The Central Idea
When an object moves at a constant rate \(r\) for a time \(t\), the distance traveled is
\[ \boxed{d=rt}. \]Here:
- \(d\) represents distance;
- \(r\) represents rate or speed;
- \(t\) represents elapsed time.
This equation is the foundation of the entire distance-rate-time problem family.
Why the Formula Makes Sense
A rate tells us how much distance is covered per unit of time.
If a vehicle travels
\[ 60\frac{\text{mi}}{\text{hr}}, \]then in \(3\) hours it travels
\[ 60\frac{\text{mi}}{\text{hr}}(3\text{ hr}) = 180\text{ mi}. \]The time units cancel, leaving distance.
Therefore,
\[ \boxed{d=rt} \]is also a unit relationship.
Solving for Distance
If rate and time are known, use
\[ d=rt. \]Suppose a train travels at \(72\) miles per hour for \(2.5\) hours.
\[ d=72(2.5). \] \[ d=180. \]Therefore,
\[ \boxed{d=180\text{ miles}}. \]Solving for Rate
Starting from
\[ d=rt, \]divide by \(t\), assuming \(t\neq0\):
\[ \boxed{r=\frac{d}{t}}. \]If a vehicle travels \(270\) miles in \(4.5\) hours, then
\[ r=\frac{270}{4.5}=60. \]Therefore,
\[ \boxed{r=60\text{ mph}}. \]Solving for Time
Starting from
\[ d=rt, \]divide by \(r\), assuming \(r\neq0\):
\[ \boxed{t=\frac{d}{r}}. \]If a cyclist travels \(42\) miles at \(14\) miles per hour, then
\[ t=\frac{42}{14}=3. \]Therefore,
\[ \boxed{t=3\text{ hours}}. \]The Three Equivalent Forms
The distance-rate-time relationship can be written in three equivalent forms:
\[ \boxed{d=rt}, \] \[ \boxed{r=\frac{d}{t}}, \]and
\[ \boxed{t=\frac{d}{r}}. \]Rather than memorizing three unrelated formulas, derive the needed form from \(d=rt\).
Units Must Be Compatible
The units of rate and time must match.
If rate is measured in miles per hour, time must be expressed in hours before using
\[ d=rt. \]If time is given in minutes, convert it first.
For example,
\[ 90\text{ min} = \frac{90}{60}\text{ hr} = 1.5\text{ hr}. \]Example: Convert Time Before Using the Formula
A car travels at \(64\) miles per hour for \(45\) minutes. Find the distance.
Convert \(45\) minutes to hours:
\[ 45\text{ min} \left( \frac{1\text{ hr}}{60\text{ min}} \right) = 0.75\text{ hr}. \]Then
\[ d=64(0.75)=48. \]Therefore,
\[ \boxed{48\text{ miles}}. \]Example: Convert Rate Before Using the Formula
A runner moves at \(3\) meters per second for \(4\) minutes. Find the distance in meters.
Convert time:
\[ 4\text{ min}=240\text{ s}. \]Then
\[ d=3(240)=720. \]Therefore,
\[ \boxed{720\text{ m}}. \]Distance-Rate-Time Tables
A table is one of the most reliable ways to organize motion problems.
\[ \begin{array}{c|c|c|c} \text{segment} & \text{rate} & \text{time} & \text{distance}\\ \hline 1 & r_1 & t_1 & r_1t_1\\ 2 & r_2 & t_2 & r_2t_2 \end{array} \]Each row satisfies
\[ d=rt. \]The verbal relationship between the rows then determines the equation used to solve the problem.
One Trip Can Have Several Segments
A journey does not need to occur at one constant rate.
Suppose a traveler moves \(2\) hours at \(50\) mph and then \(3\) hours at \(65\) mph.
The first distance is
\[ d_1=50(2)=100. \]The second distance is
\[ d_2=65(3)=195. \]Total distance is
\[ 100+195=295. \]Therefore,
\[ \boxed{295\text{ miles}}. \]Total Distance
For several travel segments, total distance is the sum of the segment distances:
\[ \boxed{D=d_1+d_2+\cdots+d_n}. \]Since each segment satisfies \(d_i=r_it_i\),
\[ \boxed{D=r_1t_1+r_2t_2+\cdots+r_nt_n}. \]This form is especially useful when one of the times or rates is unknown.
