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Skill Check 0 / 10 complete

Representing motion

A motion graph is not a picture of the road. It is a picture of one number changing over time. Once you see that, every motion graph reduces to two moves: read a slope, or read an area. The kinematic equations are the algebraic versions of those same moves. This lesson is both.

§1

What a motion graph actually is.

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An object moving along a line has three numbers at every moment: position $x$, velocity $v$, and acceleration $a$. A motion graph plots one of them against time. The shape on the page is the shape of that number, not the shape of where the object went.

Below: a cart starting at rest and accelerating east at a constant rate. Three different shapes, one motion.

POSITION VS TIME x (m) t (s) 0 VELOCITY VS TIME v (m/s) t (s) 0 ACCELERATION VS TIME a (m/s²) t (s) 0
Fig. 1.1 The same motion drawn three ways. The cart sits still, then accelerates east at a steady rate. Position curves upward (it builds as $v$ stays positive). Velocity rises in a straight line (because $a$ is constant). Acceleration just sits there flat. Three different graphs of three different numbers, all showing the same trip.

None of those three pictures looks like a cart on flat ground. They are pictures of what $x$, $v$, and $a$ are doing, not pictures of the cart.

§2

The two moves: slope and area.

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Two moves: read a slope, or read an area.

The slope says how fast the y-quantity is changing. On an $x$-$t$ graph, slope is velocity. On a $v$-$t$ graph, slope is acceleration. Same logic, one rung up.

The area is the accumulated effect of the rate. Under a $v$-$t$ graph, area is displacement (m/s · s = m). Under an $a$-$t$ graph, area is $\Delta v$ (m/s$^2$ · s = m/s).

The same $v$-$t$ graph gives acceleration (slope) or displacement (area). The graph hasn't changed; the question has.

Slope gives a v (m/s) t (s) 0 slope = ā Area gives Δx v (m/s) t (s) 0 area = Δx
Fig. 1.2 The same $v$-$t$ graph supports two different questions. Read the slope and you get the average acceleration. Read the area between the line and the $t$-axis and you get the displacement. The line on the page is one thing; what it tells you depends on which move you make.

Once you know the graph type and which move the question wants, the rest is arithmetic.

§3

Things you'll meet.

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The four moves on one card. Purple = velocity / displacement; yellow = acceleration / $\Delta v$. The colors return in the worked examples and Skill Check.

slope of $x$-$t$
gives velocity
Pick two points on the position curve. Rise over run. The number is $v$ (in m/s). On a straight line this is the average velocity over the whole interval; on a curve it is the average between your two chosen points.
slope of $v$-$t$
gives acceleration
Same move, one rung up. Rise over run on the velocity curve. The number is $a$ (in m/s²). The sign tells you direction: positive slope means $v$ is becoming more positive, negative slope means $v$ is becoming more negative.
area of $v$-$t$
gives displacement
Find the area between the velocity curve and the $t$-axis. The number is $\Delta x$ (in m). Areas above the axis are positive (east); areas below the axis are negative (west). They cancel.
area of $a$-$t$
gives change in velocity
Find the area between the acceleration curve and the $t$-axis. The number is $\Delta v$ (in m/s). Add it to the initial velocity to get the final velocity: $v_f = v_i + \Delta v$.
kinematic equations (constant $a$)
when slope and area aren't enough
When acceleration is constant, three equations relate the kinematic variables directly. They are the algebra version of the graphs above. Use them when you have numbers and want a number out.
$$v_x = v_{x0} + a_x t$$ $$x = x_0 + v_{x0}\, t + \tfrac{1}{2} a_x t^2$$ $$v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)$$
Near Earth's surface, the vertical acceleration from gravity is $a_g = g \approx 10\;\text{m/s}^2$, directed downward. Sign convention is yours to pick; stay consistent within a single problem.
§4

Worked example: reading a $v$-$t$ graph.

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A cart rolls along a track. At the start of an interval, its velocity is $v_i = +2\;\text{m/s}$; four seconds later, its velocity is $v_f = +10\;\text{m/s}$. The graph below is a straight line connecting those two readings. Find the cart's average acceleration over the interval, and its displacement over the same interval.

Velocity vs time v (m/s) t (s) 0 2 4 6 8 10 12 1 2 3 4 vi = +2 m/s area = Δx slope = ā Δt = 4 s
Fig. 1.3 The cart's velocity over a 4-second interval. The yellow line carries the slope question (what is the acceleration?), and the shaded trapezoid carries the area question (what is the displacement?).
Step 1. Identify the graph type. The vertical axis is $v$, so this is a $v$-$t$ graph. From §3: slope of $v$-$t$ is $a$; area of $v$-$t$ is $\Delta x$.
Step 2. Read endpoints from the graph. Initial: $(t_i, v_i) = (0, +2\;\text{m/s})$. Final: $(t_f, v_f) = (4\;\text{s}, +10\;\text{m/s})$.
Step 3. Slope, for the average acceleration: $$\bar{a} = \dfrac{\Delta v}{\Delta t} = \dfrac{(+10) - (+2)}{4 - 0} = \dfrac{+8\;\text{m/s}}{4\;\text{s}} = +2\;\text{m/s}^2$$ The sign is positive, so the acceleration points east (in the same direction as $v$).
Step 4. Area, for the displacement. The shaded region is a trapezoid with parallel sides of length $v_i$ and $v_f$ and width $\Delta t$: $$\Delta x = \tfrac{1}{2}(v_i + v_f)\,\Delta t = \tfrac{1}{2}(2 + 10)(4) = +24\;\text{m}$$ The whole trapezoid sits above the $t$-axis, so the area (and the displacement) is positive. The cart moved 24 m east during the interval.
Final answer
$\bar{a} = +2\;\text{m/s}^2$ (east)  ·  $\Delta x = +24\;\text{m}$ (east).

