Mechanics

1D Kinematics

Learn position, velocity, and acceleration through graph slopes, signed areas, free fall, and a stop-in-zones challenge.

2D Kinematics

Extend kinematics into vector components, projectile motion, and two-dimensional acceleration.

Video Analysis

Measure real motion from a phone video: mark the object frame by frame, set a scale, and fit the position data.

Forces

Explore common mechanics forces with Newt the physics frog, from spring-like contact forces to gravity and friction.

Collisions

Compare elastic and inelastic collisions while keeping momentum and energy bookkeeping straight.

Energy

Build a 3D roller coaster and watch kinetic, potential, and thermal energy trade as the train runs.

1D Kinematics

Kinematics describes motion. The motion of an object is characterized by its location in space, position, as a function of time. It’s convenient to then define two additional quantities:

Velocity tells us how quickly position changes, including direction.
Acceleration tells us how quickly velocity changes, including direction.

We could continue this process indefinitely, but it turns out that the underlying causes of motion can be directly tied to acceleration.

At 0.0 s the hedgehog is at x = 0.0 m, moving at 2.4 m/s at constant speed.

Once you choose a spatial and temporal coordinate system, like the gridlines and stopwatch for the above hedgehog, you can begin describing motion quantitatively. Most of the confusion surrounding kinematics comes from the colloquial use of physical terms and the ambiguity of natural language. Speed and velocity are often used interchangeably, and for this particular hedgehog, the phrase slowing down refers to both positive and negative acceleration.

Velocity

Average velocity is the rate at which position changes between two moments in time:

\htmlClass{math-hint math-hint-vbar}{\overline{v}} = \frac{\htmlClass{math-hint math-hint-delta-x}{\Delta x}}{\htmlClass{math-hint math-hint-delta-x}{\Delta t}} = \frac{x_2 - x_1}{t_2 - t_1}.

Note the first sign distinction that arises from this definition: the numerator can be positive or negative. For horizontal motion, the convention is usually that $x$ increases in the rightward direction. It follows that moving to the right corresponds to positive values of velocity, while moving to the left is negative velocity.

When we don’t have access to animations or video, it is helpful to make a graph of position and time. If all motion occurs along one spatial dimension (e.g. a hedgehog running left or right), we end up with a 2D plot. It is convention to place time along the horizontal axis.

Graphically, average velocity is the slope of a secant line on a position versus time plot. Instantaneous velocity is the slope of the tangent line at a single moment in time.

Position and Velocity

Move the markers to compare secant slopes, tangent slopes, and the matching velocity graph.

Is velocity the same as speed?

Almost, but not quite. Speed only says how fast something moves. Velocity says how fast and in which signed direction. In one dimension, the speed is the absolute value of the velocity:

\text{speed} = |v|.

An object with $v = -4\,\text{m/s}$ and an object with $v = +4\,\text{m/s}$ have the same speed, but they are moving in opposite directions.

Slope vs. Area

Reading the slopes of this plot takes you from position to velocity. To go from velocity to position, there is a similar graphical connection.

Mark off a strip of the velocity graph narrow enough that the velocity barely changes across it. Over that strip the motion is effectively constant-velocity, so the displacement is $\Delta x = v\,\Delta t$ . This is simply the area of the strip, height times width. Add up every strip between $t_1$ and $t_2$ and you have the whole displacement:

\Delta x = \text{area under } v(t) \text{ from } t_1 \text{ to } t_2.

Note these areas are signed. A strip below the horizontal axis has negative height, so it subtracts rather than adds. On a velocity graph, a trip out and back can result in a lot of shaded area with zero displacement. This illustrates the difference between displacement (change in position) and distance traveled: displacement adds the signed areas, distance traveled adds their absolute values.

Strips5Strips below the axis subtract

Δt = 2.00sΣ v Δt = 10.22mexact Δx = 10.00m

With 5 strips, the sum overshoots the exact displacement by 0.22 m.

Acceleration

Average acceleration is the rate at which velocity changes between two moments in time:

\overline{a} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}.

Note that the relationship between acceleration and velocity is analogous to the relationship between velocity and position. Velocity is the rate of change of position, and acceleration is the rate of change of velocity. This makes acceleration a rate of change of a rate of change.

Again, we often plot velocity against time. The graphical interpretations repeat: acceleration is the slope of a velocity versus time plot, and changes in velocity are indicated by the area under an acceleration versus time curve.

Velocity and Acceleration

Read average acceleration from a secant slope, instantaneous acceleration from a tangent slope, and the change in velocity from the shaded area.

How can negative acceleration speed something up?

Negative acceleration means the velocity is changing in the negative direction. If an object is already moving left with $v < 0$ , then negative acceleration makes the velocity more negative and the object speeds up. If an object is moving right with $v > 0$ , the same negative acceleration makes the object slow down until the velocity reaches zero.

Zone Challenge

Use constant acceleration controls to compete for a time on the leaderboard.

Kinematic Equations

The common constant-acceleration equations come straight from the definitions above. Let the clock start at $t_i = 0$ , and call the final time $t_f = t$ . The displacement is $\Delta x = x_f - x_i$ . Click an equation to reveal where it comes from.

v_f = v_i + at

\Delta x = \left(\frac{v_i + v_f}{2}\right)t

\Delta x = v_i t + \frac{1}{2}at^2

v_f^2 = v_i^2 + 2a\Delta x

Start with average acceleration:

a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t}.