Total Time
If a trip is divided into consecutive travel segments, total travel time is
\[ \boxed{T=t_1+t_2+\cdots+t_n}. \]If the problem includes a stop, rest period, or delay, decide whether that nonmoving interval belongs in the requested total elapsed time.
Travel time and elapsed clock time are not always the same quantity.
Average Speed
Average speed is defined by
\[ \boxed{ \text{average speed} = \frac{\text{total distance}}{\text{total travel time}} }. \]It is generally not found by simply averaging the individual speeds.
Example: Average Speed over Two Segments
A vehicle travels \(2\) hours at \(40\) mph and \(3\) hours at \(70\) mph.
Total distance:
\[ 40(2)+70(3)=80+210=290. \]Total time:
\[ 2+3=5. \]Therefore,
\[ \text{average speed} = \frac{290}{5} = 58. \]Thus,
\[ \boxed{58\text{ mph}}. \]Why the Ordinary Average of Speeds Can Fail
In the previous example, the arithmetic mean of \(40\) mph and \(70\) mph is
\[ \frac{40+70}{2}=55. \]But the actual average speed is \(58\) mph because the traveler spent more time at \(70\) mph than at \(40\) mph.
Average speed weights each rate according to the distance and time actually accumulated.
Equal Times Give the Arithmetic Mean of Two Speeds
Suppose a traveler spends the same time \(t\) at rates \(r_1\) and \(r_2\).
Total distance is
\[ r_1t+r_2t=t(r_1+r_2). \]Total time is
\[ 2t. \]Therefore,
\[ \text{average speed} = \frac{t(r_1+r_2)}{2t} = \boxed{\frac{r_1+r_2}{2}}. \]The ordinary average works here because the travel times are equal.
Equal Distances Behave Differently
Suppose the same distance \(d\) is traveled at rates \(r_1\) and \(r_2\).
Total distance is
\[ 2d. \]Total time is
\[ \frac{d}{r_1}+\frac{d}{r_2}. \]Therefore,
\[ \text{average speed} = \frac{2d}{ \frac{d}{r_1}+\frac{d}{r_2} }. \]Simplifying gives
\[ \boxed{ \text{average speed} = \frac{2r_1r_2}{r_1+r_2} }. \]This is the harmonic mean of the two positive speeds.
Example: Equal Distances at Different Speeds
A traveler covers the same distance at \(30\) mph and \(60\) mph.
The average speed is
\[ \frac{2(30)(60)}{30+60}. \] \[ \frac{3600}{90}=40. \]Therefore,
\[ \boxed{40\text{ mph}}. \]It is not \(45\) mph.
A Basic Unknown-Time Problem
A vehicle travels \(156\) miles at \(52\) mph. How long does the trip take?
Use
\[ t=\frac{d}{r}. \]Then
\[ t=\frac{156}{52}=3. \]Therefore,
\[ \boxed{3\text{ hours}}. \]A Basic Unknown-Rate Problem
A boat travels \(96\) miles in \(4\) hours at a constant speed. Find its rate.
\[ r=\frac{96}{4}=24. \]Therefore,
\[ \boxed{24\text{ mph}}. \]Unknown Time across Two Segments
A vehicle travels \(2\) hours at \(45\) mph and then continues at \(60\) mph. The total trip is \(240\) miles. How long is the second segment?
Let \(t\) be the second travel time.
The first distance is
\[ 45(2)=90. \]The second distance is
\[ 60t. \]Total distance gives
\[ 90+60t=240. \]Solve:
\[ 60t=150, \] \[ t=2.5. \]Therefore,
\[ \boxed{2.5\text{ hours}}. \]Unknown Rate across Two Segments
A traveler covers \(330\) miles in \(5\) hours. For the first \(2\) hours, the rate is \(55\) mph. During the remaining \(3\) hours, the rate is \(r\).
Write total distance:
\[ 55(2)+3r=330. \] \[ 110+3r=330. \] \[ 3r=220. \] \[ r=\frac{220}{3}. \]Therefore,
\[ \boxed{r=73\frac{1}{3}\text{ mph}}. \]Using a Total-Time Condition
A trip consists of \(120\) miles at \(40\) mph and \(180\) miles at an unknown rate \(r\). The total travel time is \(6\) hours.
