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Chapter 2: Problem 63

Dan gets on Interstate Highway \(\mathrm{I}-80\) at Seward, Nebraska, and drives due west in a straight line and at an average velocity of magnitude 88 \(\mathrm{km} / \mathrm{h}\) . After traveling 76 \(\mathrm{km}\) , he reaches the Aurora exit (Fig. \(\mathrm{P} 2.63 ) .\) Realizing he has gone too far, he turns around and drives due east 34 \(\mathrm{km}\) back to the York exit at an average velocity of magnitude 72 \(\mathrm{km} / \mathrm{h}\) . For his whole trip from Seward to the York exit, what are (a) his average speed and (b) the magnitude of his average velocity?

Short Answer

Expert verified
Average speed is approximately 82.3 km/h; average velocity is 31.4 km/h.

Step by step solution

01

Calculate Total Distance

First, determine the total distance Dan traveled on his trip. He drove 76 km west, then turned around and drove 34 km back east. Add these distances:\[ \text{Total Distance} = 76 \, \text{km} + 34 \, \text{km} = 110 \, \text{km} \]
02

Determine Total Time

Next, calculate the total time taken for the whole trip. For the westward journey, Dan took \( t_1 \) hours, calculated as:\[ t_1 = \frac{76 \, \text{km}}{88 \, \text{km/h}} = \frac{76}{88} \, \text{hours} \]For the eastward journey, the time \( t_2 \) is:\[ t_2 = \frac{34 \, \text{km}}{72 \, \text{km/h}} = \frac{34}{72} \, \text{hours} \]The total time taken is:\[ \text{Total Time} = t_1 + t_2 = \frac{76}{88} + \frac{34}{72} \]
03

Calculate Average Speed

The average speed is defined as the total distance traveled divided by the total time taken. Use the total distance and total time calculated in previous steps:\[ \text{Average Speed} = \frac{110 \, \text{km}}{ \left( \frac{76}{88} + \frac{34}{72} \right) \, \text{hours} } \]Compute the values:Convert \( t_1 \) and \( t_2 \) into decimal form and add them:- \( t_1 = 0.8636 \) hours- \( t_2 = 0.4722 \) hoursTotal Time = \( 0.8636 + 0.4722 = 1.3358 \) hoursNow, compute the average speed:\[ \text{Average Speed} = \frac{110}{1.3358} \approx 82.3 \, \text{km/h} \]
04

Calculate Average Velocity

Average velocity is defined as the total displacement divided by the total time. The displacement is the straight-line distance from the start point to the endpoint, which is 42 km east, as it is west minus east:\[ \text{Displacement} = 76 \, \text{km} - 34 \, \text{km} = 42 \, \text{km} \]Thus, the average velocity:\[ \text{Average Velocity} = \frac{42 \, \text{km}}{1.3358 \, \text{hours}} \approx 31.4 \, \text{km/h} \]

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Key Concepts

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

Displacement
Displacement is a vector quantity that refers to the change in position of an object. It is important to note that displacement considers the initial and final positions without accounting for the path taken. In the context of Dan's trip, displacement is the direct line from his starting point in Seward to his endpoint at the York exit.
  • Dan drove 76 km west initially and then drove 34 km east.
  • His displacement is calculated by subtracting the distance he drove back east from his initial westward distance.
  • The total displacement is 42 km east, which is calculated as: \[ 76 \, \text{km} - 34 \, \text{km} = 42 \, \text{km} \]
Displacement gives us a concise summary of how far Dan is from his starting point along a straight line.
Average Velocity
Average velocity is another vector quantity that considers both the magnitude and the direction of displacement over a period of time. It is calculated by dividing the total displacement by the total time of travel. Let's review Dan's case:
  • We have already established Dan's displacement is 42 km to the east.
  • The total time of the trip can be calculated by adding the time taken for each leg of the trip:- \(t_1 = \frac{76}{88} \approx 0.8636 \, \text{hours}\)- \(t_2 = \frac{34}{72} \approx 0.4722 \, \text{hours}\)- Total Time = 1.3358 hours
  • Average velocity is then:\[ \text{Average Velocity} = \frac{42 \text{ km}}{1.3358 \text{ h}} \approx 31.4 \text{ km/h} \]
This tells us how fast and in which direction Dan traveled over the entirety of his trip in a straight line manner.
Average Speed
Average speed differs from average velocity in that it is a scalar quantity; it does not have a direction and only considers the total distance traveled. It is essentially how fast something is moving regardless of direction:
  • Calculate Dan's total distance: 76 km + 34 km = 110 km
  • The total time of travel is still 1.3358 hours, as computed earlier.
  • Now, using the formula for average speed:\[ \text{Average Speed} = \frac{110 \text{ km}}{1.3358 \text{ h}} \approx 82.3 \text{ km/h} \]
Average speed tells us the over-all rate of motion including every twist and turn of the journey.
Kinematics
Kinematics is a branch of physics that describes the motion of objects. It looks at variables like displacement, velocity, speed, and acceleration, without considering the forces causing the motion. In Dan's travel scenario, kinematic principles help us understand how different calculations like displacement, average velocity, and average speed work:
  • Kinetic equations allow us to describe the linear motion Dan undergoes without discussing the underlying forces or causes.
  • By focusing on these kinematic variables, we can deeply understand the journey in terms of motion and outcomes.
Thus, kinematics gives you a broader picture of how things move, which is critical in accurately analyzing any physical activity.
Relative Motion
Relative motion describes how the position and velocity of an object are perceived from a given point of reference. It enables understanding of how different observers may see the object's motion differently:
  • In terms of Dan鈥檚 drive, another driver going in the opposite direction could perceive Dan's motion relative to their own.
  • Dan's relative velocity to the eastward-moving vehicles would be different from someone stationary on the highway.
  • Understanding relative motion allows us to interpret how motion variables change based on different perspectives.
In essence, relative motion is crucial for grasping all effects of motion from any point of view.

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Most popular questions from this chapter

Cliff Height. You are climbing in the High Sierra where you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top and 10.0 \(\mathrm{s}\) later hear the sound of it hitting the ground at the foot of the cliff. (a) Ignoring air resistance, how high is the cliff if the speed of sound is 330 \(\mathrm{m} / \mathrm{s} ?\) (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain your reasoning.

A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 \(\mathrm{m}\) above the ground. After an additional \(4.75 \mathrm{s},\) it is 1.00 \(\mathrm{km}\) above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75 -s part of its flight and (b) the first 5.90 s of its flight.

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by \(a_{y}=\left(2.80 \mathrm{m} / \mathrm{s}^{3}\right) t,\) where acceleration of the rocket is given by \(a_{y}=\left(2.80 \mathrm{m} / \mathrm{s}^{3}\right) t,\) where the \(+y\) -direction is upward. (a) What is the height of the rocket above the surface of the earth at \(t=10.0 \mathrm{s} ?\) (b) What is the speed of the rocket when it is 325 \(\mathrm{m}\) above the surface of the earth?

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