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Chapter 2: Problem 49

A school bus takes \(0.70 \mathrm{~h}\) to reach the school from your house. If the average velocity of the bus is \(56 \mathrm{~km} / \mathrm{h}\), what is the displacement?

Short Answer

Expert verified
The displacement is \(39.2 \ \mathrm{km}\).

Step by step solution

01

Identify the given variables

From the exercise, the bus journey's duration (time) is \(0.70 \mathrm{~h}\) and the average velocity of the bus is \(56 \mathrm{~km} / \mathrm{h}\).
02

Apply the formula for displacement

The formula for displacement is given by \( Displacement = Velocity \times Time \). Here, Velocity is \(56 \ \mathrm{~km/h}\) and Time is \(0.70 \ \mathrm{~h}\). Substituting these values into the formula will give: \( Displacement = 56 \mathrm{~km/h} \times 0.70 \mathrm{~h}\).
03

Perform the multiplication

Multiplying the velocity and time together gives us the displacement: \( Displacement = 56 \times 0.70 \ \mathrm{~km}\). Performing the multiplication results in \( Displacement = 39.2 \mathrm{~km}\).

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Key Concepts

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

Displacement Calculation
In physics, displacement refers to the change in position of an object. It is a vector quantity, which means it has both magnitude and direction. In the context of traveling from home to school, displacement would be the shortest path from the starting point (your house) to the endpoint (the school).
To calculate displacement, you can use the formula:
  • Displacement = Velocity \( \times \) Time
In our example, the velocity given is the average velocity of the bus, which is 56 km/h, and the time taken is 0.70 hours. By substituting these values, we can find the displacement as follows:
  • Displacement = 56 km/h \( \times \) 0.70 h = 39.2 km
This calculation shows that the bus travels a total displacement of 39.2 km from your house to the school.
Velocity-Time Relationship
The relationship between velocity and time is foundational in travel-related physics problems. Velocity is the rate of change of something鈥檚 position, and it indicates how fast something is moving and in which direction.
When discussing constant velocity as in this problem, the velocity remains unchanged over time, which simplifies calculations greatly.
  • For constant velocity, the displacement can be directly calculated using the time and velocity, as was done here.
  • This relationship assumes no changes in speed or direction, which aligns with the average velocity used in this bus travel scenario.
Understanding this relationship allows us to predict how long a journey will take or how far it will go if the velocity and time are known.
Average Velocity
Average velocity is one of the key components in determining displacement in physics problems.
It is calculated as the total displacement divided by the total time taken for the journey. It gives a straightforward way to measure the overall speed and direction over a set time frame.
  • If velocity varies during travel, average velocity serves as a useful measure of the overall speed and direction.
  • In scenarios where constant speed is involved, calculating displacement becomes directly solvable with average velocity.
  • In our school bus example, 56 km/h represents the average speed maintained by the bus throughout its journey, simplifying the calculation of displacement to a simple product of average velocity and journey time.
Average velocity provides a practical solution for determining travel outcomes, like timing and distance, on various paths.

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Most popular questions from this chapter

Alex climbs to the top of a tall tree while his friend Gary waits on the ground below. Alex throws down a ball at \(4 \mathrm{~m} / \mathrm{s}\) from \(50 \mathrm{~m}\) above the ground at the same time Gary throws a ball up. At what speed must Gary throw a ball up in order for the two balls to cross paths \(25 \mathrm{~m}\) above the ground? The starting height of the ball thrown upward is \(1.5 \mathrm{~m}\) above the ground. Ignore the effects of air resistance. SSM

A man on a railroad platform attempts to measure the length of a train car by walking the length of the train and keeping the length of his stride a constant \(82 \mathrm{~cm}\) per step. After he has paced off 12 steps from the front of the train it begins to move, in the direction opposite to his, with an acceleration of \(0.4 \mathrm{~m} / \mathrm{s}^{2}\). The end of the train passes him 10 s later, after he has walked another 20 steps. Determine the length of the train car.

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