Representing Motion: Study Guide (Physics C: Mechanics 1.3)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 1.3 called Representing Motion. This guide is the calculus-based one: slopes are derivatives and areas are integrals, so you can handle graphs of any shape, not only straight lines. The algebra-based treatment is in the Physics 1 study guide. This topic builds directly on Topic 1.2.

One motion, five representations

The same motion can be shown in five ways. Each one makes some features easy to see and hides others.

Representation	What it shows well	What it hides
Narrative (words)	the story and the conditions (“starts from rest”, “brakes evenly”)	exact values
Figure	the set-up, the axis and the origin	how things change in time
Motion diagram	where the object is at equal time steps, plus velocity and acceleration arrows	exact numbers
Graph (x–t, v_x–t, a_x–t)	how each quantity changes; slopes and areas	the physical set-up
Equation	exact values at any instant	the overall picture

A strong answer moves between them. Free-response questions often ask you to sketch a graph from words, explain a graph in words, or check that a graph agrees with an equation. Always start by stating the axis, for example “+x to the right, origin at the start”. Every sign on every representation depends on it.

Motion diagrams

A motion diagram shows the object as a dot at equal time intervals. Add a velocity arrow at each dot, with length proportional to speed. Add one acceleration arrow for each stage.

• Equal spacing: constant velocity, a = 0.
• Spacing that grows: speeding up; a points along the motion.
• Spacing that shrinks: slowing down; a points against the motion.
• If the change in spacing is the same each step, the acceleration is constant.

The graph links: slopes down, areas up

Topic 1.2 gave you the calculus. Here it becomes a way to read and draw graphs.

Link	Calculus	On the graph
position → velocity	v_x = dx/dt	slope of the tangent to the x–t graph
velocity → acceleration	a_x = dv_x/dt	slope of the tangent to the v_x–t graph
velocity → displacement	Δx = ∫ v_x dt	signed area between the v_x–t graph and the time axis
acceleration → change in velocity	Δv_x = ∫ a_x dt	signed area between the a_x–t graph and the time axis

Two more rules help when you sketch:

• Curvature of x–t. Because a_x = d²x/dt², an x–t graph that curves upward (concave up) has a_x > 0. One that curves downward has a_x < 0. A straight x–t graph has a_x = 0.
• Matching features. Where x–t has a peak or trough, v_x = 0. Where v_x–t crosses the axis, the object turns around. Where a_x jumps between stages, the v_x–t graph has a corner but no break, because velocity cannot change in zero time.

Areas give changes, not values. The area under a v_x–t graph gives Δx, so you still need x₀ to find x. The area under an a_x–t graph gives Δv_x, so you still need v_x0. For distance rather than displacement, split the v_x–t graph where it crosses the axis and add the sizes of the areas.

The constant-acceleration equations as a toolkit

When the acceleration is constant, three equations describe the motion at any instant. They were derived by integration in Topic 1.2:

1. v_x = v_x0 + a_x t (no x)
2. x = x₀ + v_x0 t + ½a_x t² (no v_x)
3. v_x² = v_x0² + 2a_x(x − x₀) (no t)

Each one leaves out one quantity. So list your knowns and your unknown, then pick the equation that leaves out the quantity you neither know nor want.

The equations are written for x, but they work in any single dimension. For vertical motion, replace x with y. They hold for one stage of motion at a time. If the acceleration changes, start a new stage, and use the final values of one stage as the initial values of the next.

Free fall near Earth

Near Earth’s surface, an object moving only under gravity has an acceleration that is downward, constant and about 9.8 m/s² in size. We call its size g. This is a model: it ignores air resistance and assumes the height changes are small compared with Earth’s radius.

With +y upward, write a_y = −g. Then the toolkit becomes:

• v_y = v_y0 − gt
• y = y₀ + v_y0 t − ½gt²
• v_y² = v_y0² − 2g(y − y₀)

The course accepts g = 10 m/s² for numerical answers and does not penalise 9.8 or 9.81 m/s². State which value you use.

Functional dependence and factors of change

Symbolic answers let you predict how a result scales without new numbers.

• Stopping distance. A car braking at a constant rate a from speed v₀ stops in d = v₀²/(2a). So d ∝ v₀². Doubling the speed makes the stopping distance 4 times larger. At a braking rate of 6.0 m/s², 12 m/s needs 12 m, and 24 m/s needs 48 m.
• Height of a vertical throw. The ball rises to h = v₀²/(2g) in time t = v₀/g. Tripling the launch speed makes the height 9 times larger and the rise time 3 times larger.
• Drop from rest. h = ½gt², so t ∝ √h. Four times the height needs only twice the time.

Write the relationship as a proportion first, then find the factor. It is quicker and safer than recalculating.

