/*! 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 25 An auto travels at the rate of \... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 25

An auto travels at the rate of \(25 \mathrm{~km} / \mathrm{h}\) for \(4.0\) minutes, then at 50 \(\mathrm{km} / \mathrm{h}\) for \(8.0\) minutes, and finally at \(20 \mathrm{~km} / \mathrm{h}\) for \(2.0\) minutes. Find (a) the total distance covered in \(\mathrm{km}\) and \((b)\) the average speed for the complete trip in \(\mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
Total distance: 9 km, Average speed: 10.71 m/s

Step by step solution

01

Convert Time to Hours

Since the speed is in km/h, we first need to convert the time from minutes to hours. For the first interval: \(4 \text{ min} = \frac{4}{60} \text{ hours}\). For the second interval: \(8 \text{ min} = \frac{8}{60} \text{ hours}\). For the third interval: \(2 \text{ min} = \frac{2}{60} \text{ hours}\).
02

Calculate Distance for Each Interval

Next, calculate the distance for each segment of the trip using the formula \(\text{distance} = \text{speed} \times \text{time}\). - First interval: \(25 \text{ km/h} \times \frac{4}{60} \text{ h} = \frac{25 \times 4}{60} \text{ km} = \frac{100}{60} \text{ km} = \frac{5}{3} \text{ km}\).- Second interval: \(50 \text{ km/h} \times \frac{8}{60} \text{ h} = \frac{50 \times 8}{60} \text{ km} = \frac{400}{60} \text{ km} = \frac{20}{3} \text{ km}\).- Third interval: \(20 \text{ km/h} \times \frac{2}{60} \text{ h} = \frac{20 \times 2}{60} \text{ km} = \frac{40}{60} \text{ km} = \frac{2}{3} \text{ km}\).
03

Total Distance Covered

Add up the distances from each interval to find the total distance:\(\text{Total distance} = \frac{5}{3} \text{ km} + \frac{20}{3} \text{ km} + \frac{2}{3} \text{ km} = \frac{27}{3} \text{ km} = 9 \text{ km}\).
04

Calculate Total Time in Seconds

Convert the total time of the trip from minutes to seconds to find the average speed in m/s. Total time: \((4 + 8 + 2) \text{ min} = 14 \text{ min}\) which is \(14 \times 60 = 840 \text{ seconds}\).
05

Calculate Average Speed in \(m/s\)

The average speed can be found using the formula: \(\text{Average speed} = \frac{\text{total distance in meters}}{\text{total time in seconds}}\).First, convert the total distance from kilometers to meters: \(9 \text{ km} = 9000 \text{ m} \).Then, calculate the average speed: \(\text{Average speed} = \frac{9000 \text{ m}}{840 \text{ s}} \approx 10.71 \text{ m/s} \).

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Key Concepts

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

Unit Conversion
One of the first important steps in solving problems related to physics, especially those involving speed and distance, is unit conversion. In this exercise, we started by converting time from minutes to hours to match the speed units given in kilometers per hour (km/h). This is crucial because when you multiply speed by time to calculate distance, both need to be in compatible units.
  • For the conversion from minutes to hours, use the formula: \( \text{Time in hours} = \frac{\text{Time in minutes}}{60} \). So, 4 minutes becomes \( \frac{4}{60} \) hours.
  • Similarly, for 8 minutes, it converts to \( \frac{8}{60} \) hours, and 2 minutes converts to \( \frac{2}{60} \) hours.
Performing these conversions accurately ensures that subsequent calculations for distance remain consistent and correct.
Distance Calculation
Calculating distances involves applying the formula \( \text{Distance} = \text{Speed} \times \text{Time} \). After converting the time to hours, we can proceed to calculate the distance covered in each segment of the journey.
  • In the first interval, the speed was 25 km/h and the time was \( \frac{4}{60} \) hours, resulting in a distance of \( \frac{100}{60} \) km or \( \frac{5}{3} \) km.
  • In the second interval, using 50 km/h and \( \frac{8}{60} \) hours, the distance is \( \frac{400}{60} \) km or \( \frac{20}{3} \) km.
  • The third interval speed was 20 km/h, with time \( \frac{2}{60} \) hours, leading to \( \frac{40}{60} \) km or \( \frac{2}{3} \) km.
Adding these intervals gives you the total distance covered along the journey.
Physics Problems
Physics problems often require blending concepts and computational skills to find solutions. In this problem, we applied several techniques to determine both the total distance and average speed over a trip with varied speeds.
  • Understanding the formula for distance and rearranging it for speed and time if necessary is crucial.
  • Converting units carefully ensures that calculations are accurate and meaningful.
  • Summing the distances from each interval gives the total distance traveled, which is a typical step in solving similar physics problems.
By breaking down these steps and focusing on one calculation at a time, complex problems become understandable, manageable, and solvable.
Kinematics Concepts
Kinematics involves describing the motion of objects using fundamental concepts such as distance, speed, and time. Understanding average speed within this context requires knowing how total distance and total time interplay.
  • For average speed, the total distance must first be in the same unit before dividing by the total time. Given total distance as 9 km or 9000 meters and total time as 840 seconds, using the average speed formula: \( \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} \).
  • This results in roughly 10.71 m/s.
Grasping these fundamental ideas allows one to analyze more complex motion scenarios, predicting how an object moves over various terrains or conditions by applying these kinematics concepts.