The first time is
\[ \frac{120}{40}=3. \]The second time is
\[ \frac{180}{r}. \]Therefore,
\[ 3+\frac{180}{r}=6. \]Solve:
\[ \frac{180}{r}=3, \] \[ r=60. \]Therefore,
\[ \boxed{60\text{ mph}}. \]Distance as a Function of Time
At constant rate \(r\), distance is a direct variation of time:
\[ \boxed{d=rt}. \]The rate \(r\) is the constant of proportionality.
If \(r=50\) mph, then
\[ d=50t. \]Every additional hour adds \(50\) miles of distance.
The Graph of Constant-Speed Motion
For
\[ d=rt, \]a graph of distance \(d\) against time \(t\) is a straight line through the origin when motion begins at distance \(0\).
Its slope is
\[ \frac{\Delta d}{\Delta t}=r. \]A larger positive slope represents a greater speed.
Starting Away from the Origin
Sometimes position rather than distance traveled is being modeled.
If an object begins at position \(p_0\) and moves with constant signed velocity \(v\), then
\[ \boxed{p=p_0+vt}. \]This is an affine relationship rather than a direct variation unless \(p_0=0\).
For elementary distance problems, however, \(d=rt\) remains the central model.
Distance and Position Are Not the Same
Distance traveled measures how much ground has been covered.
Position identifies location relative to a chosen origin.
A traveler can move \(10\) miles outward and \(10\) miles back, covering
\[ 20\text{ miles} \]of distance while ending at the original position.
Keeping these ideas separate becomes especially important in motion involving direction.
Speed and Velocity
Speed describes how fast distance is being accumulated.
Velocity includes direction and can therefore be positive or negative relative to a chosen axis.
Many introductory word problems use “rate” to mean a positive speed.
Problems involving opposing directions and relative motion require more careful treatment of direction and will be handled in the next lesson.
Elapsed Time versus Clock Time
A problem may state departure and arrival times rather than an elapsed duration.
If a trip begins at 9:20 a.m. and ends at 12:05 p.m., the elapsed time is
\[ 2\text{ hr }45\text{ min}. \]In decimal hours,
\[ 2+\frac{45}{60}=2.75. \]Thus the travel time is
\[ \boxed{2.75\text{ hr}}. \]Minutes Are Fractions of an Hour
To convert minutes to hours, divide by \(60\):
\[ m\text{ min} = \frac{m}{60}\text{ hr}. \]For example,
\[ 15\text{ min}=0.25\text{ hr}, \] \[ 30\text{ min}=0.5\text{ hr}, \]and
\[ 45\text{ min}=0.75\text{ hr}. \]Treating \(30\) minutes as \(0.30\) hours is incorrect.
Decimal Hours Back to Hours and Minutes
Suppose a calculation gives
\[ t=2.4\text{ hr}. \]The whole-number part is \(2\) hours.
Convert the decimal part:
\[ 0.4(60)=24. \]Therefore,
\[ \boxed{2.4\text{ hr}=2\text{ hr }24\text{ min}}. \]Example: A Travel-Time Constraint
A traveler must cover \(210\) miles in \(3.5\) hours. What constant speed is required?
\[ r=\frac{210}{3.5}. \] \[ r=60. \]Therefore,
\[ \boxed{60\text{ mph}}. \]Example: Required Rate after a Slow Segment
A \(300\)-mile trip must be completed in \(5\) hours. The first \(2\) hours are traveled at \(45\) mph. What average rate is required for the remaining \(3\) hours?
First segment:
\[ d_1=45(2)=90. \]Remaining distance:
\[ 300-90=210. \]Remaining time:
\[ 5-2=3. \]Required rate:
\[ r=\frac{210}{3}=70. \]Therefore,
\[ \boxed{70\text{ mph}}. \]Example: Travel with a Rest Stop
A driver travels \(120\) miles at \(60\) mph, rests for \(30\) minutes, then travels \(150\) miles at \(50\) mph.
