Lesson 1.4 · Unit 1 · HS-PS2-1
The rate at which velocity changes, and how to read it off a graph
Velocity describes motion; acceleration describes how motion is changing. It is the quantity Newton's second law connects to force, so everything in unit 2 runs through it. It is also the quantity students most often get the sign of wrong, because "negative acceleration" and "slowing down" are not the same statement.
- Acceleration is the rate of change of velocity: \( a = \dfrac{\Delta v}{\Delta t} \), in \( \text{m}/\text{s}^2 \).
- It is a vector, so it has a sign, and that sign is a direction rather than a verdict on speeding up or slowing down.
- Compare the signs of \( a \) and \( v \) to know what is happening. Same sign means speeding up; opposite signs mean slowing down.
- An object can accelerate without changing speed, which is what circular motion does in lesson 4.3.
- On a velocity-time graph, acceleration is the slope.
- On a velocity-time graph, displacement is the area under the line, counting area below the axis as negative.
- The three graphs stack: slope of position-time gives velocity, slope of velocity-time gives acceleration, and the areas run the other way.
Where students lose marks: reading a negative acceleration as "slowing down". A ball thrown upward has a constant acceleration of \( -9.8\ \text{m}/\text{s}^2 \) throughout, yet it slows on the way up and speeds up on the way down. The acceleration never changed; the velocity's sign did.
The problem. A car moving at 12 m/s speeds up steadily to 30 m/s over 6.0 s, then brakes steadily from 25 m/s to rest in 5.0 s on a later trip. (a) Find the acceleration in each case. (b) Find the distance covered while speeding up, two ways. (c) Find the braking distance. (d) Explain why a negative acceleration does not always mean slowing down.
Step one: compute the first acceleration for (a). Take the direction of travel as positive. \( a = \dfrac{\Delta v}{\Delta t} = \dfrac{30 - 12}{6.0} = \dfrac{18}{6.0} = +3.0\ \text{m}/\text{s}^2 \). The velocity gains 3.0 m/s every second, which is what the unit is saying out loud.
Step two: compute the braking acceleration. \( a = \dfrac{0 - 25}{5.0} = -5.0\ \text{m}/\text{s}^2 \). Negative, because the velocity is positive and decreasing, so the change points backward. The car is still moving forward the entire time. Only the change in its velocity points backward.
Step three: find the distance by area for (b). Sketch the velocity-time graph: a straight line from \( (0, 12) \) to \( (6.0, 30) \). The area under it is a trapezoid, whose area is the average of the two parallel sides times the width: \( \dfrac{12 + 30}{2} \times 6.0 = 21 \times 6.0 = 126\ \text{m} \).
Step four: check it by equation. Using \( \Delta x = v_0 t + \tfrac{1}{2}at^2 \): \( 12(6.0) + \tfrac{1}{2}(3.0)(6.0)^2 = 72 + 54 = 126\ \text{m} \) ✓ The two methods agree because they are the same statement. Lesson 1.5 derives the equation from the area, so a match here is a check on the arithmetic rather than on the physics.
Step five: compute (c). The velocity-time graph falls from \( (0, 25) \) to \( (5.0, 0) \), a triangle. Area \( = \dfrac{25 + 0}{2} \times 5.0 = 12.5 \times 5.0 = 62.5\ \text{m} \). The acceleration was negative but the area is above the axis, because the velocity stayed positive. The car moved forward 62.5 m while stopping.
Step six: begin (d). The sign of the acceleration tells you which way the velocity is being pushed, not whether the speed is rising. Four cases exhaust the possibilities.
| Velocity | Acceleration | What happens |
|---|---|---|
| positive | positive | moving forward, speeding up |
| positive | negative | moving forward, slowing down |
| negative | negative | moving backward, speeding up |
| negative | positive | moving backward, slowing down |
Step seven: apply it to the thrown ball. Throw a ball straight up with up positive. Gravity gives it \( a = -9.8\ \text{m}/\text{s}^2 \) for the whole flight, unchanged at every instant including the top. On the way up the velocity is positive and the acceleration negative: row two, slowing down. On the way down the velocity is negative and the acceleration negative: row three, speeding up. Same acceleration, opposite behaviors, because the velocity's sign flipped.
Step eight: settle the question about the top. At the highest point the velocity is momentarily zero. The acceleration is still \( -9.8\ \text{m}/\text{s}^2 \). Students often want it to be zero there, reasoning that nothing is moving. But acceleration measures how velocity is changing, and the velocity is changing rapidly at that instant: it is passing from positive through zero to negative. If the acceleration really were zero at the top, the ball would stay there, since zero acceleration means unchanging velocity and the velocity is zero. That it comes back down is the evidence that the acceleration never stopped.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.
- Give the SI unit of acceleration.
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\( \text{m}/\text{s}^2 \)
- What does the slope of a velocity-time graph give?