Two answers from one $v$-$t$ graph: slope and area travel together. Which you compute depends only on the question.

§5

Worked example: reading an $a$-$t$ graph.

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A train is rolling east at $v_i = +5\;\text{m/s}$. Over the next five seconds, the train's acceleration is plotted below: the acceleration is $+3\;\text{m/s}^2$ for the first two seconds, then it switches to $-2\;\text{m/s}^2$ for the remaining three seconds. What is the train's velocity at the end of the five-second interval?

Acceleration vs time a (m/s²) t (s) 0 1 2 3 4 −1 −2 −3 1 2 3 4 5 Δv1 = +6 m/s Δv2 = −6 m/s 2 s 3 s
Fig. 1.4 The acceleration sits at $+3\;\text{m/s}^2$ for two seconds, then switches to $-2\;\text{m/s}^2$ for three more. The two shaded rectangles are the changes in velocity each segment contributes. The first sits above the axis (positive); the second sits below (negative).
Step 1. Identify the graph type. The vertical axis is $a$, so this is an $a$-$t$ graph. From §3: area under an $a$-$t$ graph gives $\Delta v$. (Slope of an $a$-$t$ graph is also a thing in principle, but for this question we only need the area move.)
Step 2. Compute the area in two pieces. Each segment is a rectangle (constant $a$): $$\Delta v_1 = a_1 \cdot \Delta t_1 = (+3)(2) = +6\;\text{m/s}$$ $$\Delta v_2 = a_2 \cdot \Delta t_2 = (-2)(3) = -6\;\text{m/s}$$ The second area is negative because the rectangle is below the $t$-axis. Areas below the axis carry a negative sign; areas above carry a positive sign.
Step 3. Sum to get the total change in velocity: $$\Delta v = \Delta v_1 + \Delta v_2 = (+6) + (-6) = 0\;\text{m/s}$$
Step 4. Apply the change to the initial velocity: $$v_f = v_i + \Delta v = (+5) + 0 = +5\;\text{m/s}$$
Final answer
$v_f = +5\;\text{m/s}$ (east). The train ends the interval moving at the same velocity it started with.

Same start and end velocity does not mean nothing happened. The train sped up for $2$ s and slowed back down for $3$ s; the two areas cancelled. The graph is the only record of what happened in between.

§6

Three mistakes that cost real points.

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Three failure modes account for most of the points students who know the equations still drop on motion-graph problems.

Pitfall · 01

"The graph shows where the object went."

A motion graph is a quantity vs time, not a picture of the path. A $v$-$t$ curve arcing up and back is not a hill; an $x$-$t$ parabola dipping below zero is not a tunnel. The vertical axis is the quantity named on it, full stop.

Fix. Read the axes first, every time. If $v$ is on the vertical axis, "curve goes up" means "velocity got bigger." Shape on the page is data, not geography.

Pitfall · 02

"Slope and area give the same kind of answer."

Slope and area are two moves with two different units. On a $v$-$t$ graph: slope has units of $\dfrac{\text{m/s}}{\text{s}} = \text{m/s}^2$ (acceleration); area has units of $\text{m/s} \cdot \text{s} = \text{m}$ (displacement). Reporting one for the other is the wrong physical quantity.

Fix. Decide which move before you compute. Rate question → slope. Accumulation question → area. Use the §3 grid as a checklist.

Pitfall · 03

"I'll just read the height of the curve."

The height of an $x$-$t$ curve at time $t$ is the position then; the height of a $v$-$t$ curve is the velocity then. Neither, by itself, gives a rate or an accumulation. Reading "the curve is at 12" as the average velocity is exactly the mix-up free-response rubrics flag.

Fix. Rate question → slope (rise over run). Accumulation question → area (base × height, or trapezoid). The y-axis value at one instant is rarely the answer the question wants.

Ten scenarios. The first six are graph reads (slope or area); the last four use the kinematic equations under constant acceleration. Pick magnitude (and direction where applicable), then check.

0 OF 10 SCENARIOS COMPLETE

01 / 10Average velocity from an x-t graph

A jogger runs along a straight east-west road. Her position-time graph is shown for the first 4 seconds of the run. East is positive. Find her average velocity over the interval.

POSITION VS TIME x (m) t (s) 0 3 6 9 12 1 2 3 4 Δt = 4 s
Your answer
$\bar{v} =$  __  m/s  
Direction