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Chapter 2: Problem 15

(1I) Calculate the average speed and average velocity of a complete round trip in which the outgoing 250 \(\mathrm{km}\) is covered at 95 \(\mathrm{km} / \mathrm{h}\) , followed by a 1.0 -h lunch break, and the return 250 \(\mathrm{km}\) is covered at 55 \(\mathrm{km} / \mathrm{h}\) .

Short Answer

Expert verified
Average speed is approximately 61.16 km/h; average velocity is 0 km/h.

Step by step solution

01

Calculate the Total Distance of the Trip

The total distance of a round trip is the sum of the distance going to the destination and the distance coming back. Here, both directions cover 250 km, so the total distance is:\[ 250 \text{ km} + 250 \text{ km} = 500 \text{ km} \]
02

Calculate the Total Time for the Trip

First, calculate the time taken for the outgoing journey:\[ \text{Time (outgoing)} = \frac{\text{Distance}}{\text{Speed}} = \frac{250 \text{ km}}{95 \text{ km/h}} \approx 2.632 \text{ h} \]The return journey is:\[ \text{Time (return)} = \frac{250 \text{ km}}{55 \text{ km/h}} \approx 4.545 \text{ h} \]Add the time for the lunch break (1.0 h) to these to get the total time:\[ \text{Total time} = 2.632 \text{ h} + 4.545 \text{ h} + 1.0 \text{ h} = 8.177 \text{ h} \]
03

Calculate the Average Speed of the Trip

Average speed is the total distance divided by the total time:\[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{500 \text{ km}}{8.177 \text{ h}} \approx 61.16 \text{ km/h} \]
04

Calculate the Average Velocity of the Trip

Average velocity is defined as the total displacement divided by the total time. Since this is a round trip, the starting point is the same as the ending point, leading to a displacement of 0 km. Thus:\[ \text{Average velocity} = \frac{0 \text{ km}}{8.177 \text{ h}} = 0 \text{ km/h} \]

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Key Concepts

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

Round Trip Calculation
A round trip implies traveling to a specific destination and then returning to the original location. In this scenario, understanding the calculation involves determining both the total distance covered and the overall time taken.
  • The total distance for a round trip is simply the sum of the distances traveled in both directions. In this case, it involves traveling 250 km to the destination and then the same 250 km back, which sums up to 500 km.
  • Time calculation involves finding out how long the journey takes each way. You take the time for the outgoing trip (distance divided by speed), pause for a lunch break, and then account for the return trip time.
By calculating this total time and distance, you can find the average speed, giving a complete picture of the trip's dynamics.
Displacement in Physics
Displacement in physics refers to the change in position of an object. It's a vector quantity, meaning it has both magnitude and direction. When considering displacement, you think about the shortest path between the starting and ending points.

For a round trip, the concept of displacement becomes particularly interesting:
  • In a round trip, the starting and ending points are the same. Hence, the displacement is zero because there is no change in position.
  • Even if the journey covers a long distance, if you stop where you started, the displacement remains zero.
This zero displacement directly affects the calculation of average velocity, which depends on displacement.
Speed and Velocity Concepts
Speed and velocity are essential concepts in understanding motion. While they are related, they have different meanings and applications.
  • Speed is a scalar quantity and measures how fast an object moves, regardless of its direction. It is calculated as total distance traveled divided by total time taken.
  • Velocity, however, is a vector quantity, which means it considers direction. It is calculated as displacement over time.
In the context of this exercise, you calculate average speed by dividing the total distance by the total time. However, for average velocity, since the displacement for a round trip is zero, the average velocity is also zero, despite having traveled 500 km. This distinction highlights why both speed and velocity are crucial for describing different aspects of motion.

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Most popular questions from this chapter

(II) A horse canters away from its trainer in a straight line, moving \(116 \mathrm{~m}\) away in \(14.0 \mathrm{~s}\). It then turns abruptly and gallops halfway back in 4.8 s. Calculate \((a)\) its average speed and \((b)\) its average velocity for the entire trip, using "away from the trainer" as the positive direction.

(II) Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of \(1.5 \mathrm{~m}\) above the ground (Fig. \(2-45) .\) When you quickly turn off the nozzle, you hear the water striking the ground next to you for another \(2.0 \mathrm{~s}\). What is the water speed as it leaves the nozzle?

(II) Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk \(0.83 \mathrm{~s}\) after passing his window. Roger's room is on the third floor, \(15 \mathrm{~m}\) above the sidewalk. ( \(a\) ) How fast are the balloons traveling when they pass Roger's window? (b) Assuming the balloons are being released from rest, from what floor are they being released? Each floor of the dorm is \(5.0 \mathrm{~m}\) high.

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between stops. The more stops there are, the slower the train's average speed. To get an idea of this problem, calculate the time it takes a train to make a \(9.0-\mathrm{km}\) trip in two situations: \((a)\) the stations at which the trains must stop are \(1.8 \mathrm{~km}\) apart (a total of 6 stations, including those at the ends); and (b) the stations are \(3.0 \mathrm{~km}\) apart (4 stations total). Assume that at each station the train accelerates at a rate of \(1.1 \mathrm{~m} / \mathrm{s}^{2}\) until it reaches \(95 \mathrm{~km} / \mathrm{h},\) then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at \(-2.0 \mathrm{~m} / \mathrm{s}^{2} .\) Assume it stops at each intermediate station for \(22 \mathrm{~s}\).

(1) Estimate \((a)\) how long it took King Kong to fall straight down from the top of the Empire State Building \((380 \mathrm{m}\) high \(),\) and \((b)\) his velocity just before landing.
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