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Acceleration
- What does the area under a velocity-time graph give?
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Displacement
- A car goes from 8.0 m/s to 20 m/s in 4.0 s. Find its acceleration.
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\( \dfrac{20-8.0}{4.0} \). \( +3.0\ \text{m}/\text{s}^2 \)
- An object has positive velocity and negative acceleration. Is it speeding up or slowing
down?
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Opposite signs. Slowing down
- A velocity-time graph is a horizontal line at 15 m/s for 8.0 s. Find the acceleration
and the displacement.
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Slope is zero, so \( a = 0 \). Area is a rectangle: \( 15 \times 8.0 = 120\ \text{m} \). Constant velocity is a special case of the graph method, not a separate rule. \( a = 0 \), displacement 120 m
- A ball rolls up a ramp, stops, and rolls back down. Describe the sign of its velocity
and acceleration throughout, taking up the ramp as positive.
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Going up: velocity positive, decreasing. Acceleration negative. At the top: velocity zero. Acceleration still negative. Coming down: velocity negative, growing more negative. Acceleration still negative. The acceleration is negative and constant for the whole motion, because the component of gravity along the ramp never changes direction. Only the velocity's sign changes. Acceleration negative throughout; velocity positive, then zero, then negative
- A train brakes from 30 m/s at a steady
\( -1.5\ \text{m}/\text{s}^2 \). Find how long it takes to stop and how far it travels.
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Time: \( t = \dfrac{\Delta v}{a} = \dfrac{0 - 30}{-1.5} = 20\ \text{s} \). The negatives cancel, giving a positive time, which is the check that the setup was right. Distance: area of the triangle, \( \dfrac{30 + 0}{2} \times 20 = 300\ \text{m} \). Check by equation. \( v^2 = v_0^2 + 2a\Delta x \) gives \( 0 = 900 + 2(-1.5)\Delta x \), so \( \Delta x = \dfrac{900}{3.0} = 300\ \text{m} \) ✓ 20 s and 300 m
- Explain how an object can be accelerating while its speed stays constant.
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Because velocity is a vector, and a vector can change by turning without changing length. The definition is about velocity, not speed. Acceleration is \( \dfrac{\Delta \vec{v}}{\Delta t} \), and \( \vec{v} \) carries a direction as well as a magnitude. A change in either is a change in \( \vec{v} \). The standard case. A car going round a bend at a steady 20 m/s has constant speed and continuously changing direction, so its velocity is changing at every instant and it is accelerating the whole way round. Which way that acceleration points. Toward the center of the curve, which lesson 4.3 derives. It is perpendicular to the velocity, which is exactly why it turns the motion without changing its speed. The general rule worth carrying forward. An acceleration parallel to the velocity changes speed only; perpendicular to it changes direction only; at any other angle it does some of each. Projectile motion in lesson 1.7 is the mixed case. Why this matters for forces. Since force causes acceleration, a body moving in a circle at constant speed must have a net force on it, pointing inward. Anything moving in a curve is being pushed, and identifying what is doing the pushing is usually the whole problem. Velocity is a vector, so changing direction is a change in velocity
- A motorcycle accelerates from rest at
\( 2.5\ \text{m}/\text{s}^2 \) for 8.0 s, then holds that speed for 12 s. Find the final
speed, the total displacement, and the average velocity for the whole 20 s.
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Phase 1, the acceleration. Final speed: \( v = v_0 + at = 0 + 2.5(8.0) = 20\ \text{m/s} \). Displacement: \( \Delta x = \tfrac{1}{2}at^2 = \tfrac{1}{2}(2.5)(64) = 80\ \text{m} \). Check by area: a triangle of height 20 m/s and width 8.0 s has area \( \tfrac{1}{2}(20)(8.0) = 80\ \text{m} \) ✓ Phase 2, the steady speed. \( \Delta x = vt = 20 \times 12 = 240\ \text{m} \). Totals. Displacement: \( 80 + 240 = 320\ \text{m} \). Time: \( 8.0 + 12 = 20\ \text{s} \). Average velocity. \( \bar{v} = \dfrac{320}{20} = 16\ \text{m/s} \). Sanity check the average. It must lie between the slowest and fastest speeds reached, which are 0 and 20 m/s. It does, and it sits nearer 20 because most of the time was spent at the top speed ✓ A common wrong route. Averaging the two phase speeds, 10 and 20, to get 15 m/s. That treats the phases as equally long when the second lasts half again as long as the first. Average velocity is always total displacement over total time. Note the shape of the graph this describes. A velocity-time graph rising in a straight line from the origin to \( (8.0, 20) \), then horizontal to \( (20, 20) \). The total area is a triangle plus a rectangle, which is exactly the two phases computed separately. 20 m/s, 320 m, and an average of 16 m/s