Rate, Time, and Distance Problems – Solved Examples
The relationship between distance, rate (speed), and time, \( \text{Distance} = \text{Rate} \times \text{Time} \), is the foundation for solving uniform motion problems in mathematics. In this guide, you will find step-by-step solutions, worked examples, and clear explanations to help you master rate, time, and distance problems.
Table of Contents
Formula and Strategy Review
Core Formula:
\[ \text{Distance} = \text{Rate} \times \text{Time} \]Equivalent forms:
\[ \text{Rate} = \frac{\text{Distance}}{\text{Time}}, \quad \text{Time} = \frac{\text{Distance}}{\text{Rate}} \]Strategy by Scenario:
- Opposite directions: Distances add: \( D = D_1 + D_2 = (r_1 + r_2) t \).
- Same direction: Distances subtract: \( d = |D_1 - D_2| = |r_1 - r_2| t \).
- Right angles: Use the Pythagorean theorem: \( D = \sqrt{x^2 + y^2} \).
- Delayed start: If B starts \( h \) hours after A, then \( t_B = t_A - h \).
Question 1 — Two Cars Traveling in Opposite Directions
Two cars started from the same point at 5:00 am, traveling in opposite directions at \( 40 \text{ mph} \) and \( 50 \text{ mph} \) respectively. At what time will they be \( 450 \text{ miles} \) apart?
View Step-by-Step Solution
After \( t \) hours, the distances \( D_1 \) and \( D_2 \) (in miles) traveled by the two cars are:
\[ D_1 = 40t \quad \text{and} \quad D_2 = 50t \]Since they are traveling in opposite directions, the total distance \( D \) separating the two cars is the sum of their distances:
\[ D = D_1 + D_2 = 40t + 50t = 90t \]The distance \( D \) will be equal to 450 miles when:
\[ 90t = 450 \implies t = 5 \text{ hours} \]Therefore, if the cars start at 5:00 am, they will be 450 miles apart at:
Question 2 — Car B Passing Car A
At 9:00 am, a car \( A \) began a journey from a point, traveling at \( 40 \text{ mph} \). At 10:00 am, another car \( B \) started traveling from the same point at \( 60 \text{ mph} \) in the same direction as car \( A \). At what time will car \( B \) pass car \( A \)?
View Step-by-Step Solution
After \( t \) hours from 9:00 am, the distance \( D_1 \) traveled by car A is:
\[ D_1 = 40t \]Car B starts at 10:00 am and will therefore have traveled for one hour less than car A when it passes it. After \( (t - 1) \) hours, the distance \( D_2 \) traveled by car B is:
\[ D_2 = 60(t - 1) \]When car B passes car A, they are at the same distance from the starting point, so \( D_1 = D_2 \):
\[ 40t = 60(t - 1) \] \[ 40t = 60t - 60 \implies 20t = 60 \implies t = 3 \text{ hours} \]This means it takes 3 hours from car A's starting time (9:00 am). Car B passes car A at:
Verification: Car A travels \( 40 \times 3 = 120 \) mi by noon; car B travels \( 60 \times 2 = 120 \) mi by noon. ✓
Question 3 — Two Trains Traveling Towards Each Other
Two trains, traveling towards each other, left from two stations that are 900 miles apart at 4:00 pm. If the rate of the first train is \( 72 \text{ mph} \) and the rate of the second train is \( 78 \text{ mph} \), at what time will they pass each other?
View Step-by-Step Solution
After \( t \) hours, the two trains will have traveled distances \( D_1 \) and \( D_2 \) (in miles):
\[ D_1 = 72t \quad \text{and} \quad D_2 = 78t \]After \( t \) hours, the total distance \( D \) covered collectively by the two trains is their sum:
\[ D = D_1 + D_2 = 72t + 78t = 150t \]When the total distance \( D \) is equal to 900 miles, the two trains pass each other:
\[ 150t = 900 \implies t = 6 \text{ hours} \]Starting at 4:00 pm, they will pass each other exactly 6 hours later, which is at:
Question 4 — Total Distance Driven by John
John left home and drove at the rate of \( 45 \text{ mph} \) for 2 hours. He stopped for lunch, then drove for another 3 hours at the rate of \( 55 \text{ mph} \) to reach his destination. How many miles did John drive in total?