First travel time:
\[ \frac{120}{60}=2\text{ hr}. \]Second travel time:
\[ \frac{150}{50}=3\text{ hr}. \]Total travel time is
\[ 2+3=5\text{ hr}. \]Total elapsed time including the stop is
\[ 5+0.5=5.5\text{ hr}. \]The requested interpretation determines which total should be reported.
Average Speed with a Stop
Average speed can be defined over moving time or over total elapsed time, depending on context.
In the previous example, total distance is
\[ 120+150=270\text{ mi}. \]Using travel time only:
\[ \frac{270}{5}=54\text{ mph}. \]Using total elapsed time including the stop:
\[ \frac{270}{5.5}\approx49.09\text{ mph}. \]State clearly which time interval is being used.
Piecewise Constant Rates
When a traveler moves at different constant rates over different intervals, the trip can be modeled piece by piece.
For example, if the rate is \(40\) mph for the first \(2\) hours and \(60\) mph afterward, the accumulated distance is not described by one single constant-rate formula for the entire trip.
Instead, each segment has its own equation, and the segment distances are added.
Example: Distance after Several Stages
A traveler moves:
- \(1.5\) hours at \(48\) mph;
- \(2\) hours at \(62\) mph;
- \(0.75\) hours at \(40\) mph.
The distances are
\[ 48(1.5)=72, \] \[ 62(2)=124, \]and
\[ 40(0.75)=30. \]Therefore,
\[ D=72+124+30=226. \]Thus,
\[ \boxed{226\text{ miles}}. \]Example: Find a Missing Segment Time
A traveler covers \(310\) miles. The first \(3\) hours are traveled at \(50\) mph, and the remaining distance is traveled at \(80\) mph. Find the time spent at \(80\) mph.
First distance:
\[ 50(3)=150. \]Remaining distance:
\[ 310-150=160. \]Remaining time:
\[ t=\frac{160}{80}=2. \]Therefore,
\[ \boxed{2\text{ hours}}. \]Example: Find a Missing Segment Rate
A \(400\)-mile journey takes \(6\) hours. The first \(4\) hours are traveled at \(55\) mph. Find the rate during the final \(2\) hours.
First distance:
\[ 55(4)=220. \]Remaining distance:
\[ 400-220=180. \]Therefore,
\[ r=\frac{180}{2}=90. \]Thus,
\[ \boxed{90\text{ mph}}. \]Building an Equation from Total Distance
Suppose one segment is traveled at \(r\) mph for \(3\) hours and another at \(r+10\) mph for \(2\) hours. The total distance is \(280\) miles.
The segment distances are
\[ 3r \]and
\[ 2(r+10). \]Therefore,
\[ 3r+2(r+10)=280. \]Simplify:
\[ 5r+20=280, \] \[ 5r=260, \] \[ r=52. \]The two rates are
\[ \boxed{52\text{ mph and }62\text{ mph}}. \]Building an Equation from Total Time
A traveler covers \(100\) miles at \(r\) mph and then \(150\) miles at \(r+10\) mph. The total travel time is \(5\) hours.
The times are
\[ \frac{100}{r} \]and
\[ \frac{150}{r+10}. \]Therefore,
\[ \boxed{ \frac{100}{r} + \frac{150}{r+10} = 5 }. \]This equation can be solved using the rational-equation methods developed earlier in Algebraic Relationships.
Solving the Total-Time Example
Begin with
\[ \frac{100}{r} + \frac{150}{r+10} = 5. \]Multiply by \(r(r+10)\):
\[ 100(r+10)+150r=5r(r+10). \]Expand:
\[ 250r+1000=5r^2+50r. \]Rearrange:
\[ 5r^2-200r-1000=0. \]Divide by \(5\):
\[ r^2-40r-200=0. \]The positive solution is
\[ r=20+10\sqrt6\approx44.49. \]The negative algebraic solution is rejected because a positive speed is required in this context.
Physical Restrictions Matter
Algebraic equations may produce solutions that are mathematically valid but impossible in the modeled situation.
In ordinary travel problems:
- distance traveled is usually nonnegative;
- elapsed time must be nonnegative;
- a stated positive speed must be positive;
- denominators representing rates cannot equal zero.
