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Chapter 3: Problem 33

A bicycle travels \(3.2 \mathrm{km}\) due east in \(0.10 \mathrm{h}\), then $4.8 \mathrm{km}\( at \)15.0^{\circ}\( east of north in \)0.15 \mathrm{h},$ and finally another \(3.2 \mathrm{km}\) due east in \(0.10 \mathrm{h}\) to reach its destination. The time lost in turning is negligible. What is the average velocity for the entire trip?

Short Answer

Expert verified
Answer: The average velocity for the entire trip is approximately 31.863 km/h.

Step by step solution

01

Calculate the displacement for each part of the trip

To calculate the displacement of each part of the trip, we need to find the horizontal (x) and vertical (y) components of the displacement for each part, and then add them up. For the parts where the bicycle is moving due east, the entire displacement is in the x-direction. When moving at an angle, we can use trigonometry (sine and cosine functions) to find the x and y components of the displacement.
02

Calculate horizontal and vertical displacement components for the second part of the trip

For the second part of the trip, the bicycle is traveling 4.8 km at an angle of 15 degrees east of north. To find the x and y components of this displacement, we can use the sine and cosine functions: x = 4.8 * cos(15) = 4.8 * 0.9659 = 4.6365 km y = 4.8 * sin(15) = 4.8 * 0.2588 = 1.2422 km
03

Calculate the total displacement in the x and y directions

To find the total displacement, we can simply add up the displacements in the x and y directions for each part of the trip: Total x displacement = 3.2 + 4.6365 + 3.2 = 11.0365 km Total y displacement = 0 + 1.2422 + 0 = 1.2422 km
04

Calculate the magnitude of the total displacement

To find the magnitude of the total displacement, we can use the Pythagorean theorem: Total displacement = sqrt((total x displacement)^2 + (total y displacement)^2) Total displacement = sqrt((11.0365)^2 + (1.2422)^2) = sqrt(122.8035 + 1.5435) = sqrt(124.3471) = 11.152 km
05

Calculate the total time taken for the trip

The total time taken for the trip is the sum of the times taken for each part of the trip: Total time = 0.10 h + 0.15 h + 0.10 h = 0.35 h
06

Calculate the average velocity for the entire trip

To find the average velocity, we can divide the total displacement by the total time taken: Average velocity = Total displacement / Total time = 11.152 km / 0.35 h = 31.863 km/h The average velocity for the entire trip is approximately 31.863 km/h.

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Most popular questions from this chapter

A sailboat sails from Marblehead Harbor directly east for 45 nautical miles, then \(60^{\circ}\) south of east for 20.0 nautical miles, returns to an easterly heading for 30.0 nautical miles, and sails \(30^{\circ}\) east of north for 10.0 nautical miles, then west for 62 nautical miles. At that time the boat becomes becalmed and the auxiliary engine fails to start. The crew decides to notify the Coast Guard of their position. Using graph paper, ruler, and protractor, sketch a graphical addition of the displacement vectors and estimate their position.
John drives \(16 \mathrm{km}\) directly west from Orion to Chester at a speed of \(90 \mathrm{km} / \mathrm{h},\) then directly south for \(8.0 \mathrm{km}\) to Seiling at a speed of \(80 \mathrm{km} / \mathrm{h}\), then finally $34 \mathrm{km}\( southeast to Oakwood at a speed of \)100 \mathrm{km} / \mathrm{h}$. Assume he travels at constant velocity during each of the three segments. (a) What was the change in velocity during this trip? [Hint: Do not assume he starts from rest and stops at the end.] (b) What was the average acceleration during this trip?
Sheena can row a boat at \(3.00 \mathrm{mi} / \mathrm{h}\) in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at $1.60 \mathrm{mi} / \mathrm{h} .$ Not having her calculator ready, she guesses that to go straight across, she should head \(60.0^{\circ}\) upstream. (a) What is her speed with respect to the starting point on the bank? (b) How long does it take her to cross the river? (c) How far upstream or downstream from her starting point will she reach the opposite bank? (d) In order to go straight across, what angle upstream should she have headed?
Displacement vector \(\overrightarrow{\mathbf{A}}\) is directed to the west and has magnitude \(2.56 \mathrm{km} .\) A second displacement vector is also directed to the west and has magnitude \(7.44 \mathrm{km}\) (a) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} ?\) (b) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}} ?\) (c) What are the magnitude and direction of \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}} ?\)
A gull is flying horizontally \(8.00 \mathrm{m}\) above the ground at $6.00 \mathrm{m} / \mathrm{s} .$ The bird is carrying a clam in its beak and plans to crack the clamshell by dropping it on some rocks below. Ignoring air resistance, (a) what is the horizontal distance to the rocks at the moment that the gull should let go of the clam? (b) With what speed relative to the rocks does the clam smash into the rocks? (c) With what speed relative to the gull does the clam smash into the rocks?
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