Unit 1 · Topic 1.3 Beta

Representing Motion

Motion can be shown as a motion diagram, x-t, v-t and a-t graphs, equations and words.

Practice 1: Creating RepresentationsPractice 2: Mathematical RoutinesPractice 3: Scientific Questioning and Argumentation

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Question set for this topic

Part 1 · Hook

Why this matters

A coach's stopwatch, a phone's motion sensor and a slow-motion video describe the same sprint in different ways: a table of times, a graph that climbs and flattens, a strip of frames. The exam asks you to move between all of them, and two graph rules do most of the work: slope down, area up.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What does the slope of a graph of y against x tell you?

  1. How fast y changes per unit of x
  2. The value of y at x = 0
  3. The total of y times x
  4. The largest value of y
Show the answer

Slope = Δy/Δx, a rate, with units (y unit)/(x unit).

  • Correct: How fast y changes per unit of x:
  • The value of y at x = 0:
  • The total of y times x:
  • The largest value of y:

2. A cart's velocity changes from 2.0 m/s to 5.0 m/s in 1.5 s. What is its average acceleration?

  1. 2.0 m/s²
  2. 4.5 m/s²
  3. 3.0 m/s²
  4. 0.50 m/s²
Show the answer

a = Δv/Δt = 3.0 m/s ÷ 1.5 s = 2.0 m/s².

  • Correct: 2.0 m/s²:
  • 4.5 m/s²:
  • 3.0 m/s²:
  • 0.50 m/s²:

3. If Δx ∝ t², how does Δx change when t doubles?

  1. It becomes 4 times as large
  2. It doubles
  3. It halves
  4. It does not change
Show the answer

The factor squared: 2² = 4.

  • Correct: It becomes 4 times as large:
  • It doubles:
  • It halves:
  • It does not change:

Part 4 · See it

See it first

A cart on a track with a motion diagram of dots every 0.5 s: the dots spread out while it speeds up, are evenly spaced while it coasts, and bunch together while it slows to a stop at 4.0 m. Below, three graphs share a time axis from 0 to 6 s. Position against time: a curve bending upward from 0 to 1.0 m (0 to 2 s), a straight line to 3.0 m (2 to 4 s), then a curve bending down and leveling off at 4.0 m at 6 s; a dashed tangent line shows the slope at 4 s. Velocity against time: a line rising from 0 to 1.0 m/s, flat at 1.0 m/s, then falling to 0; the shaded rectangle from 2 to 4 s has area 1.0 m/s times 2 s = 2.0 m, the displacement. Acceleration against time: +0.5 m/s squared, then 0, then −0.5 m/s squared; the shaded area from 0 to 2 s is the change in velocity, 1.0 m/s. A side column reads: the slope of x-t is velocity, the slope of v-t is acceleration, the area under v-t is displacement, the area under a-t is the change in velocity.
One motion in four representations. Read downward with slopes (x-t → v-t → a-t) and upward with areas (a-t → v-t → x-t). LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Position changes with timethe slope of the x-t graph at any instant is the velocity
  2. Velocity changes with timethe slope of the v-t graph is the acceleration, and the area under it is the displacement
  3. The acceleration is constantthe v-t graph is a straight line and the three kinematic equations apply
  4. An object falls with air resistance negligibleits acceleration is g = 9.8 m/s² downward at every instant, even when v = 0 at the top

Part 6 · Key ideas

Key ideas

  • A motion diagram shows positions at equal times: growing gaps mean speeding up, shrinking gaps mean slowing down.
  • Slope of x-t = velocity. Slope of v-t = acceleration. Area under v-t = displacement. Area under a-t = change in velocity.
  • For constant acceleration: v = v₀ + at; x = x₀ + v₀t + ½at²; v² = v₀² + 2a(x − x₀). Pick the one that leaves out the quantity you neither know nor want.
  • Free fall: a = g = 9.8 m/s² downward (10 m/s² is accepted), on the way up, at the top and on the way down.
  • If the acceleration changes (nonuniform), the v-t graph curves. Describe it from the graph only; the equations do not apply.

Part 7 · Misconception

A common mistake

The wrong idea: At the top of its flight a ball thrown straight up has zero acceleration, because it stops for an instant.

What actually happens: Its velocity is zero for an instant, but the velocity is still changing, from upward to downward, at 9.8 m/s² each second. The acceleration is g downward at the top, as everywhere else in the flight.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

A car between two traffic lights

A car pulls away from a red light, cruises, then brakes for the next light. The graph shows its velocity, with the direction of travel positive. Each part of the motion has a constant acceleration.

0246810121402468101214Time (s)Velocity (m/s)
Data table
Time (s)Car
00
412
1012
140

1. What is the car's acceleration from t = 0 to t = 4.0 s?

Type a number and its unit.

Show the answer

The acceleration is the slope of the v-t graph: (12 m/s − 0) ÷ (4.0 s − 0) = 3.0 m/s².

