

Question. Take +x along the track in the direction of travel. A train passes a signal at 4.0 m/s. It then speeds up with constant acceleration 0.50 m/s² over the next 240 m. Find its speed at the end of the 240 m and the time taken.

1. Known: v_x0 = +4.0 m/s, a_x = +0.50 m/s², x − x₀ = +240 m. Wanted: v_x. Not known or needed: t. So use the equation that leaves out t.
2. v_x² = v_x0² + 2a_x(x − x₀) = (4.0)² + 2(0.50)(240) = 16 + 240 = 256 m²/s².
3. v_x = +√256 = +16 m/s. Take the positive root, because the train is still moving in the +x direction.
4. Now find t from v_x = v_x0 + a_x t: t = (16 − 4.0) ÷ 0.50 = 24 s.

Check with another representation. On a v_x–t graph this is a straight line from 4.0 m/s to 16 m/s over 24 s. The trapezium area is ½(4.0 + 16) × 24 = 240 m, which matches the given displacement.

Worked example 2: a ball thrown up from a balcony

Question. Take +y upward, origin at the hand. A student on a balcony throws a ball straight up at 12 m/s. The ground is 8.0 m below the hand. Ignore air resistance and use g = 9.8 m/s². Find (a) the time to reach the top, (b) the maximum height above the hand, (c) the velocity just before it hits the ground and (d) the total time in the air.

Known throughout: v_y0 = +12 m/s, a_y = −9.8 m/s².

(a) At the top v_y = 0. Using v_y = v_y0 + a_y t: 0 = 12 − 9.8t, so t = 12 ÷ 9.8 = 1.2 s (1.22 s).

(b) Using v_y² = v_y0² + 2a_y(y − y₀) with v_y = 0: 0 = 144 − 19.6y, so y = 144 ÷ 19.6 = 7.3 m above the hand.

(c) At the ground y = −8.0 m. v_y² = 144 + 2(−9.8)(−8.0) = 144 + 156.8 = 300.8 m²/s², so |v_y| = 17.3 m/s. The ball is moving down, so take the negative root: v_y = −17 m/s.

(d) Using v_y = v_y0 + a_y t: −17.34 = 12 − 9.8t, so t = 29.34 ÷ 9.8 = 3.0 s (2.99 s).

Check. From the top, the ball falls 7.35 + 8.0 = 15.35 m from rest. Using y − y₀ = ½a_y t² gives t = √(2 × 15.35 ÷ 9.8) = 1.77 s. Adding 1.22 s for the way up gives 2.99 s, which agrees.

Interpretation. The acceleration was −9.8 m/s² at every moment, including at the top where v_y = 0. With g = 10 m/s² the answers become 7.2 m, −17 m/s and 2.9 s.

Worked example 3: deriving a scaling result

Question. A stone is dropped from rest from height h. Derive an expression for the time t to fall, and use it to compare a drop from h with a drop from 4h.

1. Take +y downward for this question, so a_y = +g, v_y0 = 0 and the displacement is +h.
2. y − y₀ = v_y0 t + ½a_y t² gives h = ½gt², so t = √(2h/g).
3. Compare: t₂ / t₁ = √(2(4h)/g) ÷ √(2h/g) = √4 = 2. Four times the height takes twice the time, not four times.
4. Landing speed v = gt also doubles.

Check with numbers. For h = 5.0 m, t = √(10 ÷ 9.8) = 1.0 s and v = 9.9 m/s. For 20 m, t = 2.0 s and v = 20 m/s (19.8 m/s). The ratios are both 2.

When acceleration is not constant

Real accelerations often change. A car pulling away from traffic lights usually has a large acceleration at first and less as it speeds up. In this course you are not asked to calculate such motion. You are expected to describe it, sketch its graphs and reason about it.

For the car:

• a_x–t starts high and falls towards zero.
• v_x–t rises steeply at first, then more and more gently. It curves over and levels off. Its tangent slope at each instant is the acceleration then.
• x–t curves upward and gets steeper, but more slowly as time goes on, and ends up close to a straight line once the velocity is almost steady.

Areas still have the same meanings. The area under a curved v_x–t graph is still the displacement; you can estimate it by counting grid squares. But the equation x = x₀ + v_x0 t + ½a_x t² does not apply, because there is no single value of a_x to put in it.

Common misconceptions

• “The x–t graph shows the path.” No. It shows position against time. A ball thrown straight up and caught makes an arch on a y–t graph, but it travels in a vertical line.
• “A falling v_x–t line means moving backwards.” It means a_x is negative. The object moves backwards only when v_x itself is negative.
• “The highest point on a v_x–t graph is where the object turns around.” That is where v_x is largest, so it is moving fastest in the + direction. It turns around where v_x changes sign, which is where the v_x–t graph crosses the time axis.
• “Use any kinematic equation whenever there is motion.” They need constant acceleration. Check this first.
• “At the top of a throw, a = 0.” The acceleration is −g all the way (Worked example 2).
• “Double the height, double the fall time.” Fall time grows with the square root of height (Worked example 3).
• Taking the wrong square root. v_x² = … gives two roots. Choose the sign from the direction of motion.
• Mixing axes. If you choose +y upward, a falling object has negative velocity and a_y = −g, all in the same calculation.

Where this leads

Topic 1.4 (Reference Frames and Relative Motion) asks how these graphs and equations change when the observer is moving. Continue with the Topic 1.4 study guide. First, try the practice questions, then use the revision notes and the checklist to consolidate. You can also return to the course roadmap.