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Most popular questions from this chapter

A kid stands \(6.00 \mathrm{~m}\) from the base of a flagpole which is \(8.00 \mathrm{~m}\) tall. Determine the magnitude of the displacement of the brass eagle on top of the pole with respect to the youngster's feet. The geometry corresponds to a 3-4-5 right triangle (i.e., \(3 \times 2-4\) \(\times 2-5 \times 2\) ). Thus, the hypotenuse, which is the 5 -side, must be \(10.0 \mathrm{~m}\) long, and that's the magnitude of the displacement.

A 12-mg housefly has a maximum speed of \(4.5\) mph; what is that in \(\mathrm{m} / \mathrm{s}\) ? [ Hint: 2 significant figures. \(1 \mathrm{mph}=0.44707 \mathrm{~m} / \mathrm{s}\).] \(\mathbf{1 . 2 1}\) [I] According to its computer, a robot that left its closet and traveled \(1200 \mathrm{~m}\), had an average speed of \(20.0 \mathrm{~m} / \mathrm{s}\). How long did the trip take?

A boat can travel at a speed of \(8 \mathrm{~km} / \mathrm{h}\) in still water on a lake. In the flowing water of a stream, it can move at \(8 \mathrm{~km} / \mathrm{h}\) relative to the water in the stream. If the stream speed is \(3 \mathrm{~km} / \mathrm{h}\), how fast can the boat move past a tree on the shore when it is traveling \((a)\) upstream and \((b)\) downstream? (a) If the water was standing still, the boat's speed past the tree would be \(8 \mathrm{~km} / \mathrm{h}\). But the stream is carrying it in the opposite direction at \(3 \mathrm{~km} / \mathrm{h}\). Therefore, the boat's speed relative to the tree is \(8 \mathrm{~km} / \mathrm{h}-3 \mathrm{~km} / \mathrm{h}=5 \mathrm{~km} / \mathrm{h}\). (b) In this case, the stream is carrying the boat in the same direction the boat is trying to move. Hence, its speed past the tree is \(8 \mathrm{~km} / \mathrm{h}+3 \mathrm{~km} / \mathrm{h}=11 \mathrm{~km} / \mathrm{h}\).

Determine the displacement vector that must be added to the displacement \((25 \hat{\mathrm{i}}-16 \hat{j})\) \(\mathrm{m}\) to give a displacement of \(7.0 \mathrm{~m}\) pointing in the \(+x\) -direction?

Find the \(x\) - and \(y\) -components of a 25.0-m displacement at an angle of \(210.0^{\circ}\) The vector displacement and its components are depicted in Fig. 1-11. The scalar components are $$ \begin{array}{l} x \text { -component }=-(25.0 \mathrm{~m}) \cos 30.0^{\circ}=-21.7 \mathrm{~m} \\\ y \text { -component }=-(25.0 \mathrm{~m}) \sin 30.0^{\circ}=-12.5 \mathrm{~m} \end{array} $$ Notice in particular that each component points in the negative coordinate direction and must therefore be taken as negative.
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