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Two cars are moving in same direction with a speed of \(30 \mathrm{~km} / \mathrm{h}\). They are separated by a distance of \(5 \mathrm{~km}\). What is the speed of a car moving in opposite direction if it meets the two cars at an interval of 4 min? a. \(60 \mathrm{~km} / \mathrm{h}\) b. \(15 \mathrm{~km} / \mathrm{h}\) c. \(30 \mathrm{~km} / \mathrm{h}\) d. \(45 \mathrm{~km} / \mathrm{h}\)

Short Answer

Expert verified
The speed of the car moving in the opposite direction is 45 km/h.

Step by step solution

01

Understand the Problem

We have two cars, A and B, moving in the same direction at speeds of \(30 \mathrm{~km/h}\), with a \(5 \mathrm{~km}\) gap between them. A third car, C, is moving in the opposite direction and meets Car A first, then meets Car B after \(4\) minutes.
02

Calculate Relative Speed Between Cars A/B and Car C

Since Car C is moving in the opposite direction to Cars A and B, you add their speeds to find the relative speed. Suppose Car C's speed is \(v \mathrm{~km/h}\). Therefore, the relative speed between Car C and Car A/B is \((v + 30) \mathrm{~km/h}\).
03

Convert Timings to Hours

Car C meets Cars A and B every \(4\) minutes, which needs to be converted into hours as \(4 \text{ minutes} = \frac{4}{60} \text{ hours} = \frac{1}{15} \text{ hours}.\)
04

Use Distance-Time Relationship

The distance Car C travels in \(\frac{1}{15}\) hours is \(5 \mathrm{~km}\). Using the formula \(\text{Distance} = \text{Relative Speed} \times \text{Time}\), we have:\[5 = (v + 30) \times \frac{1}{15} \]
05

Solve for Car C's Speed

Solve the equation \[5 = (v + 30) \times \frac{1}{15}\] to find \(v\):Multiply both sides by \(15\):\[75 = v + 30\]Subtract \(30\) from both sides to solve for \(v\):\[v = 45\mathrm{~km/h}\].
06

Verify the Solution

Check if the conditions meet: If Car C's speed is \(45 \mathrm{~km/h}\) and operates the following \(True\):- The relative speed becomes \(45 + 30 = 75 \mathrm{~km/h}\).- Car C would indeed traverse \(5 \mathrm{~km}\) in \(4 \text{ minutes} = \frac{1}{15} \text{ hours}\): \[75 \times \frac{1}{15} = 5 \mathrm{~km}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed and Distance
Speed and distance are fundamental concepts in understanding motion. Speed determines how fast an object is moving, while distance is how far it travels. The relationship between these two can be expressed by the equation:
  • \( \text{Distance} = \text{Speed} \times \text{Time} \)
Any change in speed or time affects the distance covered.
In our exercise, cars A and B have a constant speed of \(30\, \text{km/h}\). Understanding that they maintain the same distance relative to each other, 5 km, helps us when calculating interactions with other moving objects, like car C.
Utilizing this understanding, we figure out how car C, moving in the opposite direction, interacts with cars A and B by considering their relative speeds and the time intervals at which car C meets them. Being able to break down these elements helps in predicting outcomes when speeds or distances change.
Problem Solving in Physics
Physics problems often require a systematic approach to arrive at a solution. This includes understanding the problem, identifying known and unknown variables, and using relevant equations.
  • Recognize the directions and speeds of the objects involved.
  • Calculate the time intervals correctly to match speed with distance.
  • Set up equations based on the distance-time relationship.
For our example, the step-by-step approach began by understanding the direction and distance challenges present.
We had to consider that car C meets the two cars at consecutive events with a time interval of 4 minutes. The relative speed is key in simplifying the problem as it combines speeds since car C is moving in the opposite direction.
Setting up our formula as \[ \text{Distance} = (v + 30) \times \text{Time} \]allowed us to solve for the unknown \(v\) efficiently.
Attention to detail is crucial in solving physics problems as every value contributes to the outcome.
Time and Motion
Time and motion are closely linked in physics problems. Measuring time accurately is as important as measuring speed or distance. In problems like our car scenario, converting units appropriately is crucial.
The time given is in minutes, but since speed is in km/h, we convert time to hours:
  • 4 minutes = \( \frac{4}{60} \text{ hours} = \frac{1}{15} \text{ hours}\).
This conversion ensures consistency in the calculations, preventing errors.
Understanding how objects move over time allows us to predict future interactions and outcomes. In our exercise, by having time correctly converted, we could precisely utilize the speed-time-distance equation to find out how quickly car C must move to meet both cars A and B within the specified intervals.
Thus, keeping a close eye on how time and motion interrelate can guide us towards precise solutions.

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Most popular questions from this chapter

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