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Subsection 2.9.6 Relative Speed

Relative speed is a simple concept that helps us understand how fast one object is moving compared to another object. It’s important because sometimes objects are moving at different speeds, and we need to figure out how quickly they are closing the gap between them or moving away from each other.
Consider a case where two bodies moving in the same direction at different speeds. Their \(\underline{relative \, speed}\) is the difference between the individual speeds.
But if they are moving toward each other, their relative speed is the sum of their speeds.

Subsubsection 2.9.6.1 When Objects Move in the Same Direction

Activity 2.9.7.

Work in Groups.
In this activity you are required to follow each step as listed.
Requirements:
  • Toy car (that can be pulled by a string).
  • String (long enough for the car to travel a sufficient distance).
  • Stopwatch or timer
  • Measuring tape or marked distance on the ground (e.g., 10 meters)
  • Pen and paper (for recording results)
  • Observers (to track the race and help evaluate results)
  • Calculator (optional, for calculations)
Steps:
  1. Find a straight line or pathway that is clear and suitable for the race ( a hallway, classroom floor, long corridor or outside on a field).
  2. Measure out a distance for the race, such as 10 meters. You can use a measuring tape or use a marked area to set the starting and finishing points.
  3. Mark the starting line and the finish line clearly, so both the toy car and the student know where to start and end.
  4. Attach one end of the string securely to the toy car. The other end of the string should be held by a participant or fixed to an anchor (like a pole or sturdy object) so that when the student pulls the string, the car moves along the line and test the toy car to ensure that it moves easily along the straight path when the string is pulled.
  5. You will have two participants: one student (who will walk) and one toy car (which will be pulled by the string).
  6. Assign at least one observer to track the race. The observer(s) will help measure the time it takes for each participant (the student and the toy car) to reach the finish line.
  7. Place the student and the toy at the starting line. The student should be ready to walk at a constant speed.
  8. On the count of "Go!", the student starts walking in a straight line towards the finish line. At the same time, the person holding the string should start pulling the toy car in the same direction (straight line) at a constant speed. The toy car should be pulled in a way that resembles a consistent movement, not too fast or too slow.
  9. The observers should start the stopwatch as soon as both participants begin moving and stop it when either the student or the toy car reaches the finish line.
  10. Record the time taken by both the student and the toy car to reach the finish line. If you have more than one observer, make sure they agree on the recorded times.
  11. After the race, you need to calculate the relative speed between the student and the toy car.
    • Let’s say the student took \(T1\) seconds to cover the distance (\(10\) meters), and the toy took \(T2\) seconds to cover same distance
    • The speed of the student is:
      \begin{align*} \frac{Distance}{Time} \amp = \frac{10}{T1 \, seconds} \end{align*}
      The speed of the toy car is:
      \begin{align*} \frac{Distance}{Time} \amp = \frac{10}{T2 \, seconds} \end{align*}
    • If the student and the toy car are moving in the same direction, the relative speed can be given as:
      \begin{align*} Relative \, speed \amp = Speed \, of \, Toy \, Car - Speed \, of \, Student \end{align*}
  12. Compare and share your fndings with other groups of both the student and the toy car.
  13. Discuss;
    1. Who moved faster? Was the student walking faster or slower than the toy car?
    2. Why one object might have moved faster than the other. Did the toy car move faster because it was pulled, or was the student faster in their walking?
    3. The relative speed: How does the speed of each participant relate to the other? Did they move away from each other, or did they move closer together?
  14. Now try the activity in different conditions or compare vraious speed by repeating the race multiple times.
    • Have the student walk faster or slower.
    • Change the length of the race to see how it affects the results.
    • Adjust how fast the toy car is pulled.

Example 2.9.20.

A cyclist is riding at a speed of \(12\) km/h, and a motorcycle is moving at a speed of \(20\) km/h on the same road in the same direction. If the cyclist starts \(100\) meters ahead of the motorcycle, how long will it take for the motorcycle to overtake the cyclist?
Solution.
Since the objects are moving in the same direction, we calculate the relative speed by subtracting the cyclist’s speed from the motorcycle’s speed:
\begin{align*} Relative Speed \amp = Speed of Motorcycle \, - \, Speed of Cyclist \end{align*}
\begin{align*} \amp = 20 \, km/h \, - \, 12 \, km/h \end{align*}
\begin{align*} \amp = 8 \, km/h \end{align*}
Now let’s convert relative speed to meters per second, so that we can work with the distance in meters, we need to convert the relative speed from km/h to m/s.
\begin{align*} Relative \, speed \amp = 8 \, km/h \times \frac{1000}{3600} \end{align*}
\begin{align*} \amp = 2.22 \, m/s \end{align*}
The initial distance between the motorcycle and the cyclist is 100 meters. To calculate the time it will take for the motorcycle to overtake the cyclist,we have:
\begin{align*} Time \amp = \frac{Distance}{Relative \, Speed} \end{align*}
\begin{align*} \amp = \frac{100 \, m}{2.22 \, m/s} \end{align*}
\begin{align*} \amp \approx 45 \, seconds \end{align*}

Example 2.9.21.

