The Central Idea
Position, velocity, and acceleration are linked by differentiation:
\[ \boxed{ v(t)=x'(t) } \]and
\[ \boxed{ a(t)=v'(t)=x''(t) }. \]They are also linked in the reverse direction by accumulation:
\[ \boxed{ \Delta x = \int v(t)\,dt } \]and
\[ \boxed{ \Delta v = \int a(t)\,dt }. \]Three Graphs, One Motion
A single motion can be represented by:
- a position-time graph;
- a velocity-time graph;
- an acceleration-time graph.
These are not three unrelated pictures. Each graph contains information that can be translated into the others.
Position-Time Graphs
A position-time graph plots
\[ x(t) \]vertically against time \(t\) horizontally.
The graph's height tells the object's coordinate position at each time.
Reading Position from Graph Height
If a position graph has
\[ x(3)=7, \]then at time
\[ t=3 \]the object is at position
\[ \boxed{x=7}. \]Position is read from the graph's vertical value, not its slope.
Velocity Is the Slope of a Position Graph
Since
\[ v(t)=x'(t), \]instantaneous velocity is the tangent slope of the position-time graph.
Therefore:
- positive position slope means positive velocity;
- negative position slope means negative velocity;
- zero position slope means zero velocity.
Average Velocity from a Position Graph
From \(t=t_1\) to \(t=t_2\),
\[ \boxed{ v_{\text{avg}} = \frac{x(t_2)-x(t_1)} {t_2-t_1} }. \]Geometrically, this is the slope of the secant line joining the two points on the position graph.
Example: Average Velocity from Two Points
Suppose
\[ x(1)=4 \]and
\[ x(5)=16. \]Then
\[ v_{\text{avg}} = \frac{16-4}{5-1} \] \[ = \frac{12}{4} \] \[ = \boxed{3}. \]Steepness and Speed
On a position-time graph, the magnitude of slope corresponds to speed.
A steep positive slope means large positive velocity.
A steep negative slope means large negative velocity.
A shallow slope means smaller speed.
A Horizontal Position Graph
If
\[ x(t)=C \]is constant over an interval, the position graph is horizontal.
Therefore,
\[ \boxed{v=0}. \]The object is at rest during that interval.
A Straight Slanted Position Graph
A straight position-time graph has constant slope.
Therefore, velocity is constant.
If
\[ x(t)=x_0+vt, \]then
\[ \boxed{x'(t)=v}. \]Curved Position Graphs
A curved position-time graph has changing slope.
Therefore, velocity is changing.
Changing velocity means the object has acceleration.
Acceleration and Concavity of Position
Since
\[ a(t)=x''(t), \]acceleration is related to the concavity of the position graph.
If the position graph is concave up,
\[ \boxed{a>0}. \]If it is concave down,
\[ \boxed{a<0}. \]What Concave Up Means
A concave-up position graph has tangent slopes that increase with time.
Those slopes are velocities.
Therefore, velocity is increasing and acceleration is positive.
What Concave Down Means
A concave-down position graph has tangent slopes that decrease with time.
Therefore, velocity is decreasing and acceleration is negative.
Position Zero Does Not Mean Velocity Zero
If
\[ x(t)=0, \]the object is at the chosen origin.
It may pass through the origin with nonzero velocity.
Position and velocity answer different questions.
Velocity Zero Does Not Mean Position Zero
If
\[ v(t)=0, \]the position graph has horizontal tangent slope.
The object may be located anywhere in the coordinate system.
Turning Points on Position Graphs
A one-dimensional object may reverse direction when velocity changes sign.
On the position graph, this often appears as a local maximum or local minimum.
At a smooth turning point,
\[ \boxed{v=0}. \]Not Every \(v=0\) Point Must Reverse Direction
Velocity can be zero without changing sign.
For example,
\[ x(t)=t^3 \]has
\[ v(t)=3t^2. \]At \(t=0\),
\[ v=0, \]but \(v\) remains nonnegative on both sides, so the object does not reverse direction.
Velocity-Time Graphs
A velocity-time graph plots
\[ v(t) \]against time.
Its vertical height tells instantaneous velocity.
Reading Direction from a Velocity Graph
In one dimension:
- \(v>0\): motion in the positive direction;
- \(v<0\): motion in the negative direction;
- \(v=0\): momentary rest.
Speed from a Velocity Graph
Speed is the magnitude of velocity:
\[ \boxed{ \text{speed}=|v| }. \]Therefore, distance from the horizontal axis on a velocity graph corresponds to speed.
Acceleration Is the Slope of a Velocity Graph
Since
\[ a(t)=v'(t), \]acceleration is the tangent slope of the velocity-time graph.
Thus:
- rising velocity graph \(\Rightarrow a>0\);
- falling velocity graph \(\Rightarrow a<0\);
- horizontal velocity graph \(\Rightarrow a=0\).
Average Acceleration from a Velocity Graph
From \(t_1\) to \(t_2\),
\[ \boxed{ a_{\text{avg}} = \frac{ v(t_2)-v(t_1) }{ t_2-t_1 } }. \]This is the secant slope of the velocity graph.
Example: Average Acceleration
Suppose
\[ v(2)=5 \]and
\[ v(6)=17. \]Then
\[ a_{\text{avg}} = \frac{17-5}{6-2} \] \[ = \frac{12}{4} \] \[ = \boxed{3}. \]Displacement Is Signed Area under a Velocity Graph
Since
\[ v(t)=\frac{dx}{dt}, \]displacement from \(t=a\) to \(t=b\) is
\[ \boxed{ \Delta x = \int_a^b v(t)\,dt }. \]Geometrically, this is signed area between the velocity graph and the time axis.
Positive and Negative Velocity Areas
Area above the time axis contributes positively to displacement.
Area below the time axis contributes negatively.
Therefore, opposite-direction motion can cancel in net displacement.
Distance from a Velocity Graph
Total distance traveled is obtained by accumulating speed:
\[ \boxed{ d = \int_a^b|v(t)|\,dt }. \]Geometrically, count every velocity-area contribution positively.
Example: Displacement versus Distance
Suppose a velocity graph has area
\[ 12 \]above the time axis and area
\[ 5 \]below the time axis.
Then displacement is
\[ \Delta x=12-5 \] \[ = \boxed{7}. \]Total distance is
\[ d=12+5 \] \[ = \boxed{17}. \]Velocity Zeros and Direction Changes
When the velocity graph crosses the time axis,
\[ v=0. \]If the sign changes across the zero, the object reverses direction.
If the graph merely touches the axis and keeps the same sign, the object may stop momentarily without reversing.
Acceleration-Time Graphs
An acceleration-time graph plots
\[ a(t) \]against time.
Its vertical height tells instantaneous acceleration.
Reading Acceleration Sign
On an acceleration graph:
- above the axis means positive acceleration;
- below the axis means negative acceleration;
- on the axis means zero acceleration.
Change in Velocity Is Signed Area under Acceleration
Since
\[ a(t)=\frac{dv}{dt}, \]the change in velocity from \(t=a\) to \(t=b\) is
\[ \boxed{ \Delta v = \int_a^b a(t)\,dt }. \]Acceleration Area Is Not Displacement
Area under an acceleration-time graph has units of velocity.
Therefore, it represents
\[ \boxed{\Delta v}, \]not displacement.
Displacement comes from area under the velocity graph.
Recovering Velocity from Acceleration
If the initial velocity \(v(t_0)\) is known, then
\[ \boxed{ v(t) = v(t_0) + \int_{t_0}^{t}a(s)\,ds }. \]An acceleration graph determines changes in velocity, but an initial velocity is needed to determine absolute velocity values.
Recovering Position from Velocity
If the initial position \(x(t_0)\) is known, then
\[ \boxed{ x(t) = x(t_0) + \int_{t_0}^{t}v(s)\,ds }. \]A velocity graph determines position changes, but an initial position is needed for absolute position values.
The Derivative Direction
Moving from position to velocity to acceleration uses slopes:
\[ \boxed{ x(t) \xrightarrow{\text{slope}} v(t) \xrightarrow{\text{slope}} a(t) }. \]The Accumulation Direction
Moving from acceleration to velocity to position uses signed areas:
\[ \boxed{ a(t) \xrightarrow{\text{signed area}} \Delta v \xrightarrow{\text{signed area}} \Delta x }. \]Initial values provide the constants needed to reconstruct the full functions.
Graph-Translation Summary
| Graph | Height Means | Slope Means | Signed Area Means |
|---|---|---|---|
| Position \(x(t)\) | Position | Velocity | Not a standard kinematic accumulation quantity |
| Velocity \(v(t)\) | Velocity | Acceleration | Displacement |
| Acceleration \(a(t)\) | Acceleration | Rate of change of acceleration | Change in velocity |
Speeding Up and Slowing Down from \(v\) and \(a\)
In one-dimensional motion:
\[ \boxed{ v\text{ and }a\text{ same sign} \Rightarrow \text{speeding up} }. \]Also,
\[ \boxed{ v\text{ and }a\text{ opposite signs} \Rightarrow \text{slowing down} }. \]Velocity Positive, Acceleration Positive
If
\[ v>0 \]and
\[ a>0, \]the object moves in the positive direction and its speed increases.
Velocity Positive, Acceleration Negative
If
\[ v>0 \]and
\[ a<0, \]the object moves in the positive direction but slows down.
Velocity Negative, Acceleration Negative
If
\[ v<0 \]and
\[ a<0, \]the object moves in the negative direction and speeds up.
Velocity Negative, Acceleration Positive
If
\[ v<0 \]and
\[ a>0, \]the object moves in the negative direction but slows down.
A Sign Table for Motion
| \(v\) | \(a\) | Direction | Speed Behavior |
|---|---|---|---|
| Positive | Positive | Positive | Increasing |
| Positive | Negative | Positive | Decreasing |
| Negative | Negative | Negative | Increasing |
| Negative | Positive | Negative | Decreasing |
Reading Speeding Up from a Position Graph
From a position graph, determine:
- the sign of slope, which gives the sign of velocity;
- the concavity, which gives the sign of acceleration.
Same signs mean speeding up.
Opposite signs mean slowing down.
Example: Increasing and Concave Up
Suppose a position graph is increasing and concave up.
Then
\[ v>0 \]and
\[ a>0. \]Therefore, the object is
\[ \boxed{\text{speeding up}}. \]Example: Increasing and Concave Down
If a position graph is increasing and concave down, then
\[ v>0 \]and
\[ a<0. \]Therefore, the object is
\[ \boxed{\text{slowing down}}. \]Example: Decreasing and Concave Down
If a position graph is decreasing and concave down, then
\[ v<0 \]and
\[ a<0. \]Therefore, the object is
\[ \boxed{\text{speeding up}}. \]Example: Decreasing and Concave Up
If a position graph is decreasing and concave up, then
\[ v<0 \]and
\[ a>0. \]Therefore, the object is
\[ \boxed{\text{slowing down}}. \]Constant Velocity Graph Family
If velocity is constant,
\[ v(t)=v_0, \]then position is linear:
\[ \boxed{ x(t)=x_0+v_0t }. \]Acceleration is
\[ \boxed{a(t)=0}. \]Thus the graph family is:
- position: straight slanted line;
- velocity: horizontal line;
- acceleration: the time axis.
Constant Acceleration Graph Family
If acceleration is constant,
\[ a(t)=a, \]then velocity is linear:
\[ \boxed{ v(t)=v_0+at }. \]Position is quadratic:
\[ \boxed{ x(t) = x_0+v_0t+\frac12at^2 }. \]Graph Shapes under Constant Positive Acceleration
If
\[ a>0, \]then:
- the acceleration graph is a horizontal line above zero;
- the velocity graph is a line with positive slope;
- the position graph is concave up.
Graph Shapes under Constant Negative Acceleration
If
\[ a<0, \]then:
- the acceleration graph is a horizontal line below zero;
- the velocity graph is a line with negative slope;
- the position graph is concave down.
Example: \(x(t)=t^2\)
Let
\[ x(t)=t^2. \]Then
\[ v(t)=2t \]and
\[ a(t)=2. \]Therefore:
- position is an upward-opening parabola;
- velocity is a line through the origin;
- acceleration is a positive horizontal line.
Interpreting \(x(t)=t^2\) for Negative and Positive Time
For
\[ t<0, \]velocity is negative.
At
\[ t=0, \]velocity is zero.
For
\[ t>0, \]velocity is positive.
Therefore, the object reverses direction at \(t=0\).
Example: \(x(t)=-t^2+4t\)
Let
\[ x(t)=-t^2+4t. \]Then
\[ v(t)=-2t+4 \]and
\[ a(t)=-2. \]The position graph is concave down, the velocity graph slopes downward, and the acceleration graph is a constant negative line.
Turning Time in the Example
Set
\[ v(t)=0. \]Then
\[ -2t+4=0, \]so
\[ \boxed{t=2}. \]The position graph reaches a local maximum at the same time.
Piecewise Velocity Graphs
Real motion is often described in stages.
A velocity graph might show:
- speeding up;
- constant velocity;
- slowing to rest;
- reversing direction;
- moving at another constant velocity.
Each segment can be analyzed separately and then combined.
Piecewise Constant Velocity
If velocity is constant on a time interval, the corresponding position graph is linear on that interval.
Different velocity values create different line slopes.
Velocity Jumps
An idealized velocity graph may have a sudden jump.
Mathematically, such a jump represents a change in velocity over zero elapsed time.
This corresponds to an idealized impulsive acceleration rather than an ordinary finite acceleration value.
Physical systems usually smooth such changes over short time intervals.
Corners in a Position Graph
A sharp corner in a position-time graph means the slope changes abruptly.
Since slope is velocity, the velocity may be discontinuous there.
Ordinary instantaneous acceleration may therefore fail to exist at that exact point.
Horizontal Tangents and Rest
A horizontal tangent on a smooth position graph gives
\[ v=0. \]To determine whether the object reverses direction, inspect the sign of the position slope on either side.
Inflection Points and Acceleration
If a smooth position graph changes concavity, then acceleration may change sign.
At a smooth inflection time, it is common to have
\[ a=0, \]though a full mathematical check depends on the function.
Maximum and Minimum Velocity
A local maximum or minimum of a smooth velocity graph occurs where its slope is zero and changes behavior.
Since the slope of velocity is acceleration, such candidate times satisfy
\[ \boxed{a=0}. \]Acceleration Zero Does Not Mean Velocity Zero
If
\[ a=0, \]the velocity graph has horizontal tangent slope at that instant.
The velocity value itself can be positive, negative, or zero.
Using Units to Check Graph Interpretations
Suppose position is measured in meters and time in seconds.
Then:
\[ [x]=\text{m}, \] \[ [v]=\frac{\text{m}}{\text{s}}, \]and
\[ [a]=\frac{\text{m}}{\text{s}^2}. \]Slope and area operations should produce these expected units.
Units of Position-Graph Slope
Position slope has units
\[ \frac{\text{m}}{\text{s}}, \]which are velocity units.
Units of Velocity-Graph Slope
Velocity slope has units
\[ \frac{ \text{m/s} }{ \text{s} } \] \[ = \frac{\text{m}}{\text{s}^2}, \]which are acceleration units.
Units of Area under Velocity
Velocity multiplied by time gives
\[ \frac{\text{m}}{\text{s}} \cdot \text{s} \] \[ = \text{m}. \]Therefore, velocity-time area has displacement units.
Units of Area under Acceleration
Acceleration multiplied by time gives
\[ \frac{\text{m}}{\text{s}^2} \cdot \text{s} \] \[ = \frac{\text{m}}{\text{s}}. \]Therefore, acceleration-time area has velocity-change units.
Average Velocity versus Average Speed from Graphs
Average velocity is
\[ \boxed{ \frac{\text{net displacement}} {\text{elapsed time}} }. \]Average speed is
\[ \boxed{ \frac{\text{total distance}} {\text{elapsed time}} }. \]If the velocity graph changes sign, these quantities generally differ.
Graphical Meaning of Returning to the Starting Position
If an object returns to its starting position from \(t=a\) to \(t=b\), then
\[ x(b)-x(a)=0. \]Therefore,
\[ \boxed{ \int_a^b v(t)\,dt=0 }. \]Positive and negative signed velocity areas cancel.
