Mechanics

1D Kinematics

Here

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

2D Kinematics

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

Collisions

Finish

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

1D Kinematics

Kinematics describes motion without asking about the underlying cause. 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.

Velocity

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

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

If the two moments are indistinguishable (calculus allows us to take the limit as $t_2 - t_1 \rightarrow 0$ ), then we have the instantaneous velocity:

v(t) = \frac{dx}{dt}.

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 point.

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.

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}.

This can be a tricky concept to grasp fully, but the good news is that you already have incredible intuition for what it feels like to experience acceleration. The challenge is to connect this everyday experience to these more precise mathematical ideas.

Note that the relationship between acceleration and velocity is analogous to the relationship between velocity and position. Velocity is a rate of change (in position). Acceleration is a rate of change [in velocity]. This means acceleration is also a rate of change [in a rate of change (in position)].

As with velocity, instantaneous acceleration is the acceleration at one moment in time:

a(t) = \frac{dv}{dt}.

The slope idea repeats: acceleration is the slope of a velocity versus time plot in the same way velocity is the slope of a position versus time plot.

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.

The challenge below turns the sign conventions into a control problem. Apply acceleration left or right, watch the position and velocity traces, and stop inside each target zone.

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 $t_i = 0$ , so the final time is just $t$ . 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.

Kinematics Checkpoint

Question 1 of 3

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

Choose an answer to get instant feedback.