Interpret every algebraic solution in context.
Rate as a Unit Ratio
A rate such as
\[ 55\frac{\text{mi}}{\text{hr}} \]is itself a ratio.
This means distance-rate-time problems are an extension of the ratio and proportion framework:
\[ r=\frac{d}{t}. \]The difference is that the quantities now have a specific physical interpretation.
Direct Proportion at Constant Rate
At a fixed rate, distance and time vary directly:
\[ \frac{d}{t}=r. \]If a vehicle travels \(150\) miles in \(3\) hours at a constant rate, then in \(5\) hours it travels \(d\) satisfying
\[ \frac{150}{3}=\frac{d}{5}. \]Since the rate is \(50\) mph,
\[ d=250. \]Therefore,
\[ \boxed{250\text{ miles}}. \]Inverse Proportion for Fixed Distance
If distance is fixed, travel time varies inversely with rate:
\[ t=\frac{d}{r}. \]For a fixed \(120\)-mile trip,
\[ t=\frac{120}{r}. \]Doubling the rate halves the travel time.
This connects distance-rate-time problems directly to the variation models from the previous lesson.
Example: Effect of Increasing Speed on Fixed-Distance Time
A \(180\)-mile trip takes
\[ \frac{180}{45}=4 \]hours at \(45\) mph.
At \(60\) mph, it takes
\[ \frac{180}{60}=3 \]hours.
Increasing the speed by
\[ \frac{60-45}{45}=\frac13=33\frac13\% \]reduces the time by
\[ \frac{4-3}{4}=25\%. \]Equal percentage changes do not appear because time varies inversely with rate for fixed distance.
Checking with Units
Suppose we calculate
\[ 70\frac{\text{mi}}{\text{hr}}(2.5\text{ hr}). \]The hours cancel:
\[ 175\text{ mi}. \]If the units fail to simplify to the requested quantity, the equation or conversion may be incorrect.
Checking with Magnitude
Estimation can expose errors quickly.
If a vehicle travels near \(60\) mph for about \(4\) hours, its distance should be near
\[ 60(4)=240\text{ miles}. \]A calculated result of \(24\) miles or \(2400\) miles would deserve immediate rechecking.
Checking by Reconstructing the Trip
For a multi-stage problem, substitute the solved values into each segment.
If the trip is supposed to total \(300\) miles, verify that
\[ d_1+d_2+\cdots=300. \]If the trip is supposed to last \(6\) hours, verify that the segment times sum to \(6\).
The original condition should be recovered exactly or to the requested rounding precision.
Common Error: Multiplying Incompatible Units
A rate of
\[ 50\frac{\text{mi}}{\text{hr}} \]cannot be multiplied directly by \(30\) minutes without first reconciling the units.
Convert:
\[ 30\text{ min}=0.5\text{ hr}. \]Then
\[ d=50(0.5)=25\text{ mi}. \]Common Error: Treating Minutes as Hundredths of an Hour
One hour contains \(60\) minutes, not \(100\).
Therefore,
\[ 20\text{ min} = \frac{20}{60}\text{ hr} = \frac13\text{ hr}, \]not
\[ 0.20\text{ hr}. \]Common Error: Averaging Speeds Without Considering Time
If a traveler moves \(1\) hour at \(30\) mph and \(4\) hours at \(60\) mph, the arithmetic mean
\[ \frac{30+60}{2}=45 \]is not the average speed.
Total distance is
\[ 30(1)+60(4)=270. \]Total time is \(5\) hours, so
\[ \boxed{\text{average speed}=54\text{ mph}}. \]Common Error: Adding Rates When the Problem Requires Distances
If one segment is traveled at \(40\) mph and another at \(60\) mph, adding the rates gives
\[ 100\text{ mph}, \]but that number is not automatically a meaningful rate for the trip.
Calculate each distance using
\[ d=rt \]and combine the distances or times according to the problem statement.
Common Error: Confusing Total Distance with Remaining Distance
If a \(300\)-mile trip already includes a \(120\)-mile first segment, the remaining distance is
\[ 300-120=180, \]not \(300\).
Labeling each segment before writing the equation helps prevent double counting.