A car travels at \(60\) km/h, and a truck travels at \(45\) km/h in the same direction on a straight road. If the car starts \(150\) meters behind the truck, how long will it take for the car to overtake the truck
Solution.
The relative speed between the car and the truck is:
\begin{align*} \text{Relative Speed} \amp = \text{Speed of Car} - \text{Speed of Truck} \end{align*}
\begin{align*} \amp =60 \, km/h - 45 \, km/h \end{align*}
\begin{align*} \amp = 15 \, km/h \end{align*}
Convert Relative Speed to Meters per Second: Convert 15 km/h to m/s:
\begin{align*} \amp = 15 \, km/h \times \frac{1000}{3600} \end{align*}
\begin{align*} \amp = 4.17 \, m/s \end{align*}
The initial distance between the car and the truck is 150 meters. Use the formula to calculate the time:
\begin{align*} \text{Times} \amp = \frac{Distance}{Relative \, Speed} \end{align*}
\begin{align*} \amp = \frac{150 m}{4.17 \, m/s} \end{align*}
\begin{align*} \amp \approx 36 \, seconds \end{align*}

Exercises Exercises

1.
A cyclist is riding at a speed of \(18\) km/h, and a motorcycle is traveling at \(30\) km/h on the same road. If the cyclist starts \(200\) meters ahead of the motorcycle, how long will it take for the motorcycle to overtake the cyclist?
2.
A toy car is being pulled along a straight path at a speed of \(5\) m/s, while a person walks at \(2\) m/s along the same path. If the toy car starts \(15\) meters ahead of the person, how much time will it take for the person to catch up with the toy car?
3.
Two cyclists are riding along the same road. Cyclist \(A\) is traveling at \(12\) km/h and Cyclist \(B\) is traveling at \(15\) km/h. If Cyclist \(B\) starts \(100\) meters behind Cyclist \(A\text{,}\) how long will it take for Cyclist \(B\) to overtake Cyclist \(A\text{?}\)
4.
A person walks at a speed of \(1.5\) m/s, and a dog runs at a speed of \(3\) m/s. If the dog starts \(10\) meters behind the person, how long will it take the dog to catch up with the person?
5.
Two buses are traveling on the same road in the same direction. Bus \(A\) moves at \(55\) km/h, and Bus \(B\) moves at \(45\) km/h. If Bus \(A\) is \(500\) meters behind Bus \(B\text{,}\) how long will it take for Bus \(A\) to overtake Bus B?

Subsubsection 2.9.6.2 When Objects Move Toward Each Other (Opposite Directions).

Activity 2.9.8.

Using Activity 2.9.7. above now try the race such t he racers move towards each other such that.
  1. Using the starting and finishing point.
    • Put the Student on the marked end, and
    • The Toy car on the other end.
  2. Measure time, record and calculate the speeds.
  3. If the student and the toy car are moving in the same direction, the reative speed is given as;
    \begin{align*} Relative \, speed \amp = Speed \, of \, Toy \, Car + Speed \, of \, Student \end{align*}
  4. Discuss;
    1. Who moved faster? Was the student walking faster or slower than the toy car?
    2. Why one object might have moved faster than the other. Did the toy car move faster because it was pulled, or was the student faster in their walking?
    3. The relative speed: How does the speed of each participant relate to the other? Did they move away from each other, or did they move closer together?
  5. Now try the activity in different conditions or compare vraious speed by repeating the race multiple times.
    • Have the student walk faster or slower.
    • Change the length of the race to see how it affects the results.
    • Adjust how fast the toy car is pulled.

Example 2.9.22.