Multiply both sides by $t$ and add $v_i$ :

at = v_f - v_i \qquad \Rightarrow \qquad v_f = v_i + at.

By definition,

\overline{v} = \frac{\Delta x}{t},

\Delta x = \overline{v}t.

When acceleration is constant, velocity changes at a steady rate, so the average velocity is the midpoint between the initial and final velocities:

\overline{v} = \frac{v_i + v_f}{2}.

Therefore,

\Delta x = \left(\frac{v_i + v_f}{2}\right)t.

Start with the average-velocity displacement equation:

\Delta x = \left(\frac{v_i + v_f}{2}\right)t.

Substitute $v_f = v_i + at$ :

\Delta x = \left(\frac{v_i + v_i + at}{2}\right)t = v_i t + \frac{1}{2}at^2.

Start with $v_f = v_i + at$ , then solve for time:

t = \frac{v_f - v_i}{a}.

Substitute that into the average-velocity displacement equation:

\Delta x = \left(\frac{v_i + v_f}{2}\right) \left(\frac{v_f - v_i}{a}\right) = \frac{v_f^2 - v_i^2}{2a}.

Rearranging gives

v_f^2 = v_i^2 + 2a\Delta x.

If the interval does not start at $t = 0$ , replace $t$ with $\Delta t$ .

Problem Solving

Average Velocity from Position Data

Problem

A cart is at $x_1 = -3.0\,\text{m}$ when $t_1 = 2.0\,\text{s}$ . Later, it is at $x_2 = 9.0\,\text{m}$ when $t_2 = 5.0\,\text{s}$ . Find the cart’s average velocity over this interval.

Given

$x_1 = -3.0\,\text{m}$
$t_1 = 2.0\,\text{s}$
$x_2 = 9.0\,\text{m}$
$t_2 = 5.0\,\text{s}$

Use

\overline{v} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}.

Solve

First find the change in position:

\Delta x = 9.0 - (-3.0) = 12.0\,\text{m}.

Then find the elapsed time:

\Delta t = 5.0 - 2.0 = 3.0\,\text{s}.

So the average velocity is

\overline{v} = \frac{12.0}{3.0} = 4.0\,\text{m/s}.

Check

The cart moved from a negative position to a positive position, so the displacement is positive. A positive average velocity is exactly what we should expect.

Average Acceleration from Velocity Data

Problem

A cyclist is moving right at $12\,\text{m/s}$ and brakes uniformly until their velocity is $4\,\text{m/s}$ to the right after $4.0\,\text{s}$ . Find the average acceleration.

Given

$v_1 = 12\,\text{m/s}$
$v_2 = 4\,\text{m/s}$
$\Delta t = 4.0\,\text{s}$

Use

\overline{a} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{\Delta t}.

Solve

The change in velocity is

\Delta v = 4 - 12 = -8\,\text{m/s}.

Therefore

\overline{a} = \frac{-8}{4.0} = -2.0\,\text{m/s}^2.

Check

The cyclist is still moving right because the velocity remains positive, but the negative acceleration means that rightward velocity is decreasing. That matches the idea of braking.

Stopping Under Constant Acceleration

Problem

In the zone challenge, suppose the cart is moving right at $8.0\,\text{m/s}$ and you apply a constant acceleration of $-2.0\,\text{m/s}^2$ . How long does it take to stop, and how far does it travel during that time?

Given

$v_i = 8.0\,\text{m/s}$
$v_f = 0$
$a = -2.0\,\text{m/s}^2$

Use

With constant acceleration,

v_f = v_i + at.

The velocity changes linearly, so the displacement is average velocity times time:

\Delta x = \overline{v}t = \left(\frac{v_i + v_f}{2}\right)t.

Solve

Solve for the stopping time:

0 = 8.0 + (-2.0)t

2.0t = 8.0 \qquad \Rightarrow \qquad t = 4.0\,\text{s}.

The average velocity during the stop is

\overline{v} = \frac{8.0 + 0}{2} = 4.0\,\text{m/s}.

So the stopping distance is

\Delta x = (4.0)(4.0) = 16\,\text{m}.

Check

The acceleration points opposite the motion, so the cart should slow to a stop. It does not stop immediately; it continues moving right while its velocity falls to zero, so a positive stopping distance makes sense.

How High Does It Go?

Problem

A ball is thrown straight up at $12\,\text{m/s}$ . How high above the launch point does it rise?

Given

$v_i = 12\,\text{m/s}$
$v_f = 0$ at the highest point
$a = -g = -9.8\,\text{m/s}^2$

Use

The question asks for a displacement and never mentions time, so reach for the equation that has no $t$ in it:

v_f^2 = v_i^2 + 2a\Delta y.

Solve

Substitute the known values:

0 = (12)^2 + 2(-9.8)\Delta y.

Rearrange and solve:

19.6\,\Delta y = 144 \qquad \Rightarrow \qquad \Delta y = 7.3\,\text{m}.

Check

A positive $\Delta y$ means the ball finished above where it started, which is right for a throw upward. Note what marks the peak: the velocity reaches zero there, not the acceleration. The acceleration is $-9.8\,\text{m/s}^2$ the entire way.

Kinematics Checkpoint

An object has positive position and negative velocity. What is it doing?