Motion in Straight Line

moving observer
train
bus
passenger
platform

Motion assumed absolute.

Motion is observer-dependent.

$$ v_{AB}=v_A-v_B $$

returns home
comes back
returns to starting point
completes loop
finishes lap

Distance assumed zero.

Displacement becomes zero.

$$ \Delta x = 0 $$ Distance may still be non-zero.

highest point
maximum height
topmost position
peak of motion
reaches greatest height

Acceleration assumed zero.

Velocity becomes zero.

$$ v = 0 $$ Acceleration still exists.

$$ a = -g $$

crosses same level again
same height during ascent and descent
passes through same point again
returns to same level
equal heights

Different speed magnitudes assumed.

Speed magnitudes are equal.

$$ |v_{up}|=|v_{down}| $$ Directions are opposite.

$v<0$ given
negative velocity
motion in negative direction
velocity shown below time axis

Negative velocity interpreted as slowing down.

Negative sign indicates direction only.

$$ v<0 $$ Motion along the negative axis.

Speeding up or slowing down depends on:

$$ v\cdot a $$

$a<0$ given
negative acceleration
acceleration opposite positive axis
decelerating mentioned

Negative acceleration interpreted as retardation.

Negative sign indicates direction only.

$$ a<0 $$ Acceleration acts along negative axis.

Check relative signs of:

$$ v $$ and $$ a $$

$$ v\cdot a>0 $$ Speed increases

$$ v\cdot a<0 $$ Speed decreases

equal distances
half journey
same path lengths
first half distance
second half distance

Arithmetic mean applied.

Use Harmonic Mean.

$$ v_{avg} = \frac{2v_1v_2} {v_1+v_2} $$

equal times
same duration
first half time
second half time
equal time intervals

Harmonic mean applied.

Use Arithmetic Mean.

$$ v_{avg} = \frac{v_1+v_2}{2} $$

acceleration constant
uniformly accelerated motion
constant retardation
free fall
acceleration does not change

Velocity assumed constant.

Activate UAM Toolkit.

$$ v=u+at $$
$$ s=ut+\frac12at^2 $$
$$ v^2=u^2+2as $$
$$ s=\frac{u+v}{2}t $$

starts from rest
initially at rest
released
dropped
$u=0$

General equations used without simplification.

Activate Rest-State Shortcuts.

$$ u=0 $$
$$ v=at $$
$$ s=\frac12at^2 $$
$$ v^2=2as $$
Successive distances:

$$ 1:3:5:7:9 $$

dropped
released
freely falling
falls under gravity
allowed to fall

Initial velocity unnecessarily introduced.

Activate Free-Fall Model.

$$ u=0 $$
$$ a=g $$
$$ v=gt $$
$$ s=\frac12gt^2 $$
$$ v^2=2gs $$

thrown upward
projected vertically upward
launched upward
ball thrown up
upward projection

Acceleration assumed upward.

Gravity always acts downward.

$$ a=-g $$
$$ v=u-gt $$
$$ s=ut-\frac12gt^2 $$
$$ v^2=u^2-2gs $$

overtaking
chase
catch up
faster body behind slower body
pursues another object

Both bodies solved separately.

Activate Relative Motion.

$$ v_{rel} = |v_A-v_B| $$
$$ t = \frac{D} {v_{rel}} $$

distance between bodies
separation asked
gap between objects
distance apart
relative position

Distance of each body calculated separately.

Activate Relative Displacement.

$$ x_{AB} = x_A-x_B $$
Separation is the difference in positions.

released later
starts after
delayed launch
second body released after
begins motion after time t

Common time origin assumed.

Use separate clocks.

For first body:
$$ t $$

For second body:
$$ t-\Delta t $$

Measure time from each body's own start.

reverses direction
changes direction
turns back
comes down after going up
reaches turning point

Motion assumed finished.

Velocity becomes zero only momentarily.

$$ v=0 $$
Motion continues afterwards in the opposite direction.

Acceleration still exists.

x–t graph
position–time graph
displacement–time graph
coordinate vs time graph

Area interpreted physically.

Primary tool = Slope.

$$ \text{Slope} = \frac{dx}{dt} = v $$
Area under x–t graph has no standard physical meaning.

v–t graph
velocity–time graph
speed–time graph
velocity plotted against time

Slope and area confused.

Slope gives acceleration.

$$ \text{Slope} = \frac{dv}{dt} = a $$
Area gives displacement.

$$ \text{Area} = \int v\,dt = \Delta x $$

a–t graph
acceleration–time graph
acceleration plotted against time

Area interpreted as displacement.

Area gives change in velocity.

$$ \text{Area} = \int a\,dt = \Delta v $$
Constant acceleration:

$$ \Delta v=at $$

half displacement
midpoint of journey
middle of path
halfway in distance
half the total displacement

Half-time formula applied.

Midpoint of displacement is not midpoint of time.

$$ v_{mid} = \sqrt{ \frac{u^2+v^2}{2} } $$
Valid only for constant acceleration.

half the time
midpoint of duration
after half the total time
halfway in time
t/2

Midpoint-of-displacement formula applied.

Half time is not half displacement.

$$ v_{t/2} = u+\frac{at}{2} $$
Valid for constant acceleration.

same acceleration
equal acceleration
both under gravity
accelerated equally
identical acceleration

Relative acceleration calculated unnecessarily.

Relative acceleration becomes zero.

$$ a_A=a_B $$
$$ a_{AB}=0 $$
Therefore,

$$ v_{AB} = \text{constant} $$

circular path
moving in a circle
turning continuously
curved motion
completes circular motion

Constant speed interpreted as zero acceleration.

Direction changes continuously.

Velocity changes continuously.

Therefore,

$$ a \ne 0 $$
Constant speed does not imply constant velocity.