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Question

The driver of a car, which is travelling at a speed of 75 km/h, locates a bus 80 m ahead of him, travelling in the same direction. After 15 seconds, he finds that the bus is 40 m behind the car. What is the speed of the bus (in km/h)?

The correct answer is

46.2

Calculating Bus Speed Using Relative Motion

This problem involves understanding relative speed, which is the speed of an object with respect to another object. When two objects are moving in the same direction, their relative speed is the difference between their individual speeds.

Let's break down the information given in the question:

  • Speed of the car = 75 km/h
  • Initial distance of bus ahead of car = 80 m
  • Time elapsed = 15 seconds
  • Final distance of bus behind car = 40 m
  • We need to find the speed of the bus in km/h.

Step-by-Step Solution

To solve this problem, it's best to work with consistent units. Let's convert the car's speed from km/h to meters per second (m/s).

We know that 1 km = 1000 m and 1 hour = 3600 seconds.

Step 1: Convert the car's speed to m/s

Speed of car (\(V_c\)) = \(75 \text{ km/h}\)

\(V_c = 75 \times \frac{1000 \text{ m}}{3600 \text{ s}} \text{ m/s}\)

\(V_c = 75 \times \frac{10}{36} \text{ m/s}\)

\(V_c = 75 \times \frac{5}{18} \text{ m/s}\)

\(V_c = \frac{375}{18} \text{ m/s}\)

\(V_c = \frac{125}{6} \text{ m/s}\)

Step 2: Determine the total relative distance covered by the car with respect to the bus

Initially, the car is 80 m behind the bus. After 15 seconds, the car is 40 m ahead of the bus.

This means that in 15 seconds, the car has closed the initial 80 m gap AND gained another 40 m on the bus.

Total relative distance covered by the car with respect to the bus = Initial distance + Final distance

Total relative distance = \(80 \text{ m} + 40 \text{ m} = 120 \text{ m}\)

Step 3: Calculate the relative speed of the car with respect to the bus

The relative speed is the speed at which the distance between the car and the bus is changing. Since the car is faster and is catching up and overtaking the bus, the relative speed is the difference between their speeds.

Let the speed of the bus be \(V_b\). Both are in m/s units for calculation.

Relative speed = Speed of car - Speed of bus (\(V_c - V_b\))

We know that Relative Distance = Relative Speed \(\times\) Time

\(120 \text{ m} = (V_c - V_b) \times 15 \text{ s}\)

Rearrange the formula to find the relative speed:

Relative speed (\(V_c - V_b\)) = \(\frac{\text{Total relative distance}}{\text{Time}}\)

Relative speed = \(\frac{120 \text{ m}}{15 \text{ s}}\)

Relative speed = \(8 \text{ m/s}\)

Step 4: Calculate the speed of the bus in m/s

We have the relative speed and the car's speed:

\(V_c - V_b = 8 \text{ m/s}\)

We know \(V_c = \frac{125}{6} \text{ m/s}\).

\(\frac{125}{6} - V_b = 8\)

\(V_b = \frac{125}{6} - 8\)

\(V_b = \frac{125}{6} - \frac{8 \times 6}{6}\)

\(V_b = \frac{125 - 48}{6}\)

\(V_b = \frac{77}{6} \text{ m/s}\)

Step 5: Convert the bus's speed back to km/h

To convert m/s to km/h, we multiply by \(\frac{18}{5}\).

\(V_b = \frac{77}{6} \times \frac{18}{5} \text{ km/h}\)

\(V_b = \frac{77}{\cancel{6}_1} \times \frac{\cancel{18}^3}{5} \text{ km/h}\)

\(V_b = \frac{77 \times 3}{5} \text{ km/h}\)

\(V_b = \frac{231}{5} \text{ km/h}\)

\(V_b = 46.2 \text{ km/h}\)

Thus, the speed of the bus is 46.2 km/h.

Summary of Results

By calculating the total relative distance covered and using the concept of relative speed, we determined the bus's velocity.

Quantity Value Unit
Car Speed (\(V_c\)) 75 km/h
Time (t) 15 s
Initial Distance (bus ahead) 80 m
Final Distance (bus behind) 40 m
Total Relative Distance 120 m
Relative Speed (\(V_c - V_b\)) 8 m/s
Bus Speed (\(V_b\)) 46.2 km/h

Revision Table: Kinematics Concepts

Concept Description Formula
Speed Rate at which an object covers distance. Speed = Distance / Time
Relative Speed (Same Direction) Difference in speeds when objects move in the same direction. \(V_{rel} = |V_1 - V_2|\)
Conversion km/h to m/s Multiply by \(\frac{5}{18}\). \(X \text{ km/h} = X \times \frac{5}{18} \text{ m/s}\)
Conversion m/s to km/h Multiply by \(\frac{18}{5}\). \(Y \text{ m/s} = Y \times \frac{18}{5} \text{ km/h}\)

Additional Information: Relative Motion

Relative motion is a fundamental concept in physics used to describe the motion of an object from the perspective of another object or frame of reference. In simple cases like this, where motion is along a straight line, relative speed is easy to calculate.

  • When two objects move towards each other, their relative speed is the sum of their speeds. The distance between them decreases at this combined speed.
  • When two objects move away from each other (in opposite directions), their relative speed is also the sum of their speeds. The distance between them increases at this combined speed.
  • When two objects move in the same direction, their relative speed is the difference between their speeds. The distance between them changes based on this difference. If the faster object is behind, the distance decreases; if the faster object is ahead, the distance increases.

Understanding relative motion helps solve many problems involving vehicles, boats in rivers, or planes in wind, by simplifying the analysis from one moving frame of reference to another.

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Important Questions from Relative Speed

  1. The distance between two station A and B is 800 km. A train X starts from A and moves towards B at 40 km/h and another trains Y starts from B and moves towards A at 60 km/h. how far from A will they cross each other?

  2. A and B are travelling towards each other from the points P and Q respectively. After crossing each other, A and B take \(6\frac{1}{8}\) hours and 8 hours, respectively, to reach their destinations Q and P, respectively. If the speed of B is 16.8 km/h, then the speed (in km/hr) of A is: 

  3. A train leaves station A at 8 am and reaches station B at 12 noon. A car leaves station B at 8:30 am and reaches station A at the same time when the train reaches station B. At what time do they meet?

  4. A and B start moving towards each other from places X and Y respectively, at the same time. The speed of A is 20% more than that of B. After meeting on the way, A and B take \(2\frac{1}{2}\) hours and x hours now to reach Y and X respectively. What is the value of x?

  5. X and Y are two stations that are 280 km apart. A train starts at a certain time from X and travels towards Y at 60 km/h. After 2 hours, another train starts from Y and travels towards X at 20 km/h After how many hours does the train leaving from X meet the train which left from Y?

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