PHY131H1/Unit 2

Unit 2 · Introduction to Physics I

Kinematics in One Dimension

Describing motion without asking what causes it. Three quantities — position, velocity, acceleration — each the rate of change of the one before, and four equations that solve every constant-acceleration problem you will meet this term.

By the end you can

• Distinguish distance from displacement and speed from velocity.
• Read position, velocity and acceleration off each other's graphs.
• Choose the right constant-acceleration equation for a problem.
• Handle free fall, including the moment at the top of a throw.

Why it matters & where it comes from

Why it exists

Before you can explain motion you need a precise language for it. Galileo's insight was that this description is separable from the causes: the equations here hold whatever produces the acceleration.

In the real world

Stopping distance on a highway is a constant-acceleration problem, and the quadratic dependence on speed is why doubling your speed quadruples the distance. Airbag timing, elevator comfort limits and sprint analysis are all the same three quantities.

Story & open questions

Galileo could not measure short times accurately, so he rolled balls down inclines to slow the motion and timed them by weighing water. He established that distance grows as the square of time without ever being able to time a falling object directly.

The three quantities

Definition

Displacement $\Delta x = x_f - x_i$ is the change in position — a vector, so it carries a sign. Distance is the total path length travelled and is never negative.

Average velocity is $\Delta x / \Delta t$; instantaneous velocity $v$ is its limit as $\Delta t \to 0$, the slope of the position graph.

Acceleration $a$ is the rate of change of velocity, the slope of the velocity graph.

Distance and displacement differ

Walk $10$ m east then $10$ m west: distance $20$ m, displacement zero. Average speed is distance over time and stays positive; average velocity is displacement over time and here is zero. Exam questions exploit this difference constantly.

Rule Sign conventions

Choose a positive direction at the start and hold it. A negative velocity means moving the other way; a negative acceleration does not mean slowing down.

Speeding up: $v$ and $a$ have the same sign.

Slowing down: $v$ and $a$ have opposite signs.

A ball thrown up has $a = -g$ throughout, but it speeds up only on the way down.

Graphs

Rule Slopes go down, areas go up

The slope of a position-time graph is velocity. The slope of a velocity-time graph is acceleration.

The area under a velocity-time graph is displacement. The area under an acceleration-time graph is change in velocity.

Areas below the axis count as negative, which is exactly how a graph encodes a return trip.

Example Reading a graph

A velocity-time graph rises linearly from $0$ to $20$ m/s over $4$ s, then stays flat for $6$ s.

Acceleration in the first phase is the slope, $20/4 = 5$ m/s$^2$; in the second it is zero.

Displacement in the first phase is the triangle, $\tfrac12 (4)(20) = 40$ m; in the second the rectangle, $(6)(20) = 120$ m. Total $160$ m.

Note the position graph would be a parabola then a straight line — curved where $a \ne 0$, straight where it is zero.

The constant-acceleration equations

Rule Four equations, five quantities

When $a$ is constant, with $v_0$ the initial velocity:

$v = v_0 + at$ — no $\Delta x$

$\Delta x = v_0 t + \tfrac12 a t^2$ — no $v$

$v^2 = v_0^2 + 2a\Delta x$ — no $t$

$\Delta x = \tfrac12 (v_0 + v) t$ — no $a$

Each omits exactly one of the five quantities $\{v_0, v, a, t, \Delta x\}$. List what you know, note what you want, and pick the equation missing the one you neither know nor want.

Example Choosing well

A car at $30$ m/s brakes at $6$ m/s$^2$. How far to stop?

Known: $v_0 = 30$, $v = 0$, $a = -6$. Wanted: $\Delta x$. Not known and not wanted: $t$.

So use the equation without $t$: $0 = 30^2 + 2(-6)\Delta x$, giving $\Delta x = 900/12 = 75$ m.

Finding $t$ first would have worked but taken two steps. Choosing the right equation is the whole skill.

Example Why stopping distance is quadratic

From $v^2 = v_0^2 + 2a\Delta x$ with $v = 0$, the stopping distance is $\Delta x = v_0^2 / (2|a|)$.

The dependence is on $v_0$ squared. Doubling from $50$ to $100$ km/h quadruples the distance, not doubles it.

This is the single most useful result in the unit outside an exam hall.

Free fall

Definition

Neglecting air resistance, every object near Earth's surface accelerates downward at $g = 9.8$ m/s$^2$ regardless of mass. With upward positive, $a = -g$ throughout the motion — rising, at the top, and falling.

At the top, $v = 0$ but $a \ne 0$

The most common misconception in the unit. At the peak of a throw the velocity is momentarily zero, but the acceleration is still $-g$. If it were zero the object would hang there permanently.

Example A throw, all the way up and back

A ball is thrown up at $20$ m/s. Find the peak height and the total flight time.

