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Honors Physics Unit 3 Review: Acceleration

This unit is about acceleration, the rate at which an object's velocity changes, and how to describe it with numbers, equations, and graphs. By the end, you should be able to calculate acceleration, tell whether an object is speeding up or slowing down, solve constant-acceleration problems with the kinematic equations, and read motion graphs using slope and area.

Read the overview first, then work through each topic's notes and try the checkpoint question before checking the answer.

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What is Honors Physics unit 3?

In earlier units you described motion with constant velocity. Now velocity can change, and acceleration measures how quickly it does. This unit gives you the tools to predict where an object will be and how fast it will move when acceleration stays constant.

Honors Physics Unit 3 covers acceleration in one dimension. You learn that average acceleration is aavg=Δv/Δta_{avg} = \Delta v / \Delta t and that acceleration is measured in m/s². You also learn how the signs of velocity and acceleration show whether an object speeds up or slows down, and how free fall works with g≈9.8g \approx 9.8 m/s² downward. The second half covers the three constant-acceleration kinematic equations and how to choose between them. It also covers position-time, velocity-time, and acceleration-time graphs, where slopes and areas connect one graph to the next.

Acceleration is the rate of change of velocity, so it is a vector with a direction.

An acceleration of 4 m/s² means velocity changes by 4 m/s every second. Because velocity has a direction, acceleration happens whenever speed changes, direction changes, or both. In one-dimensional problems, the sign (+ or −) carries the direction.

Comparing the signs of velocity and acceleration tells you whether an object speeds up or slows down.

When velocity and acceleration have the same sign, the object speeds up. When they have opposite signs, it slows down. A negative acceleration alone does not mean slowing down.

Constant acceleration can be described with three equations and three connected graphs.

The kinematic equations link displacement, initial velocity, final velocity, acceleration, and time, and each one leaves out a different variable. On graphs, the slope of position-time is velocity, and the slope of velocity-time is acceleration. Area under velocity-time is displacement, and area under acceleration-time is change in velocity.

Equations and graphs are two views of the same constant-acceleration motion.

Every kinematics problem in this unit can be solved with an equation or read from a graph, and both give the same answer. For example, the area under a velocity-time graph equals the displacement you would get from Δx=v0t+12at2\Delta x = v_0 t + \frac{1}{2}at^2. If you get comfortable moving between the two, you can check your own work.

Honors Physics unit 3 topics

Unit 3 review notes

3.1

Acceleration

Acceleration is the rate at which velocity changes, measured in m/s². Average acceleration over an interval is aavg=Δv/Δta_{avg} = \Delta v / \Delta t. If acceleration is constant at 3 m/s², velocity changes by 3 m/s every second. Acceleration is a vector with a direction.

The sign of acceleration tells you its direction. It does not tell you whether the object is speeding up. To find that out, compare the acceleration's sign with the velocity's sign. If the signs match, the object speeds up, and if they are opposite, it slows down. A car moving left (negative velocity) with negative acceleration is speeding up.

Average acceleration covers a whole time interval. Instantaneous acceleration is the value at one moment, which you find from the slope of a tangent line on a velocity-time graph. When acceleration is constant, the two values are equal.

Near Earth's surface, an object in free fall with negligible air resistance has a=g≈9.8a = g \approx 9.8 m/s² downward, whatever its mass. Newton's second law, a=Fnet/ma = F_{net}/m, previews why forces cause acceleration. You'll use it fully in the forces unit.

Worked example. You throw a ball straight up at 15 m/s. Take up as positive, so a=−9.8a = -9.8 m/s². After 2.0 s, v=v0+at=15+(−9.8)(2.0)=15−19.6=−4.6v = v_0 + at = 15 + (-9.8)(2.0) = 15 - 19.6 = -4.6 m/s. The negative sign means the ball has already passed its highest point and is moving down at 4.6 m/s.

  • Average acceleration: The total change in velocity divided by the time interval, aavg=(vf−v0)/Δta_{avg} = (v_f - v_0)/\Delta t.
  • Instantaneous acceleration: The acceleration at a single moment, equal to the slope of the tangent line on a velocity-time graph at that time.
  • Free fall: Motion where gravity is the only significant force. Near Earth's surface, the acceleration is approximately constant at 9.8 m/s² downward.

A bike moving at −6 m/s has a constant acceleration of +2 m/s². Is it speeding up or slowing down, and what is its velocity 2 s later? (Answer: It is slowing down because the signs are opposite. Its velocity is v=−6+(2)(2)=−2v = -6 + (2)(2) = -2 m/s.)