  • Answer: 3.0 m/s^2

2. How far does the car travel from t = 0 to t = 14 s?

Type a number in m.

Show the answer

Displacement is the area under the v-t graph: triangle ½(4.0 s)(12 m/s) = 24 m, rectangle (6.0 s)(12 m/s) = 72 m, triangle ½(4.0 s)(12 m/s) = 24 m. Total = 120 m (two significant figures).

  • Answer: 120 m

3. During which interval is the car's acceleration negative?

  1. From 0 to 4 s
  2. From 4 s to 10 s
  3. At no time, since the velocity stays positive
  4. From 10 s to 14 s
Show the answer

Acceleration is the slope of the v-t graph. The slope is negative from 10 s to 14 s, while the car is still moving forward but slowing.

  • From 0 to 4 s: The graph rises there, so the slope is positive: the car speeds up.
  • From 4 s to 10 s: The graph is flat there, so the slope is zero: steady velocity, no acceleration.
  • At no time, since the velocity stays positive: Velocity and acceleration have separate signs. The velocity is positive while it decreases, so a is negative.
  • Correct: From 10 s to 14 s: Right: the v-t graph slopes downward there, so the slope (the acceleration) is negative.

Diagram

A ball tossed up from a balcony

A student on a balcony 15 m above the ground throws a ball straight up at 12 m/s. The ball rises, stops for an instant, then falls past the balcony to the ground. Air resistance is negligible. Take up as positive and g = 9.8 m/s².

highest pointv₀ = 12 m/s upward15 m (balcony to ground)path: straight up,then straight downground

Key: gold arrow, the initial velocity; dashed line, the path of the ball.

4. How long does the ball take to reach its highest point? Use g = 9.8 m/s².

Type a number in s.

Show the answer

At the top v = 0. Use v = v₀ + at with a = −9.8 m/s²: 0 = 12 m/s − (9.8 m/s²)t, so t = 12 ÷ 9.8 = 1.2 s.

  • Answer: 1.2 s

5. What is the ball's acceleration at its highest point?

  1. Zero, because the ball is momentarily at rest
  2. 9.8 m/s² downward
  3. 9.8 m/s² upward, because the ball was thrown up
  4. Smaller than 9.8 m/s², because the ball has slowed down
Show the answer

Free fall means a = g downward at every point of the flight: on the way up, at the top and on the way down.

  • Zero, because the ball is momentarily at rest: v = 0 is not a = 0: the velocity is changing from upward to downward at that instant.
  • Correct: 9.8 m/s² downward: Right: in free fall the acceleration is g downward the whole time, including the instant the velocity is zero.
  • 9.8 m/s² upward, because the ball was thrown up: The throw sets the initial velocity, not the acceleration; once released, the acceleration points down.
  • Smaller than 9.8 m/s², because the ball has slowed down: The acceleration does not depend on the speed; it stays 9.8 m/s² down throughout the flight.

6. A cart starts with a known velocity and speeds up with a known constant acceleration over a measured distance. The time is not known and not asked for. Which equation gives its final velocity most directly?

  1. v = v₀ + at
  2. v² = v₀² + 2a(x − x₀)
  3. x = x₀ + v₀t + ½at²
  4. v(avg) = Δx/Δt
Show the answer

Choose the equation that contains the unknown and the knowns and leaves out the quantity you do not have. Each constant-acceleration equation leaves out one quantity; v² = v₀² + 2aΔx leaves out t.

  • v = v₀ + at: This needs the time, which is not known.
  • Correct: v² = v₀² + 2a(x − x₀): Right: it links v, v₀, a and the displacement without t.
  • x = x₀ + v₀t + ½at²: This needs t, and it does not contain the final velocity.
  • v(avg) = Δx/Δt: This needs the time, and an average velocity is not the final velocity.

7. A motion diagram shows a cart's positions at equal time intervals. The dots get closer together from left to right while the cart moves right. What are the directions of its velocity and acceleration?

  1. Velocity to the right, acceleration to the right
  2. Velocity to the right, acceleration to the left
  3. Velocity to the left, acceleration to the left
  4. Velocity to the right, zero acceleration
Show the answer

Spacing shows speed: bigger gaps mean faster. Shrinking gaps while moving right mean v right and a left.

  • Velocity to the right, acceleration to the right: That would make the gaps grow: speeding up.
  • Correct: Velocity to the right, acceleration to the left: Right: it moves right, and shrinking gaps mean it is slowing, so a points opposite to v.
  • Velocity to the left, acceleration to the left: The cart moves right, so its velocity points right.
  • Velocity to the right, zero acceleration: Zero acceleration would give equal gaps.

Part 9 · Summary

Summary

Motion can be shown as a motion diagram, x-t, v-t and a-t graphs, equations and words. Slopes go down the chain (x-t to v-t to a-t) and areas go up it. For constant acceleration, three kinematic equations apply, each missing one quantity; free fall is constant acceleration g downward.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections