The Central Idea
The previous lesson distinguished distance from displacement.
The same distinction produces two different rates of motion:
- speed is the rate at which distance is traveled;
- velocity is the rate at which displacement changes.
In one dimension, velocity can be positive, negative, or zero because direction is encoded by sign.
Speed is ordinarily nonnegative.
What Is a Rate?
A rate compares the change in one quantity with the change in another.
A motion rate commonly compares length with time:
\[ \frac{\text{length}}{\text{time}}. \]Examples include:
\[ 60\frac{\text{mi}}{\text{h}}, \qquad 12\frac{\text{m}}{\text{s}}, \qquad 5\frac{\text{km}}{\text{min}}. \]What Is Speed?
Speed measures how quickly distance is traveled.
Over a time interval, average speed is
\[ \boxed{ \text{average speed} = \frac{\text{total distance traveled}}{\text{elapsed time}} }. \]If the total distance is \(d\) and elapsed time is \(\Delta t\), then
\[ \boxed{ \text{average speed} = \frac{d}{\Delta t} }. \]Example: Average Speed
A car travels 180 kilometers in 3 hours.
Its average speed is
\[ \frac{180\text{ km}}{3\text{ h}} = 60\frac{\text{km}}{\text{h}}. \]Therefore,
\[ \boxed{ \text{average speed} = 60\frac{\text{km}}{\text{h}} }. \]Elapsed Time
If motion begins at time \(t_i\) and ends at time \(t_f\), then elapsed time is
\[ \boxed{ \Delta t=t_f-t_i }. \]For example, if motion begins at \(t=2\text{ s}\) and ends at \(t=8\text{ s}\), then
\[ \Delta t=8-2=6\text{ s}. \]What Is Velocity?
Velocity measures the rate at which position changes.
In one-dimensional motion, average velocity is
\[ \boxed{ v_{\text{avg}} = \frac{\Delta x}{\Delta t} }. \]Since
\[ \Delta x=x_f-x_i, \]we may also write
\[ \boxed{ v_{\text{avg}} = \frac{x_f-x_i}{t_f-t_i} }. \]Example: Average Velocity
Suppose an object moves from
\[ x_i=4\text{ m} \]at
\[ t_i=1\text{ s} \]to
\[ x_f=22\text{ m} \]at
\[ t_f=7\text{ s}. \]The displacement is
\[ \Delta x=22-4=18\text{ m}. \]The elapsed time is
\[ \Delta t=7-1=6\text{ s}. \]Therefore,
\[ v_{\text{avg}} = \frac{18}{6} \] \[ \boxed{ v_{\text{avg}} = 3\frac{\text{m}}{\text{s}} }. \]Velocity Can Be Negative
In one dimension, negative velocity indicates motion in the chosen negative direction.
Suppose
\[ x_i=10\text{ m}, \qquad x_f=2\text{ m}, \]over
\[ \Delta t=4\text{ s}. \]Then
\[ \Delta x=2-10=-8\text{ m}, \]so
\[ v_{\text{avg}} = \frac{-8}{4} \] \[ \boxed{ v_{\text{avg}} = -2\frac{\text{m}}{\text{s}} }. \]The negative sign indicates direction, not a negative amount of motion.
Speed Is Not Negative
Speed measures the rate at which path length accumulates.
Therefore, ordinary speed satisfies
\[ \boxed{\text{speed}\ge 0}. \]An object may have velocity
\[ -5\frac{\text{m}}{\text{s}}, \]while its speed is
\[ 5\frac{\text{m}}{\text{s}}. \]Speed and the Magnitude of Velocity
At a particular instant in one-dimensional motion, speed is the magnitude of velocity:
\[ \boxed{\text{speed}=|v|}. \]Thus if
\[ v=-12\frac{\text{m}}{\text{s}}, \]then the speed is
\[ \boxed{ 12\frac{\text{m}}{\text{s}} }. \]Average Speed and Average Velocity Can Differ
Average speed uses total distance.
Average velocity uses net displacement.
Therefore,
\[ \boxed{ \text{average speed} = \frac{\text{distance}}{\Delta t} } \]while
\[ \boxed{ v_{\text{avg}} = \frac{\text{displacement}}{\Delta t} }. \]Example: Out and Back
A runner travels 100 meters east and then 100 meters west, returning to the starting point in 40 seconds.
Total distance:
\[ d=100+100=200\text{ m}. \]Displacement:
\[ \Delta x=0. \]Average speed:
\[ \frac{200}{40} = 5\frac{\text{m}}{\text{s}}. \]Average velocity:
\[ \frac{0}{40}=0. \]Therefore,
\[ \boxed{ \text{average speed}=5\frac{\text{m}}{\text{s}}, \qquad v_{\text{avg}}=0 }. \]When Average Speed Equals the Magnitude of Average Velocity
In one-dimensional motion, average speed equals the magnitude of average velocity when the object moves without reversing direction.
Then
\[ \text{distance}=|\text{displacement}|, \]so
\[ \boxed{ \text{average speed} = |v_{\text{avg}}| }. \]Units of Speed and Velocity
Speed and velocity have units of length divided by time.
Common units include:
- meters per second, \(\text{m/s}\);
- kilometers per hour, \(\text{km/h}\);
- miles per hour, \(\text{mi/h}\);
- feet per second, \(\text{ft/s}\).
Units should be carried through the calculation rather than added only at the end.
Converting Speed Units
Suppose a speed is
\[ 72\frac{\text{km}}{\text{h}}. \]Use
\[ 1\text{ km}=1000\text{ m} \]and
\[ 1\text{ h}=3600\text{ s}. \]Then
\[ 72\frac{\text{km}}{\text{h}} \cdot \frac{1000\text{ m}}{1\text{ km}} \cdot \frac{1\text{ h}}{3600\text{ s}} \] \[ = 20\frac{\text{m}}{\text{s}}. \]Therefore,
\[ \boxed{ 72\frac{\text{km}}{\text{h}} = 20\frac{\text{m}}{\text{s}} }. \]Velocity Is Directional
In one-dimensional motion, sign carries direction.
If east is positive, then
\[ +15\frac{\text{m}}{\text{s}} \]may mean 15 meters per second east, while
\[ -15\frac{\text{m}}{\text{s}} \]means 15 meters per second west.
Both motions have the same speed.
Average Velocity from a Position Function
If position is
\[ x=x(t), \]then average velocity from time \(t_1\) to time \(t_2\) is
\[ \boxed{ v_{\text{avg}} = \frac{x(t_2)-x(t_1)}{t_2-t_1} }. \]This is a difference quotient.
Example: Average Velocity from \(x(t)\)
Let
\[ x(t)=t^2+2t. \]Find the average velocity from \(t=1\) to \(t=4\).
First,
\[ x(1)=1+2=3. \]Next,
\[ x(4)=16+8=24. \]Therefore,
\[ v_{\text{avg}} = \frac{24-3}{4-1} \] \[ = \frac{21}{3} \] \[ \boxed{ v_{\text{avg}}=7 }. \]Average Velocity as a Slope
On a position-time graph, average velocity is the slope of the secant line through two points on the graph.
If the points are
\[ (t_1,x(t_1)) \]and
\[ (t_2,x(t_2)), \]then the slope is
\[ \frac{x(t_2)-x(t_1)}{t_2-t_1}. \]That is exactly the formula for average velocity.
Positive Slope and Positive Velocity
If a position-time graph rises as time increases, then position is increasing.
Over that interval, the velocity is positive.
In one dimension, this indicates motion in the chosen positive direction.
Negative Slope and Negative Velocity
If a position-time graph falls as time increases, then position is decreasing.
The velocity is negative.
The object is moving in the chosen negative direction.
Zero Slope and Rest
A horizontal segment on a position-time graph means the position does not change while time passes.
Therefore,
\[ \Delta x=0 \]over that interval, and the velocity is
\[ \boxed{v=0}. \]The object is at rest during that interval.
Steeper Position Graphs Represent Greater Speed
On a position-time graph, the magnitude of the slope indicates how rapidly position changes.
A slope of
\[ 8 \]has greater magnitude than a slope of
\[ -3. \]Therefore, the motion associated with slope \(8\) has greater speed at that comparison.
Average versus Instantaneous Velocity
Average velocity describes motion over a time interval.
Instantaneous velocity describes the rate of change of position at one particular instant.
Informally, it answers:
“How fast, and in which direction, is the object moving right now?”
From Secant Slope to Tangent Slope
Average velocity over an interval is a secant slope.
To estimate instantaneous velocity at time \(t\), we can use smaller and smaller time intervals around that instant.
This leads to the limiting expression
\[ v(t) = \lim_{\Delta t\to 0} \frac{x(t+\Delta t)-x(t)}{\Delta t}. \]This limit is the derivative of position.
Velocity as a Derivative
When \(x(t)\) is differentiable, instantaneous velocity is
\[ \boxed{ v(t)=x'(t)=\frac{dx}{dt} }. \]This equation connects kinematics directly with calculus.
The derivative measures the instantaneous rate at which position changes with respect to time.
Example: Instantaneous Velocity
Let
\[ x(t)=t^2+3t-2. \]Differentiate:
\[ v(t)=x'(t)=2t+3. \]At \(t=4\),
\[ v(4)=2(4)+3=11. \]Therefore,
\[ \boxed{v(4)=11}. \]Instantaneous Speed
Instantaneous speed is the magnitude of instantaneous velocity.
\[ \boxed{\text{speed}(t)=|v(t)|}. \]If
\[ v(t)=-7\frac{\text{m}}{\text{s}}, \]then
\[ \boxed{ \text{speed}=7\frac{\text{m}}{\text{s}} }. \]Example: Velocity Changes Sign
Let
\[ x(t)=t^2-6t+5. \]Then
\[ v(t)=2t-6. \]At \(t=2\),
\[ v(2)=-2. \]At \(t=4\),
\[ v(4)=2. \]The sign change indicates that the object changes direction between those times.
Turning Points in One-Dimensional Motion
A direction reversal can occur when velocity passes through zero.
For
\[ v(t)=2t-6, \]set
\[ v(t)=0. \]Then
\[ 2t-6=0 \] \[ \boxed{t=3}. \]This is the candidate turning time.
Why Turning Points Matter for Distance
To compute total distance from a position function, we must account for intervals in which the object changes direction.
Suppose an object moves from position \(x_1\) to a turning point \(x_T\) and then to final position \(x_2\).
The total distance is
\[ |x_T-x_1|+|x_2-x_T|. \]Simply computing
\[ |x_2-x_1| \]would miss the extra travel caused by the reversal.
Example: Distance from a Position Function
Let
\[ x(t)=t^2-6t+8 \]for
\[ 0\le t\le 5. \]Velocity is
\[ v(t)=2t-6. \]Set \(v(t)=0\):
\[ 2t-6=0 \] \[ t=3. \]Positions:
\[ x(0)=8, \] \[ x(3)=9-18+8=-1, \] \[ x(5)=25-30+8=3. \]Distance:
\[ |{-1}-8|+|3-(-1)| \] \[ =9+4 \] \[ \boxed{13}. \]Displacement is only
\[ 3-8=-5. \]Constant Velocity
If velocity is constant,
\[ v(t)=v, \]then position changes linearly with time:
\[ \boxed{ x(t)=x_0+vt }. \]Here \(x_0\) is the position at \(t=0\).
Example: Constant Velocity Motion
Suppose
\[ x_0=5\text{ m} \]and
\[ v=3\frac{\text{m}}{\text{s}}. \]Then
\[ x(t)=5+3t. \]At \(t=4\text{ s}\),
\[ x(4)=5+12=17\text{ m}. \]Therefore,
\[ \boxed{x(4)=17\text{ m}}. \]Negative Constant Velocity
If
\[ v=-4\frac{\text{m}}{\text{s}}, \]then the object moves 4 meters per second in the negative direction.
If
\[ x_0=10\text{ m}, \]then
\[ x(t)=10-4t. \]The position decreases linearly with time.
Solving for Time with Constant Speed
For motion at constant speed,
\[ \text{speed} = \frac{\text{distance}}{\text{time}}. \]Rearranging,
\[ \boxed{ \text{time} = \frac{\text{distance}}{\text{speed}} }. \]If a vehicle travels 150 miles at a constant speed of \(50\text{ mi/h}\), then
\[ t=\frac{150}{50}=3\text{ h}. \]Solving for Distance with Constant Speed
Rearranging the same rate equation,
\[ \boxed{ \text{distance} = \text{speed}\times\text{time} }. \]At
\[ 12\frac{\text{m}}{\text{s}} \]for
\[ 8\text{ s}, \]the distance is
\[ 12\cdot8=96\text{ m}. \]Average Speed Is Not Usually the Arithmetic Mean of Two Speeds
A common mistake is to average two speeds directly without considering how long or how far the object travels at each speed.
Average speed must always be computed from
\[ \boxed{ \frac{\text{total distance}}{\text{total time}} }. \]Example: Two Different Speeds for Equal Times
A car travels for 1 hour at \(40\text{ mi/h}\) and then for 1 hour at \(60\text{ mi/h}\).
Distances:
\[ 40\text{ mi} \]and
\[ 60\text{ mi}. \]Total distance:
\[ 100\text{ mi}. \]Total time:
\[ 2\text{ h}. \]Therefore,
\[ \boxed{ \text{average speed} = 50\text{ mi/h} }. \]In this equal-time case, the arithmetic mean happens to work.
Example: Two Different Speeds for Equal Distances
A car travels 60 miles at \(30\text{ mi/h}\) and another 60 miles at \(60\text{ mi/h}\).
First time:
\[ \frac{60}{30}=2\text{ h}. \]Second time:
\[ \frac{60}{60}=1\text{ h}. \]Total distance:
\[ 120\text{ mi}. \]Total time:
\[ 3\text{ h}. \]Therefore,
\[ \boxed{ \text{average speed} = 40\text{ mi/h} }. \]The arithmetic mean of 30 and 60 is 45, which is incorrect here.
Velocity in Two Dimensions
In two-dimensional motion, position is represented by
\[ \mathbf{r}(t) = \langle x(t),y(t)\rangle. \]Average velocity over a time interval is
\[ \boxed{ \mathbf{v}_{\text{avg}} = \frac{\Delta\mathbf{r}}{\Delta t} }. \]The division by \(\Delta t\) applies to each component.
Example: Average Velocity Vector
Suppose an object moves from
\[ \mathbf{r}_i=\langle 2,1\rangle \]to
\[ \mathbf{r}_f=\langle 10,5\rangle \]in 4 seconds.
Displacement:
\[ \Delta\mathbf{r} = \langle 8,4\rangle. \]Therefore,
\[ \mathbf{v}_{\text{avg}} = \frac{\langle 8,4\rangle}{4} \] \[ \boxed{ \mathbf{v}_{\text{avg}} = \langle 2,1\rangle }. \]Instantaneous Velocity Vector
If
\[ \mathbf{r}(t) = \langle x(t),y(t)\rangle, \]then instantaneous velocity is obtained by differentiating each component:
\[ \boxed{ \mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t),y'(t)\rangle }. \]Example: Velocity Vector from Position
Let
\[ \mathbf{r}(t) = \langle t^2,3t-1\rangle. \]Differentiate:
\[ \mathbf{v}(t) = \langle 2t,3\rangle. \]At \(t=2\),
\[ \boxed{ \mathbf{v}(2) = \langle 4,3\rangle }. \]Speed from a Velocity Vector
In two or more dimensions, speed is the magnitude of the velocity vector.
In two dimensions,
\[ \boxed{ \text{speed} = |\mathbf{v}| = \sqrt{v_x^2+v_y^2} }. \]In three dimensions,
\[ \boxed{ |\mathbf{v}| = \sqrt{v_x^2+v_y^2+v_z^2} }. \]Example: Speed from Components
Suppose
\[ \mathbf{v} = \langle 6,8\rangle \frac{\text{m}}{\text{s}}. \]Then
\[ |\mathbf{v}| = \sqrt{6^2+8^2} \] \[ = \sqrt{100} \] \[ \boxed{ \text{speed} = 10\frac{\text{m}}{\text{s}} }. \]Direction of a Velocity Vector
A velocity vector contains both magnitude and direction.
If
\[ \mathbf{v} = \langle v_x,v_y\rangle, \]then, when appropriate,
\[ \tan\theta = \frac{v_y}{v_x}. \]Quadrant information must be used to determine the correct direction angle.
Velocity Is Tangent to the Path
For smooth motion along a curve, the instantaneous velocity vector points tangent to the path in the direction of motion.
This geometric fact connects:
- the path traced by the object;
- the derivative of the position vector;
- the direction in which the object is moving at that instant.
Later lessons will develop this connection further.
A Moving Object Can Have Constant Speed but Changing Velocity
Velocity depends on direction as well as magnitude.
Therefore, an object can move at constant speed while its velocity changes continuously.
Uniform circular motion is an important example.
The speed can remain constant while the velocity vector continually turns.
A Moving Object Can Have Zero Average Velocity
Zero average velocity does not imply that the object remained at rest.
It means only that the net displacement over the interval was zero.
A runner who completes a lap and returns to the starting point may have substantial speed throughout the trip while having
\[ \boxed{ v_{\text{avg}}=0 }. \]A Zero Instantaneous Velocity Does Not Always Mean Long-Term Rest
An object can have
\[ v(t)=0 \]at one instant and still be moving immediately before and after that instant.
This often occurs at a turning point.
For example, an object thrown straight upward briefly has zero vertical velocity at the top of its motion before beginning to descend.
Average Speed from Several Segments
Suppose an object travels distances
\[ d_1,d_2,\ldots,d_n \]over time intervals
\[ \Delta t_1,\Delta t_2,\ldots,\Delta t_n. \]Then
\[ \boxed{ \text{average speed} = \frac{d_1+d_2+\cdots+d_n} {\Delta t_1+\Delta t_2+\cdots+\Delta t_n} }. \]Average Velocity from Several Segments
For several motion segments, add displacements rather than distances:
\[ \Delta x_{\text{total}} = \Delta x_1+\Delta x_2+\cdots+\Delta x_n. \]Then
\[ \boxed{ v_{\text{avg}} = \frac{\Delta x_{\text{total}}} {\Delta t_{\text{total}}} }. \]Example: Multi-Segment Motion
A person walks 60 meters east in 30 seconds, then 20 meters west in 10 seconds.
Distance:
\[ d=60+20=80\text{ m}. \]Displacement:
\[ \Delta x=60-20=40\text{ m}. \]Total time:
\[ 40\text{ s}. \]Average speed:
\[ \frac{80}{40} = 2\frac{\text{m}}{\text{s}}. \]Average velocity:
\[ \frac{40}{40} = 1\frac{\text{m}}{\text{s}}. \]Therefore,
\[ \boxed{ \text{average speed}=2\frac{\text{m}}{\text{s}}, \qquad v_{\text{avg}}=1\frac{\text{m}}{\text{s}}\text{ east} }. \]Velocity-Time Information
A velocity function
\[ v=v(t) \]tells the velocity at each time.
Positive values indicate motion in the positive direction, negative values indicate motion in the negative direction, and zero values indicate instantaneous rest.
Distance from Velocity
In continuous motion, displacement over an interval can be accumulated from velocity:
\[ \Delta x = \int_{t_1}^{t_2}v(t)\,dt. \]Total distance instead accumulates speed:
\[ \boxed{ d = \int_{t_1}^{t_2}|v(t)|\,dt }. \]These formulas will become more important when calculus-based kinematics is developed further.
Average Value of Velocity
When velocity is integrable,
\[ v_{\text{avg}} = \frac{1}{t_2-t_1} \int_{t_1}^{t_2}v(t)\,dt. \]Since the integral of velocity gives displacement,
\[ v_{\text{avg}} = \frac{\Delta x}{\Delta t}. \]Thus the algebraic and calculus definitions agree.
Speedometers and Velocity
A vehicle speedometer normally reports speed, not full velocity.
A reading of
\[ 55\text{ mi/h} \]tells how fast the vehicle is moving but not whether it is traveling north, south, east, west, or along some other direction.
Velocity requires directional information.
Common Error: Using Distance in the Velocity Formula
Average velocity uses displacement:
\[ \boxed{ v_{\text{avg}} = \frac{\Delta x}{\Delta t} }. \]Replacing displacement with total distance changes the calculation into average speed.
Common Error: Using Displacement in the Speed Formula
Average speed uses total distance traveled:
\[ \boxed{ \text{average speed} = \frac{d}{\Delta t} }. \]If an object reverses direction, using only the magnitude of net displacement can underestimate the actual average speed.
Common Error: Treating Negative Velocity as Negative Speed
If
\[ v=-9\frac{\text{m}}{\text{s}}, \]the speed is not
\[ -9\frac{\text{m}}{\text{s}}. \]It is
\[ \boxed{ 9\frac{\text{m}}{\text{s}} }. \]Common Error: Forgetting Elapsed Time
Suppose motion begins at \(t=3\) and ends at \(t=11\).
The elapsed time is not 11.
It is
\[ \Delta t=11-3=8. \]Rates require the duration of the interval, not merely the final clock reading.
Common Error: Averaging Speeds Without Weighting
The expression
\[ \frac{v_1+v_2}{2} \]gives the average of two numerical speed values, but it is not automatically the average speed of a trip.
