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Chapter 2: Problem 12

A car makes a trip due north for three-fourths of the time and due south one- fourth of the time. The average northward velocity has a magnitude of \(27 \mathrm{m} / \mathrm{s},\) and the average southward velocity has a magnitude of \(17 \mathrm{m} / \mathrm{s}\). What is the average velocity (magnitude and direction) for the entire trip?

Short Answer

Expert verified
The average velocity is 16 m/s north.

Step by step solution

01

Understand the Problem

The problem involves a car traveling in two different directions with different velocities for different amounts of time. We need to determine the average velocity for the entire trip, which considers both the magnitude and direction.
02

Define Total Time

Let the total time for the trip be \(T\). The car travels north for three-fourths of this time, which is \(\frac{3}{4}T\), and south for one-fourth of the time, which is \(\frac{1}{4}T\).
03

Calculate Distance Traveled North and South

The distance traveled north is velocity times time: \(d_{north} = 27\,\text{m/s} \times \frac{3}{4}T\). The distance traveled south is \(d_{south} = 17\,\text{m/s} \times \frac{1}{4}T\).
04

Determine Net Displacement

Net displacement is total distance traveled north minus total distance traveled south: \(D = d_{north} - d_{south} = 27\cdot\frac{3}{4}T - 17\cdot\frac{1}{4}T\).
05

Simplify the Expression for Displacement

Substituting the values, we get: \[D = \left(27 \cdot \frac{3}{4} \right)T - \left(17 \cdot \frac{1}{4} \right)T = \left(20.25\,T - 4.25\,T\right) = 16T\].
06

Calculate Average Velocity

Average velocity is net displacement divided by total time: \(v_{avg} = \frac{D}{T} = \frac{16T}{T} = 16\,\text{m/s}\). The positive result indicates the direction is north.
07

Conclude with the Magnitude and Direction

The magnitude of the average velocity is \(16\,\text{m/s}\) and the direction is north.

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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 how far an object has moved from its initial position in a specific direction. It is different from distance, which is only concerned with how much ground is covered, regardless of direction. For example, if you're walking, the the path you take forms the distance, but displacement only considers your start and end points.
In the context of a car trip, displacement is crucial because it allows us to determine the car's change in position, not just the distance it traveled. For instance, if a car drives north and then south, its net displacement is the length of the journey north minus the length of the journey south.
  • Net displacement can help calculate average velocity.
  • Displacement has both magnitude (how far) and direction (which way).
Velocity
Velocity is another vector quantity, like displacement, but it specifically refers to how fast an object is moving and in what direction. Unlike speed, which is a scalar and only measures how fast something moves, velocity provides information on direction too.
For instance, a car could be traveling at a velocity of 50 meters per second north. Here, the north indicates the direction, and 50 meters per second represents the magnitude of the velocity.
  • Velocity can be affected by changes in speed or direction.
  • Understanding velocity helps determine how quickly displacement is achieved.
Direction
The direction is a crucial aspect of any vector quantity, including displacement and velocity. It helps define the path an object follows as it moves from one place to another. In our example of the car traveling north and south, direction allows us to correctly assess the car's overall movement.
Direction becomes particularly important when calculating net results, like net displacement or overall velocity. When movements in opposite directions occur, such as north and south, directional indicators help determine net outcomes.
  • Direction is essential for correctly calculating vectors like displacement and velocity.
  • It helps differentiate movements in a multidirectional space.
Net Displacement
Net displacement combines multiple displacement movements into a single vector quantity that reflects the total change in an object's position considering every move it made. In the car's journey, net displacement occurs by weighing how far it traveled north against how far it traveled south.
Mathematically, net displacement can be calculated by subtracting the distance moved in the opposite direction from the total distance in the primary direction. In our example:
  • The northward displacement is 27 m/s for three-quarters of the time.
  • The southward displacement is 17 m/s for the remaining quarter.
  • Net displacement, therefore, becomes the result of sorting these opposing vectors.
Overall, net displacement is a key concept in finding average velocity, since it shows the actual change in position.

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Most popular questions from this chapter

The Kentucky Derby is held at the Churchill Downs track in Louisville, Kentucky. The track is one and one-quarter miles in length. One of the most famous horses to win this event was Secretariat. In 1973 he set a Derby record that would be hard to beat. His average acceleration during the last four quarter-miles of the race was \(+0.0105 \mathrm{m} / \mathrm{s}^{2}\). His velocity at the start of the final mile \((x=+1609 \mathrm{m})\) was about \(+16.58 \mathrm{m} / \mathrm{s}\). The acceleration, although small, was very important to his victory. To assess its effect, determine the difference between the time he would have taken to run the final mile at a constant velocity of \(+16.58 \mathrm{m} / \mathrm{s}\) and the time he actually took. Although the track is oval in shape, assume it is straight for the purpose of this problem.

A diver springs upward with an initial speed of \(1.8 \mathrm{m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. / Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{m}\) (measured from the board), assuming that the dowmward direction is chosen as the negative direction./ (b) What is the highest point he reaches above the water?

At the beginning of a basketball game, a referee tosses the ball straight up with a speed of \(4.6 \mathrm{m} / \mathrm{s} .\) A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball?

A jogger accelerates from rest to \(3.0 \mathrm{m} / \mathrm{s}\) in \(2.0 \mathrm{s}\). A car accelerates from 38.0 to \(41.0 \mathrm{m} / \mathrm{s}\) also in \(2.0 \mathrm{s}\). (a) Find the acceleration (magnitude only of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the \(2.0 \mathrm{s} ?\) If so, how much farther?

The greatest height reported for a jump into an airbag is \(99.4 \mathrm{m}\) by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of \(39 \mathrm{m} / \mathrm{s}(88 \mathrm{mi} / \mathrm{h}) .\) To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.
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