View Step-by-Step Solution
The total distance \( D \) traveled by John is the sum of the distance traveled before lunch and the distance traveled after lunch:
\[ D = (\text{Rate}_1 \times \text{Time}_1) + (\text{Rate}_2 \times \text{Time}_2) \] \[ D = (45 \times 2) + (55 \times 3) \] \[ D = 90 + 165 = 255 \text{ miles} \]Question 5 — Linda's Driving Rate Before Lunch
Linda left home and drove for 2 hours. She stopped for lunch, then drove for another 3 hours at a rate that is \( 10 \text{ mph} \) higher than her rate before lunch. If the total distance Linda traveled is 230 miles, what was her rate before lunch?
View Step-by-Step Solution
If \( x \) is the rate at which Linda drove before lunch, the rate after lunch is \( x + 10 \). The total distance \( D \) traveled by Linda is:
\[ D = 2x + 3(x + 10) \]Since the total distance is 230 miles:
\[ 2x + 3(x + 10) = 230 \] \[ 2x + 3x + 30 = 230 \implies 5x = 200 \implies x = 40 \text{ mph} \]Verification: \( 2(40) + 3(50) = 80 + 150 = 230 \) ✓
Question 6 — Two Cars at Right Angles
Two cars leave at 8:00 am from the same point. One travels east at \( 50 \text{ mph} \) and the other travels south at \( 60 \text{ mph} \). At what time will they be 300 miles apart?
View Step-by-Step Solution
The two cars are traveling in directions that form a right angle. Let \( x \) and \( y \) be the distances traveled by the two cars in \( t \) hours:
\[ x = 50t \quad \text{and} \quad y = 60t \]Since the two directions are at right angles, the Pythagorean theorem gives the direct distance \( D \) between the two cars:
\[ D = \sqrt{x^2 + y^2} \]We want \( D = 300 \) miles:
\[ \sqrt{(50t)^2 + (60t)^2} = 300 \]Square both sides:
\[ (50t)^2 + (60t)^2 = 300^2 \] \[ 2500t^2 + 3600t^2 = 90000 \] \[ 6100t^2 = 90000 \implies t^2 = \frac{90000}{6100} \approx 14.754 \] \[ t \approx 3.84 \text{ hours} \]Convert 0.84 hours into minutes: \( 0.84 \times 60 \approx 50.4 \) minutes. So the time is approximately 3 hours and 51 minutes.
Question 7 — Distance Between Two Cities
By car, John traveled from city \( A \) to city \( B \) in 3 hours. At a rate that was \( 20 \text{ mph} \) higher than John's, Peter traveled the exact same distance in 2 hours. Find the distance between the two cities.
View Step-by-Step Solution
Let \( x \) be John's rate. Peter's rate is \( x + 20 \). Using the rate-time-distance formula, the distance \( D \) for each traveler is:
\[ D = 3x \quad \text{(John)} \] \[ D = 2(x + 20) \quad \text{(Peter)} \]Since the distances are equal:
\[ 3x = 2(x + 20) \] \[ 3x = 2x + 40 \implies x = 40 \text{ mph} \]Substituting \( x = 40 \) into either formula:
\[ D = 3(40) = 120 \text{ miles} \]Verification: Peter's rate = 60 mph, so \( 60 \times 2 = 120 \) mi ✓
Question 8 — Two Cars Crossing Each Other
Gary started driving at 9:00 am from city \( A \) towards city \( B \) at a rate of \( 50 \text{ mph} \). At a rate that is \( 15 \text{ mph} \) higher than Gary's, Thomas started driving at the same time as Gary from city \( B \) towards city \( A \) along the same route. If Gary and Thomas crossed each other at 11:00 am, what is the distance between the two cities?
View Step-by-Step Solution
Both drivers traveled for exactly 2 hours (from 9:00 am to 11:00 am). Let \( D \) be the distance between the two cities. When Gary and Thomas cross, the sum of their distances equals the total distance between the cities.
Gary's distance:
\[ D_1 = 2 \times 50 = 100 \text{ miles} \]Thomas's rate is \( 50 + 15 = 65 \text{ mph} \). Thomas's distance:
\[ D_2 = 2 \times 65 = 130 \text{ miles} \]The total distance between the cities is the sum:
\[ D = D_1 + D_2 = 100 + 130 = 230 \text{ miles} \]Question 9 — Distance Between Two Cars (Same Direction)
Two cars started at the same time from the same point, driving along the same road. The speed of the first car is \( 50 \text{ mph} \) and the speed of the second car is \( 60 \text{ mph} \). How long will it take for the distance between the two cars to be 30 miles?