Common Error: Using Total Time for Every Segment
If a \(5\)-hour trip contains one segment lasting \(2\) hours, the second segment lasts
\[ 5-2=3 \]hours.
Using \(5\) hours in both segment equations would count the same time twice.
Common Error: Ignoring Stops in Elapsed-Time Questions
A rest stop adds elapsed time even though it adds no travel distance.
If a traveler drives \(4\) hours and stops for \(45\) minutes, the elapsed time is
\[ 4.75\text{ hr}, \]while the moving time remains \(4\) hours.
Read the requested quantity carefully.
Common Error: Accepting an Impossible Algebraic Solution
A rational or quadratic motion equation may produce more than one algebraic solution.
If one solution gives a negative travel time or an inadmissible positive speed denominator, it does not fit the physical model.
Algebra produces candidates. Context determines which candidates are meaningful.
A Reliable Procedure for Basic Distance-Rate-Time Problems
- Identify the known distance, rate, and time.
- Choose a variable for the unknown quantity.
- Convert units so rate and time are compatible.
- Write \(d=rt\).
- Substitute the known values.
- Solve for the unknown.
- Attach the correct unit.
- Check the result by substitution and estimation.
A Reliable Procedure for Multi-Stage Travel
- Separate the journey into distinct segments.
- Create a distance-rate-time row for each segment.
- Write \(d_i=r_it_i\) for every segment.
- Identify the total-distance or total-time condition.
- Translate that condition into an equation.
- Solve the equation.
- Reconstruct each segment using the solution.
- Verify the total distance and total time.
A Reliable Procedure for Average-Speed Problems
- Find the distance traveled during each segment.
- Add the segment distances.
- Find the time spent during each segment.
- Add the relevant times.
- Use total distance divided by total time.
- Do not average speeds directly unless the structure justifies it.
- State whether stops are included in the time denominator.
Practice 1: Find Distance
A vehicle travels at \(68\) mph for \(3.5\) hours. Find the distance.
Practice 2: Find Rate
A train travels \(315\) miles in \(4.5\) hours. Find its average rate over the trip.
Practice 3: Find Time
A cyclist travels \(54\) miles at \(18\) mph. Find the travel time.
Practice 4: Convert Minutes
A car travels at \(72\) mph for \(35\) minutes. Find the distance traveled.
Practice 5: Two-Segment Distance
A traveler moves \(2\) hours at \(45\) mph and \(1.5\) hours at \(70\) mph. Find the total distance.
Practice 6: Find a Missing Segment Time
A \(280\)-mile trip begins with \(3\) hours at \(60\) mph. The remaining distance is traveled at \(50\) mph. Find the time spent on the second segment.
Practice 7: Find a Missing Segment Rate
A \(350\)-mile trip takes \(5\) hours. The first \(2\) hours are traveled at \(55\) mph. Find the average rate required during the final \(3\) hours.
Practice 8: Average Speed
A traveler moves \(2\) hours at \(50\) mph and \(4\) hours at \(65\) mph. Find the average speed.
Practice 9: Equal-Distance Average Speed
The same distance is traveled at \(40\) mph and \(60\) mph. Find the average speed for the complete trip.
Practice 10: Travel with a Stop
A driver travels \(150\) miles at \(50\) mph, stops for \(30\) minutes, then travels \(120\) miles at \(60\) mph. Find both the total moving time and the total elapsed time.
Practice 11: Average Speed Including a Stop
Using Practice 10, find the average speed over the total elapsed time.
Practice 12: Related Rates in Two Segments
A traveler drives \(3\) hours at \(r\) mph and \(2\) hours at \(r+15\) mph. The total distance is \(315\) miles. Find both rates.
Practice 13: Fixed-Distance Time
How long does a \(240\)-mile trip take at \(64\) mph?
Practice 14: Decimal Hours
Convert \(3.65\) hours to hours and minutes.
Practice 15: Identify the Error
A student travels \(120\) miles at \(40\) mph and \(120\) miles at \(60\) mph. The student says the average speed is
\[ \frac{40+60}{2}=50\text{ mph}. \]Explain the error and find the correct average speed.
