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Kinematics describes motion using position, displacement, velocity, acceleration, and time. This cheat sheet helps students choose the right equation, identify known and unknown quantities, and keep vector directions consistent. It is especially useful for one-dimensional motion, free fall, and interpreting motion graphs.

Students need these formulas because many physics problems use the same core relationships in different situations.

The main ideas are based on changes in position and velocity over time. Average velocity is vavg=ΔxΔtv_{avg} = \frac{\Delta x}{\Delta t}, and average acceleration is aavg=ΔvΔta_{avg} = \frac{\Delta v}{\Delta t}. For constant acceleration, the equations v=v0+atv = v_0 + at, Δx=v0t+12at2\Delta x = v_0t + \frac{1}{2}at^2, and v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x connect motion quantities.

Motion graphs show the same relationships visually, where slope and area often give important physical meaning.

Key Facts

  • Displacement is change in position, so Δx=xfxi\Delta x = x_f - x_i and direction matters.
  • Average velocity is vavg=ΔxΔtv_{avg} = \frac{\Delta x}{\Delta t}, while average speed is total distanceΔt\frac{\text{total distance}}{\Delta t}.
  • Average acceleration is aavg=ΔvΔt=vv0ta_{avg} = \frac{\Delta v}{\Delta t} = \frac{v - v_0}{t}.
  • For constant acceleration, final velocity is found with v=v0+atv = v_0 + at.
  • For constant acceleration, displacement is found with Δx=v0t+12at2\Delta x = v_0t + \frac{1}{2}at^2.
  • When time is not known, use v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x.
  • For free fall near Earth, acceleration is usually a=9.8 m/s2a = -9.8\ \text{m/s}^2 if upward is chosen as positive.
  • On a velocity-time graph, the slope equals acceleration and the area under the graph equals displacement.

Vocabulary

Position
Position is an object's location relative to a chosen origin, often written as xx.
Displacement
Displacement is the change in position, written Δx=xfxi\Delta x = x_f - x_i, and includes direction.
Velocity
Velocity is the rate of change of position, so average velocity is vavg=ΔxΔtv_{avg} = \frac{\Delta x}{\Delta t}.
Acceleration
Acceleration is the rate of change of velocity, written a=ΔvΔta = \frac{\Delta v}{\Delta t} for average acceleration.
Free Fall
Free fall is motion under the influence of gravity alone, with a=g9.8 m/s2a = g \approx 9.8\ \text{m/s}^2 downward near Earth.
Constant Acceleration
Constant acceleration means the value of aa does not change, allowing the standard kinematics equations to be used.

Common Mistakes to Avoid

  • Confusing distance with displacement is wrong because distance has no direction, while displacement can be positive, negative, or zero.
  • Using the wrong sign for acceleration is wrong because the direction of aa must match the chosen coordinate system, such as a=9.8 m/s2a = -9.8\ \text{m/s}^2 when upward is positive.
  • Mixing up initial and final velocity is wrong because v0v_0 is the velocity at the start and vv is the velocity at the end of the time interval.
  • Using constant-acceleration equations when acceleration changes is wrong because formulas like v=v0+atv = v_0 + at assume aa stays constant.
  • Forgetting units is wrong because quantities such as m\text{m}, s\text{s}, m/s\text{m/s}, and m/s2\text{m/s}^2 identify what the number physically means.

Practice Questions

  1. 1 A cyclist starts from rest and accelerates at 2.5 m/s22.5\ \text{m/s}^2 for 6.0 s6.0\ \text{s}. Find the final velocity using v=v0+atv = v_0 + at.
  2. 2 A car moving at 20 m/s20\ \text{m/s} slows down at 4.0 m/s2-4.0\ \text{m/s}^2 for 3.0 s3.0\ \text{s}. Find its displacement using Δx=v0t+12at2\Delta x = v_0t + \frac{1}{2}at^2.
  3. 3 A ball is dropped from rest off a building and falls for 2.0 s2.0\ \text{s}. Using a=9.8 m/s2a = 9.8\ \text{m/s}^2 downward, find the distance it falls.
  4. 4 A velocity-time graph is a horizontal line above the time axis. Explain what this shows about the object's velocity, acceleration, and displacement.

Understanding Kinematics Equations

The constant acceleration equations are models, not rules that fit every moving object. They work when acceleration stays the same during the time interval being studied. A dropped ball is often close to this condition for a short fall.

A car in city traffic usually is not, because the driver brakes, turns, or changes pedal pressure. In a problem, words such as steady, uniform, or constant signal that these equations may apply.

If acceleration changes, split the motion into smaller sections or use a graph. Knowing the limits of a model is part of doing physics correctly.

A clear choice of positive direction prevents many errors. Choose one direction at the start, then give every displacement, velocity, and acceleration a sign based on that choice. For vertical motion, students often choose upward as positive.

Gravity then has a negative value throughout the flight. A ball thrown upward slows because its velocity is positive while its acceleration is negative. At the highest point, its velocity is zero for an instant, but gravity is still acting.

On the way down, both velocity and acceleration are negative under this convention. The signs describe direction, not whether an object is speeding up or slowing down by themselves.

Graphs can reveal details that a single equation may hide. A horizontal line on a position versus time graph means the object remains at one location. A straight sloping line means constant velocity.

A curved position graph means velocity is changing. On a velocity versus time graph, a line that crosses zero shows a change in direction. Areas below the time axis count as negative displacement, so they can cancel areas above the axis.

This matters when an object goes out from its starting point and later returns. Data from phone motion sensors, bicycle trackers, and video analysis often produce uneven graphs because real measurements contain noise. Look for the overall pattern rather than expecting every point to lie perfectly on a line.

Good problem solving starts before any calculation. Draw a simple motion diagram with arrows showing direction at important moments. List known quantities with units, identify the time interval, and decide what quantity is actually needed.

Convert units early, especially kilometres per hour to metres per second. Check the final answer against the situation. A negative displacement can be correct when the object ends on the negative side of the chosen origin.

A negative time usually signals an incorrect setup. Estimate whether the size of an answer makes sense.

A short fall should not produce a distance like several kilometres, and a walking speed should not resemble the speed of a racing car. These checks catch errors that algebra alone cannot find.