Graphical Meaning of Returning to the Initial Velocity
If
\[ v(b)=v(a), \]then
\[ \Delta v=0. \]Therefore,
\[ \boxed{ \int_a^b a(t)\,dt=0 }. \]Position Graph versus Literal Path
A position-time graph is not usually a picture of the physical path.
Its horizontal axis is time.
Its vertical axis is position.
A curved position-time graph does not necessarily mean the object moves on a curved path in physical space.
Example: Straight-Line Motion with a Curved Position Graph
Consider one-dimensional motion
\[ x(t)=t^2. \]The object moves only along one straight coordinate axis.
Yet its position-time graph is a parabola.
The graph shape describes how position changes with time, not the literal spatial path.
Multi-Dimensional Motion and Component Graphs
For
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle, \]each coordinate can have its own position-time graph.
Likewise,
\[ \mathbf{v}(t) = \langle v_x(t),v_y(t),v_z(t)\rangle \]and
\[ \mathbf{a}(t) = \langle a_x(t),a_y(t),a_z(t)\rangle. \]Component graphs can be analyzed using the same slope and area principles.
Projectile Motion Graph Pattern
For ideal projectile motion:
- \(x(t)\) is linear;
- \(v_x(t)\) is constant;
- \(a_x(t)=0\);
- \(y(t)\) is quadratic and concave down;
- \(v_y(t)\) is linear with negative slope;
- \(a_y(t)=-g\) is constant.
Uniform Circular Motion and Component Graphs
Uniform circular motion may be written
\[ x(t)=r\cos(\omega t) \]and
\[ y(t)=r\sin(\omega t). \]Each coordinate oscillates sinusoidally even though the object moves around a circle in physical space.
Velocity Components in Circular Motion
Differentiating,
\[ v_x(t) = -r\omega\sin(\omega t) \]and
\[ v_y(t) = r\omega\cos(\omega t). \]These component graphs are also sinusoidal and shifted relative to the position components.
Acceleration Components in Circular Motion
Differentiate again:
\[ a_x(t) = -r\omega^2\cos(\omega t) \]and
\[ a_y(t) = -r\omega^2\sin(\omega t). \]Therefore,
\[ \boxed{ a_x=-\omega^2x, \qquad a_y=-\omega^2y }. \]Common Error: Reading Position as Velocity
The vertical height of a position graph gives position.
Velocity comes from the graph's slope.
A large positive position does not necessarily mean large positive velocity.
Common Error: Reading Velocity as the Height of a Position Graph
Velocity is not the position value.
It is the rate of change:
\[ \boxed{ v=x' }. \]Common Error: Treating Negative Position as Negative Velocity
Negative position means the object lies on the negative side of the chosen origin.
Negative velocity means position is decreasing.
These are independent ideas.
Common Error: Treating Negative Velocity as Slowing Down
Negative velocity describes direction.
Whether speed increases or decreases depends on acceleration as well.
If
\[ v<0 \]and
\[ a<0, \]the object is speeding up.
Common Error: Treating Negative Acceleration as Slowing Down
Negative acceleration means velocity is decreasing.
If velocity is already negative, its magnitude can increase.
Always compare the signs of \(v\) and \(a\).
Common Error: Using Area under Position to Find Displacement
Displacement is signed area under the velocity graph:
\[ \boxed{ \Delta x=\int v\,dt }. \]Area under a position graph does not have the standard meaning of displacement.
Common Error: Using Area under Acceleration to Find Position
One accumulation of acceleration gives change in velocity:
\[ \boxed{ \Delta v=\int a\,dt }. \]Position requires another accumulation through velocity.
Common Error: Counting Negative Velocity Area as Negative Distance
Signed area below the velocity axis contributes negatively to displacement.
Distance, however, is nonnegative.
Therefore, use
\[ \boxed{ d=\int|v|\,dt }. \]Common Error: Assuming a Zero of Velocity Always Means a Turnaround
A turnaround requires a velocity sign change.
If velocity merely touches zero and keeps the same sign, the object does not reverse direction.
Common Error: Assuming \(a=0\) Means the Object Is Stopped
Zero acceleration means velocity is locally constant.
That constant velocity may be nonzero.
Common Error: Ignoring Graph Scale
A graph can appear steep or flat because of axis scaling.
