A motorcycle traveling on the highway at a velocity of 120 km/h passes a car traveling at a velocity of 90 km/h in the same direction. From the point of view (or frame of reference... A motorcycle traveling on the highway at a velocity of 120 km/h passes a car traveling at a velocity of 90 km/h in the same direction. From the point of view (or frame of reference) of a passenger in the car, how much time will the car take to overtake the motorcycle in this case?

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Understand the Problem

The question is asking how long it will take for a car traveling at 90 km/h to overtake a motorcycle traveling faster at 120 km/h from the perspective of a passenger in the car. We need to analyze their relative speeds to determine the time it will take for the car to catch up.

Answer

$0$ hours
Answer for screen readers

The time taken for the car to overtake the motorcycle is $0$ hours.

Steps to Solve

  1. Identify the speeds of both vehicles The motorcycle is traveling at a speed of $120 \text{ km/h}$ and the car at $90 \text{ km/h}$.

  2. Calculate the relative speed Since both vehicles are moving in the same direction, we find the relative speed by subtracting the speed of the car from the motorcycle: $$ \text{Relative speed} = 120 \text{ km/h} - 90 \text{ km/h} = 30 \text{ km/h} $$

  3. Understanding the time to overtake From the passenger's perspective in the car, the motorcycle is moving away. Thus, the car cannot overtake the motorcycle since it is slower.

  4. Conclusion Since the car cannot catch up to or overtake the motorcycle, the time taken to overtake is essentially zero.

The time taken for the car to overtake the motorcycle is $0$ hours.

More Information

In this scenario, the car is not able to overtake the motorcycle due to its slower speed. From the passenger's viewpoint, the motorcycle is only getting farther away.

Tips

  • Confusing relative speed: It's important to see that overtaking can only happen if the car is faster than the motorcycle.
  • Miscalculating the scenario: Assuming that the car can eventually catch up without considering its slower speed.

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