The correct motion calculation is
\[ \boxed{ \frac{\text{total distance}}{\text{total time}} }. \]Common Error: Reading a Position Graph as a Velocity Graph
On a position-time graph, the vertical coordinate is position.
Velocity is represented by the graph's slope, not by its height.
A large positive position does not automatically mean a large positive velocity.
Common Error: Assuming Zero Position Means Zero Velocity
An object can pass through the origin
\[ x=0 \]with substantial velocity.
Position tells where the object is. Velocity tells how its position is changing.
Common Error: Assuming Zero Velocity Means Zero Position
An object can be momentarily at rest at any position.
For example, an object might have
\[ x=20\text{ m} \]while
\[ v=0. \]The two quantities measure different aspects of motion.
A Reliable Procedure for Average Speed
- Identify every segment of the trip.
- Find the distance traveled on each segment.
- Add the segment distances.
- Find the elapsed time on each segment.
- Add the elapsed times.
- Divide total distance by total time.
- Attach units of length per time.
A Reliable Procedure for Average Velocity
- Choose or identify the positive direction.
- Find the initial position.
- Find the final position.
- Compute displacement using final minus initial.
- Compute elapsed time using final time minus initial time.
- Divide displacement by elapsed time.
- Interpret the sign or direction.
- Attach the correct units.
A Reliable Procedure from a Position Function
- Evaluate the position at the beginning of the interval.
- Evaluate the position at the end of the interval.
- Subtract to find displacement.
- Divide by elapsed time for average velocity.
- Differentiate \(x(t)\) to obtain instantaneous velocity when appropriate.
- Use \(|v(t)|\) for instantaneous speed.
- Check for sign changes when total distance is required.
Example: Full Average-Rate Workflow
A cyclist travels 12 kilometers east in 30 minutes, then 4 kilometers west in 10 minutes.
Total distance:
\[ 12+4=16\text{ km}. \]Net displacement:
\[ 12-4=8\text{ km east}. \]Total time:
\[ 40\text{ min}. \]Average speed:
\[ \frac{16}{40} = 0.4\frac{\text{km}}{\text{min}}. \]Average velocity:
\[ \frac{8}{40} = 0.2\frac{\text{km}}{\text{min}}\text{ east}. \]Therefore,
\[ \boxed{ \text{average speed} = 0.4\frac{\text{km}}{\text{min}} } \]and
\[ \boxed{ v_{\text{avg}} = 0.2\frac{\text{km}}{\text{min}}\text{ east} }. \]Example: Full Function Workflow
Let
\[ x(t)=t^3-6t^2+9t. \]The velocity is
\[ v(t)=3t^2-12t+9. \]Factor:
\[ v(t)=3(t^2-4t+3) \] \[ =3(t-1)(t-3). \]Therefore velocity is zero at
\[ t=1 \]and
\[ t=3. \]These times separate intervals on which the direction of motion may differ. They become important whenever total distance is requested.
Practice 1: Average Speed
A vehicle travels 240 kilometers in 4 hours. Find its average speed.
Practice 2: Average Velocity
An object moves from \(x=-3\text{ m}\) at \(t=2\text{ s}\) to \(x=13\text{ m}\) at \(t=6\text{ s}\). Find its average velocity.
Practice 3: Negative Average Velocity
An object moves from \(x=9\text{ m}\) to \(x=-6\text{ m}\) in 5 seconds. Find its average velocity.
Practice 4: Round Trip
A runner travels 300 meters out and 300 meters back in 120 seconds. Find average speed and average velocity.
Practice 5: Two-Segment Motion
A person walks 80 meters east in 40 seconds and then 20 meters west in 10 seconds. Find average speed and average velocity.
Practice 6: Position Function
Let
\[ x(t)=t^2-2t+5. \]Find the average velocity from \(t=1\) to \(t=5\).
Practice 7: Instantaneous Velocity
Let
\[ x(t)=2t^3-3t^2+4. \]Find \(v(t)\) and then find \(v(2)\).
Practice 8: Speed from Velocity
If
\[ v=-18\frac{\text{m}}{\text{s}}, \]find the speed.
Practice 9: Velocity Vector
Suppose
\[ \mathbf{r}(t) = \langle t^2,4t-1\rangle. \]Find \(\mathbf{v}(t)\) and \(\mathbf{v}(3)\).
