/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A hospital nurse walks \(25 \mat... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 6

A hospital nurse walks \(25 \mathrm{~m}\) to a patient's room at the end of the hall in 0.50 min. She talks with the patient for 4.0 min, and then walks back to the nursing station at the same rate she came. What was the nurse's average speed?

Short Answer

Expert verified
The nurse's average speed was 0.167 m/s.

Step by step solution

01

Understanding the Problem

The problem involves calculating the average speed of the nurse over a series of actions: walking to a patient's room, staying for a conversation, and walking back. We need to consider both the walking and stationary times.
02

Calculate Total Distance

The nurse walks to the room covering a distance of 25 m and then back, covering an additional 25 m. The total distance she travels is thus:\[25 \text{ m} + 25 \text{ m} = 50 \text{ m}\]
03

Calculate Total Time

The nurse spends 0.50 min walking to the room and 0.50 min walking back. She talks for 4.0 min with the patient. Total time is:\[0.50 \text{ min} + 4.0 \text{ min} + 0.50 \text{ min} = 5.0 \text{ min}\]Converting this into seconds (since speed is typically in m/s):\[5.0 \text{ min} \times 60 = 300 \text{ s}\]
04

Compute Average Speed

Average speed is computed by dividing the total distance by the total time:\[\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{50 \text{ m}}{300 \text{ s}} = \frac{1}{6} \text{ m/s}\]This simplifies to approximately 0.167 m/s.

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Key Concepts

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

Distance and Time
Understanding distance and time is essential when discussing motion and speed. Distance refers to the total path length covered during motion. In this exercise, the nurse walks to a patient鈥檚 room, covering a distance of 25 meters, and then back the same distance to the starting point, resulting in a total distance of 50 meters. Similarly, time is the duration during which the motion occurs. Here, the time taken includes both the walking time and the time spent stationary while talking to the patient, totaling 5.0 minutes. When calculating average speed, you always need both the total distance traveled and the total time taken.
Unit Conversion
Unit conversion is an important skill in physics, especially when calculating speed, which is usually expressed in meters per second (m/s). In the given problem, the time is initially provided in minutes: 0.50 minutes for each leg of the journey and an additional 4.0 minutes talking, resulting in 5.0 minutes total. To convert this to seconds, since 1 minute equals 60 seconds, you multiply 5.0 by 60. This gives us 300 seconds. By ensuring that both distance and time have compatible units鈥攈ere, meters and seconds鈥攜ou can accurately compute the speed using the formula: speed = distance/time.
Motion Analysis
Motion analysis involves breaking down various movements to assess speed and other kinematic quantities. In the nurse's scenario, her motion consists of walking, standing still, and walking back. By evaluating each segment of motion separately, you can determine the overall average speed. Average speed is obtained by dividing the total distance by the total time spent moving and stationary:
  • Total Distance: 50 meters (25 meters each way)
  • Total Time: 300 seconds
Thus, the average speed calculation becomes: \( \frac{50 \text{ m}}{300 \text{ s}} \). The final result is approximately 0.167 m/s. This means that on average, the nurse covers about one-sixth of a meter every second during her entire trip, including the time spent talking.

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Most popular questions from this chapter

An object initially at rest experiences an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) on a level surface. Under these conditions, it travels \(6.00 \mathrm{~m}\). Let's designate the first \(3.00 \mathrm{~m}\) as phase 1 with a subscript of 1 for those quantities, and the second \(3.00 \mathrm{~m}\) as phase 2 with a subscript of \(2 .\) (a) The times for traveling each phase should be related by which condition: (3) \(t_{1}>t_{2} ?\) (b) Now calculate the (1) \(t_{1}

A car initially at rest experiences loss of its parking brake and rolls down a straight hill with a constant acceleration of \(0.850 \mathrm{~m} / \mathrm{s}^{2},\) traveling a total of \(100 \mathrm{~m}\) Let's designate the first half of the distance as phase 1 with a subscript of 1 for those quantities, and the second half as phase 2 with a subscript of \(2 .\) (a) The car's speeds at the end of each phase should be related by which condition (1) \(v_{1}<\frac{1}{2} v_{2},\) (2) \(v_{1}=\frac{1}{2} v_{2},\) or (3) \(v_{1}>\frac{1}{2} v_{2} ?\) (b) Now calculate the two speeds and compare them quantitatively.

You throw a stone vertically upward with an initial speed of \(6.0 \mathrm{~m} / \mathrm{s}\) from a third-story office window. If the window is \(12 \mathrm{~m}\) above the ground, find (a) the time the stone is in flight and (b) the speed of the stone just before it hits the ground.

A car traveling at a speed of \(v\) can brake to an emergency stop in a distance \(x\). Assuming all other driving conditions are similar, if the traveling speed of the car doubles, the stopping distance will be (1) \(\sqrt{2 x,}\) (2) \(2 x\) (3) \(4 x\). (b) A driver traveling at \(40.0 \mathrm{~km} / \mathrm{h}\) in a school zone can brake to an emergency stop in \(3.00 \mathrm{~m}\). What would be the braking distance if the car were traveling at \(60.0 \mathrm{~km} / \mathrm{h} ?\)

You are driving slowly in the right lane of a straight country road. For a while, a car to your left has lagged \(50.0 \mathrm{~m}\) behind you at the same speed of \(25.0 \mathrm{mi} / \mathrm{h}\) Suddenly that car speeds up and passes you, traveling at a constant acceleration until it is \(40.0 \mathrm{~m}\) in front of you 7.00 s later. (a) Qualitatively sketch the location-versustime graphs for both cars on the same axes, letting \(t=0\) be the start of the acceleration, and \(x=0\) be the location of the other car at that time. (b) Determine the other car's acceleration. (c) How far did each of you travel during the passing procedure? (d) What is the other car's speed at the end of the passing procedure?
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