View Step-by-Step Solution
Let \( D_1 \) and \( D_2 \) be the distances traveled by the two cars in \( t \) hours:
\[ D_1 = 50t \quad \text{and} \quad D_2 = 60t \]Since they are traveling in the same direction, the distance \( d \) separating the two cars is the difference of their distances:
\[ d = 60t - 50t = 10t \]For \( d = 30 \) miles:
\[ 10t = 30 \implies t = 3 \text{ hours} \]Note: The relative speed is \( 60 - 50 = 10 \) mph, so the time to gain 30 miles is \( 30/10 = 3 \) hours.
Question 10 — Two Trains Traveling in Opposite Directions
Two trains started at 10:00 pm from the same point. The first train traveled north at a speed of \( 80 \text{ mph} \) and the second train traveled south at a speed of \( 100 \text{ mph} \). At what time were they 450 miles apart?
View Step-by-Step Solution
Let \( D_1 \) and \( D_2 \) be the distances traveled by the two trains in \( t \) hours:
\[ D_1 = 80t \quad \text{and} \quad D_2 = 100t \]Since the trains are traveling in opposite directions, the total distance \( D \) between them is the sum:
\[ D = D_1 + D_2 = 80t + 100t = 180t \]For this distance to be 450 miles:
\[ 180t = 450 \implies t = 2.5 \text{ hours} \]2.5 hours = 2 hours 30 minutes. If the trains leave at 10:00 pm:
Question 11 — Two Trains with One-Hour Delay
Two trains started from the same point. At 8:00 am, the first train traveled East at the rate of \( 80 \text{ mph} \). At 9:00 am, the second train traveled West at the rate of \( 100 \text{ mph} \). At what time were they 530 miles apart?
View Step-by-Step Solution
Let \( t \) be the number of hours the first train has traveled. The second train started 1 hour later, so it has traveled for \( (t - 1) \) hours. The distances are:
\[ D_1 = 80t \quad \text{and} \quad D_2 = 100(t - 1) \]Since the trains are traveling in opposite directions, the total distance \( D \) between them is:
\[ D = D_1 + D_2 = 80t + 100(t - 1) = 180t - 100 \]We want \( D = 530 \) miles:
\[ 180t - 100 = 530 \] \[ 180t = 630 \implies t = 3.5 \text{ hours} \]3.5 hours = 3 hours 30 minutes. Adding to the start time of the first train (8:00 am):
Verification: First train travels \( 80 \times 3.5 = 280 \) mi; second travels \( 100 \times 2.5 = 250 \) mi; total = 530 mi ✓
Summary Table of Answers
| # | Scenario | Answer |
|---|---|---|
| 1 | Two cars, opposite directions | 10:00 am |
| 2 | Car B passing Car A | 12:00 pm (noon) |
| 3 | Two trains, towards each other | 10:00 pm |
| 4 | Total distance driven | 255 miles |
| 5 | Finding Linda's rate | 40 mph |
| 6 | Two cars at right angles | ≈ 11:51 am |
| 7 | Distance between cities | 120 miles |
| 8 | Two cars crossing | 230 miles |
| 9 | Same-direction separation | 3 hours |
| 10 | Trains, opposite directions | 12:30 am |
| 11 | Trains with 1-hour delay | 11:30 am |
Frequently Asked Questions
What is the formula for rate, time, and distance problems?
The fundamental formula is \( \text{Distance} = \text{Rate} \times \text{Time} \). Equivalent forms are \( \text{Rate} = \frac{\text{Distance}}{\text{Time}} \) and \( \text{Time} = \frac{\text{Distance}}{\text{Rate}} \). All rate-time-distance problems are solved by applying one of these three forms.
How do you solve uniform motion problems when two objects travel towards each other?
When two objects travel towards each other, add their individual distances to get the total separating distance: \( D_1 + D_2 = \text{Total} \). Alternatively, add their rates to find the closing speed: \( (r_1 + r_2) \times t = \text{Total} \).
How do you solve problems where two objects travel in the same direction?
When two objects travel in the same direction, the separation distance equals the difference of their distances: \( d = (r_2 - r_1) \times t \), where \( r_2 \) is the faster rate. The faster object gains on the slower at the rate of \( r_2 - r_1 \).
How do you solve problems where two objects travel at right angles?
When two objects travel at right angles (e.g., one east and one south), their paths form a right triangle. Use the Pythagorean theorem: \( D = \sqrt{x^2 + y^2} \), where \( x \) and \( y \) are the distances each object has traveled.
How do you handle problems where one object starts later than the other?
Let \( t \) be the time the first object has traveled. The second object has traveled for \( (t - \text{delay}) \) hours. Write the distance expressions with these times and set up an equation based on the problem's condition.