Worked example 1: from words to three matching graphs

Question. Take +x forward along a corridor, origin at the start. A delivery robot starts from rest. It speeds up evenly at 0.50 m/s² for 4.0 s, travels at a steady speed for 6.0 s, then slows evenly to rest in 2.0 s. Sketch the a_x–t, v_x–t and x–t graphs and find the total displacement.

1. Stage speeds. After stage 1: v_x = 0 + 0.50 × 4.0 = 2.0 m/s. Stage 2 keeps 2.0 m/s. In stage 3, a_x = (0 − 2.0)/2.0 = −1.0 m/s².
2. a_x–t graph. Three flat lines: +0.50, 0 and −1.0 m/s². Check the areas: +0.50 × 4.0 = +2.0 m/s and −1.0 × 2.0 = −2.0 m/s. They cancel, so the robot ends at rest, as stated.
3. v_x–t graph. A straight rise from 0 to 2.0 m/s, a flat line, then a straight fall to 0. Each slope equals the a_x value of that stage.
4. Displacement from areas. Stage 1: ½ × 4.0 × 2.0 = 4.0 m. Stage 2: 2.0 × 6.0 = 12 m. Stage 3: ½ × 2.0 × 2.0 = 2.0 m. Total: 18 m.
5. x–t graph. It curves upward from 0 to 4.0 m (a_x > 0), is a straight line from 4.0 m to 16 m (a_x = 0), then curves downward and levels off at 18 m (a_x < 0). There are no corners, because the velocity changes smoothly.

Check. Average velocity = 18 m ÷ 12 s = 1.5 m/s, less than the cruising 2.0 m/s, as expected with slow starts and finishes.

Worked example 2: free fall in equations and symbols

Question. Take +y upward, origin at a footbridge rail 8.0 m above a river. A student tosses a stone straight up at 6.0 m/s from the rail. It misses the bridge on the way down. Ignore air resistance and use g = 9.8 m/s². Find (a) the greatest height above the water, (b) the velocity just before it hits the water, (c) the time in the air. Then (d) derive a symbolic expression for the time.

1. Knowns. y₀ = 0, v_y0 = +6.0 m/s, a_y = −9.8 m/s². The water is at y = −8.0 m.
2. (a) Top of the path. v_y = 0 there, and t is not needed, so use the equation with no t: 0 = 6.0² − 2(9.8)(y_top). So y_top = 36 ÷ 19.6 = 1.84 m. Height above water = 8.0 + 1.84 = 9.8 m.
3. (b) Impact velocity. Again no t is needed: v_y² = 6.0² − 2(9.8)(−8.0 − 0) = 36 + 156.8 = 192.8. So v_y = −14 m/s (13.9 m/s downward). Take the negative root: the stone is moving down.
4. (c) Time. Now that v_y is known, the first equation avoids a quadratic: v_y = v_y0 − gt gives t = (6.0 − (−13.9)) ÷ 9.8 = 2.0 s. Check with the position equation: −8.0 = 6.0t − 4.9t² has the positive root t = 2.03 s. The other root, −0.80 s, is before the throw, so reject it.
5. (d) Symbol form. Let the launch speed be v₀ and the height of the rail above the water be h. Then −h = v₀t − ½gt². Solve the quadratic and keep the positive root:

t = [v₀ + √(v₀² + 2gh)] / g

Check. Put v₀ = 6.0 m/s, h = 8.0 m, g = 9.8 m/s²: t = (6.0 + 13.9)/9.8 = 2.0 s. Units: (m/s)/(m/s²) = s. If h = 0, t = 2v₀/g, the familiar time to go up and come back to the launch level. With g = 10 m/s² you get 14 m/s and 2.0 s, the same to 2 significant figures.

Common misconceptions

• “The graph is a picture of the path.” An x–t graph that rises then falls does not mean the object went up a hill. It shows position along one axis against time.
• “At the top, a = 0.” At the top of a throw, v_y = 0 but a_y = −g. The velocity is still changing, from upward to downward. The v_y–t graph crosses zero there with the same slope, −g, as everywhere else (see the arrow in Figure 1).
• Reading the value instead of the slope. The velocity is the slope of x–t, not its height. A large x can go with zero velocity.
• “The area under a_x–t is the velocity.” It is the change in velocity. Add v_x0.
• Using one set of constant-a equations across stages. In Worked example 1, the acceleration changes twice. Treat each stage separately.
• Sign of g. g is a size, about 9.8 m/s². With +y upward, a_y = −g. Do not write a_y = −(−9.8).
• Corners on an x–t graph. A corner would mean an instant change in velocity. Real x–t graphs bend smoothly; corners appear on v_x–t graphs when a_x jumps.

Where this leads

Topic 1.4, Reference Frames and Relative Motion, asks how the same motion looks to different observers, and Topic 1.5 extends these representations to two and three dimensions. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.