Practice 10: Average Speed Trap
A car travels 90 miles at \(30\text{ mi/h}\) and then 90 miles at \(60\text{ mi/h}\).
Find the actual average speed for the entire trip.
Practice Answers
Practice 1
\[ \frac{240}{4} = \boxed{ 60\frac{\text{km}}{\text{h}} }. \]Practice 2
Displacement:
\[ 13-(-3)=16\text{ m}. \]Elapsed time:
\[ 6-2=4\text{ s}. \]Therefore,
\[ \boxed{ v_{\text{avg}} = 4\frac{\text{m}}{\text{s}} }. \]Practice 3
\[ \Delta x=-6-9=-15\text{ m}. \] \[ v_{\text{avg}} = \frac{-15}{5} \] \[ \boxed{ v_{\text{avg}} = -3\frac{\text{m}}{\text{s}} }. \]Practice 4
Distance:
\[ 300+300=600\text{ m}. \]Average speed:
\[ \frac{600}{120} = 5\frac{\text{m}}{\text{s}}. \]Displacement:
\[ 0. \]Average velocity:
\[ 0. \]Therefore,
\[ \boxed{ \text{average speed} = 5\frac{\text{m}}{\text{s}}, \qquad v_{\text{avg}}=0 }. \]Practice 5
Distance:
\[ 80+20=100\text{ m}. \]Displacement:
\[ 80-20=60\text{ m east}. \]Total time:
\[ 50\text{ s}. \]Average speed:
\[ \frac{100}{50} = 2\frac{\text{m}}{\text{s}}. \]Average velocity:
\[ \frac{60}{50} = 1.2\frac{\text{m}}{\text{s}}\text{ east}. \]Therefore,
\[ \boxed{ \text{average speed} = 2\frac{\text{m}}{\text{s}}, \qquad v_{\text{avg}} = 1.2\frac{\text{m}}{\text{s}}\text{ east} }. \]Practice 6
\[ x(1)=1-2+5=4. \] \[ x(5)=25-10+5=20. \] \[ v_{\text{avg}} = \frac{20-4}{5-1} \] \[ = \frac{16}{4} \] \[ \boxed{v_{\text{avg}}=4}. \]Practice 7
\[ x(t)=2t^3-3t^2+4. \] \[ v(t)=x'(t)=6t^2-6t. \]At \(t=2\),
\[ v(2)=6(4)-12=12. \]Therefore,
\[ \boxed{ v(t)=6t^2-6t, \qquad v(2)=12 }. \]Practice 8
\[ \text{speed}=|v| \] \[ =|-18| \] \[ \boxed{ 18\frac{\text{m}}{\text{s}} }. \]Practice 9
\[ \mathbf{r}(t) = \langle t^2,4t-1\rangle. \] \[ \mathbf{v}(t) = \langle 2t,4\rangle. \] \[ \mathbf{v}(3) = \langle 6,4\rangle. \]Therefore,
\[ \boxed{ \mathbf{v}(t) = \langle 2t,4\rangle, \qquad \mathbf{v}(3) = \langle 6,4\rangle }. \]Practice 10
First segment time:
\[ \frac{90}{30}=3\text{ h}. \]Second segment time:
\[ \frac{90}{60}=1.5\text{ h}. \]Total distance:
\[ 180\text{ mi}. \]Total time:
\[ 4.5\text{ h}. \]Therefore,
\[ \boxed{ \text{average speed} = \frac{180}{4.5} = 40\text{ mi/h} }. \]What to Remember
Average speed is
\[ \boxed{ \text{average speed} = \frac{\text{total distance}}{\Delta t} }. \]Average velocity is
\[ \boxed{ v_{\text{avg}} = \frac{\Delta x}{\Delta t} }. \]Instantaneous velocity is the derivative of position:
\[ \boxed{ v(t)=x'(t) }. \]Instantaneous speed is the magnitude of velocity:
\[ \boxed{ \text{speed}(t)=|v(t)| }. \]In vector motion,
\[ \boxed{ \mathbf{v}(t) = \mathbf{r}'(t) }. \]The most important distinction is:
- speed measures how rapidly distance accumulates;
- velocity measures how rapidly position changes;
- speed has magnitude but no directional sign;
- velocity includes direction.
Where We Go Next
Velocity describes the rate at which position changes. But velocity itself can change.
The next lesson, Acceleration, measures how quickly velocity changes and connects changes in speed, direction, and motion over time.