Practice Answers
Practice 1
\[ d=68(3.5)=238. \]Therefore,
\[ \boxed{238\text{ miles}}. \]Practice 2
\[ r=\frac{315}{4.5}=70. \]Therefore,
\[ \boxed{70\text{ mph}}. \]Practice 3
\[ t=\frac{54}{18}=3. \]Therefore,
\[ \boxed{3\text{ hours}}. \]Practice 4
Convert time:
\[ 35\text{ min} = \frac{35}{60}\text{ hr} = \frac{7}{12}\text{ hr}. \]Then
\[ d=72\left(\frac{7}{12}\right)=42. \]Therefore,
\[ \boxed{42\text{ miles}}. \]Practice 5
\[ d_1=45(2)=90, \] \[ d_2=70(1.5)=105. \]Therefore,
\[ \boxed{D=195\text{ miles}}. \]Practice 6
First distance:
\[ 60(3)=180. \]Remaining distance:
\[ 280-180=100. \]Second time:
\[ t=\frac{100}{50}=2. \]Therefore,
\[ \boxed{2\text{ hours}}. \]Practice 7
First distance:
\[ 55(2)=110. \]Remaining distance:
\[ 350-110=240. \]Remaining time:
\[ 5-2=3. \]Therefore,
\[ r=\frac{240}{3}=80. \] \[ \boxed{80\text{ mph}}. \]Practice 8
Total distance:
\[ 50(2)+65(4)=100+260=360. \]Total time:
\[ 2+4=6. \]Therefore,
\[ \boxed{\text{average speed}=60\text{ mph}}. \]Practice 9
For equal distances,
\[ \text{average speed} = \frac{2(40)(60)}{40+60}. \] \[ =\frac{4800}{100}=48. \]Therefore,
\[ \boxed{48\text{ mph}}. \]Practice 10
First travel time:
\[ \frac{150}{50}=3\text{ hr}. \]Second travel time:
\[ \frac{120}{60}=2\text{ hr}. \]Total moving time:
\[ \boxed{5\text{ hr}}. \]Including the \(0.5\)-hour stop:
\[ \boxed{5.5\text{ hr elapsed}}. \]Practice 11
Total distance is
\[ 150+120=270. \]Using total elapsed time,
\[ \text{average speed} = \frac{270}{5.5} \approx49.09. \]Therefore,
\[ \boxed{49.09\text{ mph approximately}}. \]Practice 12
Write total distance:
\[ 3r+2(r+15)=315. \] \[ 3r+2r+30=315. \] \[ 5r=285. \] \[ r=57. \]Therefore, the two rates are
\[ \boxed{57\text{ mph and }72\text{ mph}}. \]Practice 13
\[ t=\frac{240}{64}=3.75. \]Therefore,
\[ \boxed{3.75\text{ hr}=3\text{ hr }45\text{ min}}. \]Practice 14
The decimal part is
\[ 0.65\text{ hr}. \]Convert to minutes:
\[ 0.65(60)=39. \]Therefore,
\[ \boxed{3\text{ hr }39\text{ min}}. \]Practice 15
Equal distances do not imply equal travel times, so the two speeds cannot be averaged directly.
Let each one-way distance be \(120\) miles.
Total distance:
\[ 120+120=240. \]Total time:
\[ \frac{120}{40}+\frac{120}{60} = 3+2 = 5. \]Therefore,
\[ \boxed{\text{average speed}=\frac{240}{5}=48\text{ mph}}. \]What to Remember
The fundamental distance-rate-time relationship is
\[ \boxed{d=rt}. \]From it,
\[ \boxed{r=\frac{d}{t}} \]and
\[ \boxed{t=\frac{d}{r}}. \]For multi-stage travel,
\[ \boxed{D=\sum d_i=\sum r_it_i}. \]Average speed is
\[ \boxed{ \text{average speed} = \frac{\text{total distance}}{\text{relevant total time}} }. \]The most important habits are to keep units compatible, separate the trip into meaningful segments, distinguish moving time from elapsed time, and reconstruct the original travel conditions after solving.
Where We Go Next
The problems in this lesson treated each travel segment largely on its own. Motion becomes more interesting when two objects move at the same time, toward one another, away from one another, or in the same direction.
The next lesson, Relative and Opposing Motion, develops the algebra of combined and relative rates.