Velocity signAcceleration signWhat happens
PositivePositiveSpeeds up
PositiveNegativeSlows down
NegativeNegativeSpeeds up
NegativePositiveSlows down
AnyZeroConstant velocity
3.2

Representing Acceleration with Equations and Graphs

Three kinematic equations describe motion with constant acceleration:

  • v=v0+atv = v_0 + at (no displacement)
  • Δx=v0t+12at2\Delta x = v_0 t + \frac{1}{2}at^2 (no final velocity)
  • v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x (no time)

To pick one, list what you know and what you want. Look for an equation that contains the unknown you want and values for all its other variables. If none fits, you may need more than one equation.

Graphs carry the same information. On a position-time graph, slope is velocity. Nonzero constant acceleration makes a parabola, while zero acceleration makes a straight line. On a velocity-time graph, slope is acceleration and area is displacement. Area below the time axis counts as negative. On an acceleration-time graph, area is the change in velocity.

Worked example. A car moving at 20 m/s brakes at a constant −5.0 m/s² until it stops. How far does it travel? You know v0=20v_0 = 20 m/s, v=0v = 0, and a=−5.0a = -5.0 m/s², and you want Δx\Delta x. Time is missing, so use v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x. Then 0=400+2(−5.0)Δx0 = 400 + 2(-5.0)\Delta x, so Δx=400/10=40\Delta x = 400/10 = 40 m.

Check with a graph. The stop takes t=(0−20)/(−5.0)=4.0t = (0 - 20)/(-5.0) = 4.0 s. The velocity-time graph is a triangle with area 12(4.0)(20)=40\frac{1}{2}(4.0)(20) = 40 m, which matches.

  • Kinematic equations: Three equations linking displacement, initial velocity, final velocity, acceleration, and time, valid only when acceleration is constant.
  • Slope: Rise over run on a graph. On position-time and velocity-time graphs, slope over an interval gives average velocity and average acceleration, respectively. A tangent slope gives the instantaneous value.
  • Area under the curve: The signed area between a graph and the time axis. It gives displacement on a velocity-time graph and change in velocity on an acceleration-time graph. Areas below the time axis count as negative.

A cart starts from rest and accelerates at 3.0 m/s² for 4.0 s. How far does it go? (Answer: Use Δx=v0t+12at2=0+12(3.0)(4.0)2=24\Delta x = v_0 t + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3.0)(4.0)^2 = 24 m.)

GraphSlope givesArea givesConstant acceleration looks like
Position vs. timeVelocityNothing useful in this courseA parabola for nonzero acceleration, or a straight line for zero acceleration
Velocity vs. timeAccelerationDisplacementA straight line, horizontal if acceleration is zero
Acceleration vs. timeNot used hereChange in velocityA horizontal line

Key terms

TermDefinition
AccelerationThe rate of change of velocity, a vector measured in m/s². Average acceleration over an interval is $a_{avg} = \Delta v / \Delta t$.
VelocityThe rate of change of position, a vector that includes both speed and direction.
DisplacementThe change in position, $\Delta x = x_f - x_0$, which can be positive or negative.
Constant accelerationAcceleration that does not change over time, the condition the kinematic equations require.
Instantaneous accelerationThe acceleration at one moment, found from the tangent slope on a velocity-time graph.
Free fallMotion under gravity alone. Near Earth's surface, acceleration is about 9.8 m/s² downward for any mass when air resistance is negligible.
Acceleration due to gravity (g)The magnitude of free-fall acceleration near Earth's surface, about 9.8 m/s² (sometimes rounded to 10 m/s²).
Velocity-time graphA graph whose slope gives acceleration and whose signed area gives displacement. Area below the time axis counts as negative.
VectorA quantity with both magnitude and direction, such as displacement, velocity, or acceleration.

Mistakes to avoid in Acceleration

Thinking negative acceleration always means slowing down

The word "deceleration" makes it feel that way, but the sign only shows direction. Compare signs instead. If velocity and acceleration match, the object speeds up, and if they differ, it slows down.

Saying acceleration is zero at the top of a throw

The ball's velocity is zero for an instant, but its velocity is still changing, so it is still accelerating. Acceleration stays −9.8 m/s² (with up positive) the whole flight, including the top.

Reading a velocity-time graph like a position-time graph

A line sloping down on a velocity-time graph means velocity is decreasing. It does not mean the object is moving backward or downhill. Check the axis label first, then ask what the slope and area mean for that graph.

Mixing signs or using the equations when acceleration isn't constant

Pick a positive direction at the start and give every vector (displacement, velocity, acceleration) a sign that matches it. The kinematic equations only work for constant acceleration. If acceleration is constant within each stage, you can solve each stage separately. For continuously changing acceleration, a graph can help you find changes in velocity or position from signed areas.

Study unit 3 other ways

Ready to review Unit 3?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.