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Chapter 2: Problem 57

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. 2.44\()\) . 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 82.33 km/h; average velocity magnitude is 31.44 km/h.

Step by step solution

01

Calculate the Total Distance Traveled

Dan first drives 76 km west and then 34 km east. Therefore, the total distance traveled is the sum of these two distances: \[ 76 + 34 = 110 \, \text{km} \]
02

Calculate the Total Time for the Trip

To find the time for each segment of the trip, use the formula \[ \text{time} = \frac{\text{distance}}{\text{velocity}} \]For the westbound trip:\[ \text{time}_{west} = \frac{76 \, \text{km}}{88 \, \text{km/h}} = 0.864 \, \text{hours} \]For the eastbound trip:\[ \text{time}_{east} = \frac{34 \, \text{km}}{72 \, \text{km/h}} = 0.472 \, \text{hours} \]Adding these gives the total time for the trip:\[ 0.864 + 0.472 = 1.336 \, \text{hours} \]
03

Calculate the Average Speed

Average speed is calculated as the total distance traveled divided by the total time of the trip:\[ \text{Average Speed} = \frac{110 \, \text{km}}{1.336 \, \text{hours}} = 82.33 \, \text{km/h} \]
04

Determine the Displacement

Displacement is the straight-line distance from the starting point to the ending point along with the direction. Since Dan starts at Seward and ends at York, his displacement is:\[ 76 \, \text{km} - 34 \, \text{km} = 42 \, \text{km} \, \text{(west)} \]
05

Calculate the Magnitude of Average Velocity

The magnitude of average velocity is determined by dividing the total displacement by the total time of the trip:\[ \text{Average Velocity} = \frac{42 \, \text{km}}{1.336 \, \text{hours}} = 31.44 \, \text{km/h} \]

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Key Concepts

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

Displacement
When we talk about displacement, we mean the straight-line distance from the starting point to the ending point, taking into account the direction. Unlike total distance, which considers the entire path traveled, displacement focuses on the shortcut, the most direct path. In the exercise, Dan starts at Seward, drives to the Aurora exit, and then to the York exit. So, the displacement considers only the distance between Seward and York, ignoring the detour to Aurora.

To find Dan's displacement, we subtract the distance he drove back east from the distance he drove west. His path west was 76 km and then he drove back 34 km east. Thus the displacement is:
  • Displacement = 76 km (west) - 34 km (east)
  • This results in a displacement of 42 km west.
Displacement is always a vector quantity, which means it has both magnitude and direction (in this case, west). Remember, while total distance traveled was 110 km, the displacement was only 42 km west.
Average Speed
Average speed helps us understand how fast someone travels on average over a trip. It's calculated by dividing the total distance traveled by the total time taken. In Dan's trip, he covered a total distance of 110 km, which includes both his westward and eastward journeys.

Let's lay out the calculation for average speed:
  • Total Distance Traveled = 110 km (76 km west + 34 km east)
  • Total Time Taken = 1.336 hours (0.864 hours west + 0.472 hours east)
  • Average Speed = Total Distance / Total Time = 110 km / 1.336 hours = 82.33 km/h
Note that average speed doesn't take into account the direction of travel 鈥� it's all about how much ground was covered during the trip, irrespective of the path taken.
Velocity and Speed Calculations
Calculating velocity and speed are fundamental skills in physics. While average speed tells us how fast someone went overall, average velocity takes into consideration both the speed and the direction of travel.

For Dan's journey:
  • Average Speed = 82.33 km/h (includes both directions)
  • Average Velocity focuses on displacement direction, i.e., west.
Average velocity depends on the total displacement, not the total distance. Dan's displacement is 42 km (west), and his total time is 1.336 hours.

This leads to the average velocity calculation:
  • Average Velocity = Displacement / Total Time = 42 km (west) / 1.336 hours = 31.44 km/h (west)
Average velocity is a vector quantity, which means it must have a direction, unlike average speed, which is scalar. Recognizing these differences helps in truly understanding motion dynamics.

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Most popular questions from this chapter

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 \(\mathrm{s}\) , she hears the echo of her shout from the valley floor below. The speed of sound is 340 \(\mathrm{m} / \mathrm{s}\) . (a) How tall is the cliff? \((b)\) If air resistance is neglected, bow fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)

An astronaut has left the Intemational Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a 10 -s interval. What are the magniude, the algebraic sign, and the direction of the average acceleration in each interval? Assume that the positive direction is to the right. (a) At the beginning of the interval the astronaut is moving toward the right along the \(x\) -axis at \(15.0 \mathrm{m} / \mathrm{s},\) and at the end of the interval she is moving toward the right at 5.0 \(\mathrm{m} / \mathrm{s}\) . (b) At the beginning she is moving toward the left at \(5.0 \mathrm{m} / \mathrm{s},\) and at the end she is moving toward the left at 15.0 \(\mathrm{m} / \mathrm{s}\) . (c) At the beginning she is moving toward the right at \(15.0 \mathrm{m} / \mathrm{s},\) and at the end she is moving toward the left at 15.0 \(\mathrm{m} / \mathrm{s}\) .

The acceleration of a bus is given by \(a_{x}(t)=\alpha t,\) where \(\alpha=1.2 \mathrm{m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{s}\) is 5.0 \(\mathrm{m} / \mathrm{s}\) , what is its velocity at time \(t=2.0 \mathrm{s} ?(\mathrm{b})\) If the bus's position at time \(t=1.0 \mathrm{s}\) is \(6.0 \mathrm{m},\) what is its position at time \(t=2.0 \mathrm{s} ?\) (c) Sketch \(a_{x}-t, v_{x}-t,\) and \(x-\) graphs for the motion.

The catapult of the aireraft carrier USS Abraham Lincoln accelerates an FIA-18 Homet jet fighter from rest to a takeoff speed of 173 \(\mathrm{mi} / \mathrm{h}\) in a distance of 307 \(\mathrm{ft}\) . Assume constant acceleration. (a) Calculate the acceleration of the fighter in \(\mathrm{m} / \mathrm{s}^{2} .(\mathrm{b})\) Calculate the time required for the fighter to accelerate to takeoff speed.

The position of the front bumper of a test car under micro-processor control is given by \(x(t)=2.17 \mathrm{m}+\left(4.80 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}-\) \(\left(0.100 \mathrm{m} / \mathrm{s}^{6}\right) t^{6} .\) (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw \(x-t, v_{x}-t,\) and \(a_{x}-t\) graphs for the motion of the bumper between \(t=0\) and \(t=2.00 \mathrm{s}\)
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