When numerical values matter, use the labeled coordinates and actual slope calculation rather than visual steepness alone.
A Reliable Procedure for Reading a Position Graph
- Read vertical height for position.
- Read slope sign for direction of motion.
- Read slope magnitude for speed tendency.
- Find horizontal tangents for \(v=0\).
- Check whether slope changes sign at those times.
- Use concavity for acceleration sign.
- Combine slope and concavity signs for speeding or slowing.
A Reliable Procedure for Reading a Velocity Graph
- Read graph height for velocity.
- Use sign to determine direction.
- Use distance from the axis for speed.
- Use slope for acceleration.
- Find axis crossings for possible reversals.
- Use signed area for displacement.
- Use absolute area for total distance.
A Reliable Procedure for Reading an Acceleration Graph
- Read graph height for acceleration.
- Use sign to determine whether velocity is increasing or decreasing.
- Use signed area to find change in velocity.
- Add that change to the initial velocity.
- Combine the reconstructed velocity sign with acceleration sign to interpret speed change.
A Reliable Procedure for Translating between Graphs
- From position to velocity, sketch or calculate slopes.
- From velocity to acceleration, sketch or calculate slopes again.
- From acceleration to velocity, accumulate signed area and use the initial velocity.
- From velocity to position, accumulate signed area and use the initial position.
- Check zeros, extrema, and sign changes for consistency.
- Check units at every stage.
Example: Full Function-to-Graph Workflow
Let
\[ x(t)=t^3-3t^2. \]Then
\[ v(t)=3t^2-6t \] \[ =3t(t-2). \]Acceleration is
\[ a(t)=6t-6. \]Therefore:
- velocity is zero at \(t=0\) and \(t=2\);
- acceleration is zero at \(t=1\);
- position is concave down for \(t<1\);
- position is concave up for \(t>1\).
Direction Intervals in the Example
Since
\[ v(t)=3t(t-2), \]
for \(0
For \(t>2\),
Thus the object moves in the negative direction between \(0\) and \(2\),
then reverses direction at \(t=2\).
Speed Behavior in the Example
For
\[ 0so the object speeds up.
For
\[ 1so the object slows down.
After the Turning Point
For
\[ t>2, \]velocity is positive.
Since
\[ a(t)=6t-6>0 \]there,
the object moves in the positive direction and speeds up.
Example: Area from a Piecewise Constant Velocity
Suppose
\[ v(t)=4 \]from \(t=0\) to \(t=3\), and
\[ v(t)=-2 \]from \(t=3\) to \(t=7\).
Positive displacement:
\[ 4(3)=12. \]Negative displacement:
\[ (-2)(4)=-8. \]Net displacement:
\[ \boxed{\Delta x=4}. \]Distance in the Same Piecewise Example
Total distance is
\[ 12+8. \]Therefore,
\[ \boxed{d=20}. \]Example: Acceleration Area and Final Velocity
Suppose initial velocity is
\[ v(0)=3 \]and acceleration is constant:
\[ a(t)=2 \]from \(t=0\) to \(t=5\).
Change in velocity is the rectangle area:
\[ \Delta v=2(5)=10. \]Therefore,
\[ v(5)=3+10 \] \[ = \boxed{13}. \]Practice 1: Position-Graph Slope
A position-time graph passes through \((1,3)\) and \((5,15)\) as a straight segment. Find the velocity on that segment.
Practice 2: Position Concavity
A position graph is increasing and concave down. Is the object speeding up or slowing down?
Practice 3: Velocity Direction
If
\[ v(t)=-7, \]what does the sign tell you about the direction of motion?
Practice 4: Displacement from Velocity Area
A velocity graph encloses area \(10\) above the time axis and area \(4\) below the time axis. Find the net displacement.
Practice 5: Distance from Velocity Area
Using the same areas as Practice 4, find total distance traveled.
Practice 6: Change in Velocity from Acceleration
Acceleration is constant at
\[ 3\frac{\text{m}}{\text{s}^2} \]for 4 seconds. Find the change in velocity.
Practice 7: Constant Acceleration Functions
Suppose
\[ a(t)=4, \qquad v(0)=2, \qquad x(0)=5. \]Find \(v(t)\) and \(x(t)\).