A train left town X at \(10:00\) AM and traveled towards town \(Y\) at a speed of \(90\) km/h. A second train left town \(Y\) at \(11:00\) AM and traveled towards town \(X\) at \(120\) km/h. The distance between town \(X\) and town \(Y\) is \(360\) \(km\text{.}\)
  1. At what time will the two trains meet?
  2. How far from town X will they meet?
Solution.
The first train departs at \(10:00\) AM, and the second train departs at \(11:00\) AM. Therefore, the first train travels for \(1\) hour before the second train starts.
The distance traveled by the first train in 1 hour:
\begin{align*} \text{Distance} \amp = 90 \, km/h \times \ \, hour \end{align*}
\begin{align*} \amp = 90 \, km \end{align*}
Thus, by \(11:00\) AM, the remaining distance between the two trains is:
\begin{align*} 360\, km \, - \, 90\,km \amp = 270\,km \end{align*}
The time it takes for the two trains to meet after 11:00 AM is given by the formula:
\begin{align*} \text{Time} \amp = \frac{\text{Distnace}}{\text{Relative Speed}} \end{align*}
\begin{align*} \amp = \frac{270\,km}{210\,km/h} \end{align*}
\begin{align*} \amp = 1.2857\,hours \end{align*}
This is approximately 1 hr and 17 minutes.
Therefore, the trains will meet at:
\begin{align*} 11:00 \text{AM} + \text{1 hour and 17 minutes} \amp = 12:17 \text{PM} \end{align*}
So the two trains meet at \(12:17 \text{PM}\)
The distance traveled by the first train before the meeting is:
\begin{align*} \text{Distance} \amp = 90\,km/h \times 1.2857\, hours \end{align*}
\begin{align*} \amp = 115.714 \, km \end{align*}
Thus, the meeting point is approximately \(116\,km \) from town \(X.\)

Example 2.9.23.

Two cyclists start from the same point and travel in opposite directions. One cyclist rides at 20 km/h, and the other rides at 30 km/h. After 2 hours, they are 100 km apart.
  1. How long did it take for the cyclists to be 100 km apart?
  2. How far did each cyclist travel?
Solution.
The cyclists are moving in opposite directions. The total distance between them after \(2\) hours is \(100\) km.
Since the cyclists are moving in opposite directions, their relative speed is the sum of their individual speeds:
\begin{align*} \text{Relative speed} \amp = 20\,km/h + 30\,km/h. \end{align*}
In 2 hours, the total distance traveled by both cyclists is:
\begin{align*} \text{Total distance} \amp = 50\,km/h \times 2\,hours \end{align*}
\begin{align*} \amp = 100\,km \end{align*}
So, the cyclists will be 100 km apart after 2 hours. This confirms the given information.
The first cyclist, riding at 20 km/h, travels:
\begin{align*} \text{Distance covered by first cyclist} \amp = 20\,km/h \times 2\,hours \end{align*}
\begin{align*} \amp = 40\,km. \end{align*}
The second cyclist, riding at 30 km/h, travels:
\begin{align*} \text{Distance covered by second cyclist} \amp = 30\,km/h \times 2\,hours \end{align*}

Exercises Exercises

1.
A motorist left Nakuru for Nairobi, a distance of \(240\) km, at 8.00 am. and travelled at an average speed of \(90\) km/h. Another motorist left Nairobi for Nakuru at \(8.30\) am. and travelled at \(100\) km/h. Find:
  1. The time they met.
  2. How far from Nairobi they met.
2.
A train travels from Mombasa to Nairobi, a distance of \(500\) km, at a speed of \(90\) km/h. If a second train starts \(1\) hour later from Mombasa and travels at a speed of \(120\) km/h, after how much time will the second train overtake the first one?
3.
A matatu left town \(A\) at \(7\) am and travelled towards a town \(B\) at an average speed of \(60\) km/h. A second matatu left town \(B\) at \(8\) am and travelled towards town \(A\) at \(60\) km/h. If the distance between the two towns is \(400\) km, find:
  1. The time at which the two matatus met.
  2. The distance of the meeting point from town \(A\text{.}\)
4.
A cyclist is riding towards a motorcyclist on a straight road. The cyclist travels at \(15\) km/h and the motorcyclist at \(45\) km/h. If they are initially \(100\) meters apart, how long will it take for the motorcyclist to overtake the cyclist?
5.
Two cars, \(A\) and \(B\text{,}\) are traveling on parallel roads. Car \(A\) moves at a speed of \(50\) km/h and car \(B\) moves at \(70\) km/h. If car \(B\) is \(150\) meters behind car \(A\text{,}\) how long will it take for car \(B\) to overtake car \(A\text{?}\)

Checkpoint 2.9.24.

Checkpoint 2.9.25.

Checkpoint 2.9.26.