At the peak $v = 0$, so $0 = 20^2 + 2(-9.8)h$ gives $h = 400/19.6 = 20.4$ m.

Time to the peak: $0 = 20 - 9.8t$ gives $t = 2.04$ s. By symmetry the fall takes the same, so the total is $4.1$ s.

It returns at $-20$ m/s: same speed, opposite direction. That symmetry only holds without air resistance.

Common pitfalls

• Believing negative acceleration always means slowing down. Compare the signs of $v$ and $a$.
• Setting $a = 0$ at the top of a throw.
• Using the constant-acceleration equations when $a$ is not constant. They are worthless then.
• Confusing distance with displacement in an average.
• Losing the sign of $g$ by switching conventions mid-problem.

Practice

Work each fully before opening the answer. Original exercises written for this guide in the style of the course's problem sets.

Practice 1 · Distance vs displacement

A runner does one lap of a $400$ m track in $60$ s. Find average speed and average velocity.

Answer

Speed $400/60 = 6.7$ m/s. Velocity zero, since the displacement is zero for a complete lap.

Practice 2 · Pick the equation

A car accelerates from rest at $3$ m/s$^2$ for $5$ s. How far does it travel?

Answer

No $v$ needed, so use $\Delta x = v_0t + \tfrac12at^2 = 0 + \tfrac12(3)(25) = 37.5$ m.

Practice 3 · Stopping distance

By what factor does stopping distance change when speed triples?

Answer

Nine times, since distance goes as $v_0^2$.

Practice 4 · Top of the throw

What are $v$ and $a$ at the highest point of a vertical throw?

Answer

$v = 0$ and $a = -9.8$ m/s$^2$. The acceleration never changes during free fall.

Practice 5 · Graph reading

A position-time graph is a downward parabola. Describe the motion.

Answer

Velocity starts positive, decreases steadily through zero, then goes negative: constant negative acceleration. It is a throw straight up.

Practice 6 · Two-phase motion

A train accelerates from rest at $1$ m/s$^2$ for $20$ s, then brakes to rest at $2$ m/s$^2$. Total distance?

Solution

Phase one reaches $20$ m/s over $\tfrac12(1)(400) = 200$ m. Phase two needs $20^2/(2 \times 2) = 100$ m. Total $300$ m. Solve each phase separately — the equations require constant $a$ within a phase.

Practice 7 · Dropped versus thrown

Ball A is dropped; ball B is thrown down at $5$ m/s from the same height. Which hits first, and do they have the same acceleration?

Answer

B hits first, having started with downward speed. Both have acceleration $9.8$ m/s$^2$ downward the whole way — initial velocity does not change the acceleration.

Practice 8 · Reaction time

A driver at $25$ m/s takes $0.7$ s to react, then brakes at $8$ m/s$^2$. Total stopping distance?

Solution

Reaction phase is constant velocity: $25 \times 0.7 = 17.5$ m. Braking phase: $25^2/16 = 39.1$ m. Total $56.6$ m. Forgetting the reaction distance is the standard error, and it is a third of the answer here.

Expanded topic checklist

Course coverage. This unit is part of the complete PHY131H1 topic map. It should cover the following items before you move on:

• Definitions. Displacement versus distance, velocity versus speed, acceleration as a rate of change.
• Graphs. Slopes give the next derivative; areas give the previous quantity.
• The four equations. Which quantity each omits, and how to choose.
• Free fall. $a = -g$ throughout, and the behaviour at the top of a throw.

Textbook/source alignment

The calendar's "classical kinematics" begins here. Check the syllabus on Quercus for the assigned problems; this material is normally covered in the first two weeks.

Core idea map

Position, velocity and acceleration form a chain of rates of change. Constant acceleration makes that chain solvable in closed form, which is why four equations cover so much. Unit 3 applies the same equations independently to each axis.

Study routine. Before solving, write three lists: what you know, what you want, and what you do not care about. The right equation is the one missing the third list.

Deep practice extensions

Use these after the main practice set. They push on definitions and edge cases rather than routine calculation.

Extension 1 · Sign of acceleration

Does negative acceleration mean slowing down?

Guide

Not necessarily. It means slowing only if the velocity is positive. Same signs means speeding up.

Extension 2 · Graph areas

What does the area under a velocity-time graph represent?

Guide

Displacement, with area below the axis counting negative.

Extension 3 · Equation choice

Which equation has no time in it?

Guide

$v^2 = v_0^2 + 2a\Delta x$ — the one to reach for whenever time is neither given nor wanted.

Extension 4 · Validity

When do the four equations fail?

Guide

Whenever acceleration is not constant — with air resistance, or a spring force. Split the motion into constant-$a$ phases, or use calculus.

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