Practice 8: Speeding Up or Slowing Down
At one instant,
\[ v=-5 \]and
\[ a=2. \]Is the object speeding up or slowing down?
Practice 9: Turning Point
Let
\[ v(t)=t-3. \]At what time does the object reverse direction, assuming the model is valid on both sides of the zero?
Practice 10: Identify the Error
A student says:
“The position graph is below the time axis, so the object must be moving in the negative direction.”
Explain the error.
Practice Answers
Practice 1
Velocity is the slope:
\[ v = \frac{15-3}{5-1} \] \[ = \frac{12}{4} \] \[ = \boxed{3}. \]Practice 2
Increasing means
\[ v>0. \]Concave down means
\[ a<0. \]The signs are opposite, so the object is
\[ \boxed{\text{slowing down}}. \]Practice 3
Since
\[ v<0, \]the object moves in the negative coordinate direction.
The speed is
\[ \boxed{7}. \]Practice 4
\[ \Delta x=10-4. \] \[ \boxed{\Delta x=6}. \]Practice 5
\[ d=10+4. \] \[ \boxed{d=14}. \]Practice 6
\[ \Delta v = a\Delta t \] \[ = 3(4) \] \[ = \boxed{ 12\frac{\text{m}}{\text{s}} }. \]Practice 7
Integrate acceleration:
\[ v(t)=2+4t. \]Therefore,
\[ \boxed{ v(t)=2+4t }. \]Integrate velocity and use \(x(0)=5\):
\[ x(t) = 5+2t+2t^2. \]Therefore,
\[ \boxed{ x(t)=5+2t+2t^2 }. \]Practice 8
Velocity is negative while acceleration is positive.
The signs are opposite.
Therefore, the object is
\[ \boxed{\text{slowing down}}. \]Practice 9
Set velocity equal to zero:
\[ t-3=0. \]Therefore,
\[ \boxed{t=3}. \]Since \(v\) changes from negative to positive across \(t=3\), the object reverses direction there.
Practice 10
A negative position value means the object lies on the negative side of the chosen origin.
Direction of motion is determined by the slope of the position graph:
\[ \boxed{ v=x' }. \]The object can have negative position while moving either positively or negatively.
What to Remember
Position, velocity, and acceleration are connected by
\[ \boxed{ v=x' } \]and
\[ \boxed{ a=v'=x'' }. \]The reverse relationships are
\[ \boxed{ \Delta x = \int v\,dt } \]and
\[ \boxed{ \Delta v = \int a\,dt }. \]On a position graph:
- height gives position;
- slope gives velocity;
- concavity gives acceleration sign.
On a velocity graph:
- height gives velocity;
- slope gives acceleration;
- signed area gives displacement;
- absolute area gives distance.
On an acceleration graph:
- height gives acceleration;
- signed area gives change in velocity.
For one-dimensional motion:
\[ \boxed{ v\text{ and }a\text{ same sign} \Rightarrow \text{speeding up} } \]and
\[ \boxed{ v\text{ and }a\text{ opposite signs} \Rightarrow \text{slowing down} }. \]Section Synthesis
Motion and Kinematics began with position, path, displacement, and distance. It then developed speed, velocity, acceleration, parametric and vector-valued motion, relative motion, intersecting paths, circular motion, projectile motion, reference frames, and path-relative acceleration components.
Position, velocity, and acceleration graphs bring those ideas together in a single interpretive framework:
\[ \boxed{ \text{position} \xrightarrow{\text{slope}} \text{velocity} \xrightarrow{\text{slope}} \text{acceleration} } \]and
\[ \boxed{ \text{acceleration} \xrightarrow{\text{area}} \text{velocity change} \xrightarrow{\text{area}} \text{position change} }. \]These relationships provide a bridge from geometric motion to calculus, modeling, physics, numerical methods, and higher-dimensional analysis.
Where We Go Next
This lesson completes Motion and Kinematics. The ideas in this section will continue to reappear whenever mathematics studies changing systems: slopes describe instantaneous change, integrals describe accumulation, vectors organize multidimensional motion, and graphs connect formulas with geometric behavior.
The resulting toolkit is now ready to support later work in continuous change, fields, differential models, optimization, and applied mathematical analysis.