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Chapter 2: Problem 3

A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for \(30.0 \mathrm{~min}\) at \(80.0 \mathrm{~km} / \mathrm{h}, 12.0 \mathrm{~min}\) at \(100 \mathrm{~km} / \mathrm{h}\), and \(45.0 \mathrm{~min}\) at \(40.0 \mathrm{~km} / \mathrm{h}\) and spends \(15.0 \mathrm{~min}\) eating lunch and buying gas. (a) Determine the average speed for the trip. (b) Determine the distance between the initial and final cities along the route.

Short Answer

Expert verified
The average speed for the trip (part a) is calculated by dividing the total distance of the trip by the total time of the trip (including breaks), and the total distance covered in the trip (part b) is calculated by adding up the distances covered in each part of the trip.

Step by step solution

01

Convert Minutes to Hours

Firstly, convert the time from minutes to hours since the speed is in km/hr. Divide the minutes by 60: Time1 = 30/60 hr, Time2 = 12/60 hr, Time3 = 45/60 hr, TimeLunch = 15/60 hr.
02

Calculate the Distances for Each Segment

Next, calculate the distance covered in each segment using the formula Distance = Speed x Time: Distance1 = Speed1 x Time1, Distance2 = Speed2 x Time2, Distance3 = Speed3 x Time3.
03

Calculate Total Distance and Total Time

Add the distances to get the total distance from starting to the end point. Also add the times spent in each segment plus the lunch break to get the total time spent: Total Distance = Distance1 + Distance2 + Distance3, Total Time = Time1 + Time2 + Time3 + TimeLunch.
04

Calculate Average Speed

Divide the total distance by the total time to find the average speed: Average Speed = Total Distance / Total Time.
05

Determine Average Speed and Distance Between the Cities

Lastly, write down your final answers for the average speed and total distance.

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Key Concepts

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

Constant Speed Travel
When we talk about constant speed travel, it means the person is driving at the same speed over a certain period of time. Unlike a variable speed, where the speed changes over time, constant speed is uniform and predictable. This makes calculations simpler as the formula to find the distance remains straightforward.
During the trip, each segment between cities was driven at a different constant speed which allows for individual distance calculations using the formula:
  • Distance = Speed 脳 Time
Hence, knowing the speed and the time allows us to easily find how far the person traveled during each segment. Understanding this concept helps in breaking down the problem into manageable pieces.
Distance Calculation
In the context of physics problems, especially those involving speed and time, distance calculation is an essential skill. Here, we utilized the formula:
  • Distance = Speed 脳 Time
From the problem, there are three segments, each with a unique constant speed and measured time in minutes. The first step is translating these minute values into hours to match the speed units given in kilometers per hour (km/h).
Once we have the time in hours, we multiply it by the speed of that segment to find out the distance. For example, for the first segment with a speed of 80 km/h and a time of 0.5 hours, the distance would be:
  • Distance1 = 80 脳 0.5 = 40 km
Adding all segment distances together gives the total journey distance.
Time Conversion
Time conversion is crucial when working with speed and distance, as it ensures uniform units across calculations. In most physics problems like this, speed is expressed in kilometers per hour (km/h), so time must be converted to hours too.
We often start with time given in minutes, as in this problem. To convert minutes to hours, simply divide the number of minutes by 60:
  • For 30 minutes, it becomes \( \frac{30}{60} = 0.5 \) hours.
  • 12 minutes becomes \( \frac{12}{60} = 0.2 \) hours.
  • And 45 minutes converts to \( \frac{45}{60} = 0.75 \) hours.
Considering these conversions is necessary to perform accurate distance calculations and subsequently determine the total travel time, including the lunch break.
Physics Problem Solving
Tackling physics problems involves systematic thinking and understanding of relationships between variables, such as speed, distance, and time. In this problem, we followed structured steps to deduce average speed鈥攁 central concept in physics involving motion.
The average speed is computed by dividing the total distance traveled by the total time taken, including all activities like driving and lunch breaks. It's a summary of overall trip efficiency.
  • Total Distance: Add up all individual segment distances.
  • Total Time: Sum up all driving times plus any additional breaks, converted into hours.
  • Average Speed: \( \frac{\text{Total Distance}}{\text{Total Time}} \)
Simplifying complex information into such a coherent set of steps aids in tackling most motion-related physics problems effectively.

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Most popular questions from this chapter

A speedboat moving at \(30.0 \mathrm{~m} / \mathrm{s}\) approaches a no-wake buoy marker \(100 \mathrm{~m}\) ahead. The pilot slows the boat with a constant acceleration of \(-3.50 \mathrm{~m} / \mathrm{s}^{2}\) by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy?

A ball is thrown upward from the ground with an initial speed of \(25 \mathrm{~m} / \mathrm{s}\); at the same instant, another ball is dropped from a building \(15 \mathrm{~m}\) high. After how long will the balls be at the same height?

A bullet is fired through a board \(10.0 \mathrm{~cm}\) thick in such a way that the bullet's line of motion is perpendicular to the face of the board. If the initial speed of the bullet is \(400 \mathrm{~m} / \mathrm{s}\) and it emerges from the other side of the board with a speed of \(300 \mathrm{~m} / \mathrm{s}\), find (a) the acceleration of the bullet as it passes through the board and (b) the total time the bullet is in contact with the board.

An object moves with constant acceleration \(4.00 \mathrm{~m} / \mathrm{s}^{2}\) and over a time interval reaches a final velocity of \(12.0 \mathrm{~m} / \mathrm{s}\). (a) If its original velocity is \(6.00 \mathrm{~m} / \mathrm{s}\), what is its displacement during the time interval? (b) What is the distance it travels during this interval? (c) If its original velocity is \(-6.00 \mathrm{~m} / \mathrm{s}\), what is its displacement during this interval? (d) What is the total distance it travels during the interval in part (c)?

A person sees a lightning bolt pass close to an airplane that is flying in the distance. The person hears thunder \(5.0 \mathrm{~s}\) after seeing the bolt and sees the airplane overhead \(10 \mathrm{~s}\) after hearing the thunder. The speed of sound in air is \(1100 \mathrm{ft} / \mathrm{s}\). (a) Find the distance of the airplane from the person at the instant of the bolt. (Neglect the time it takes the light to travel from the bolt to the eye.) (b) Assuming the plane travels with a constant speed toward the person, find the velocity of the airplane. (c) Look up the speed of light in air and defend the approximation used in part (a).
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