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Chapter 2: Problem 5

A person walks first at a constant speed of \(5.00 \mathrm{m} / \mathrm{s}\) along a straight line from point \(A\) to point \(B\) and then back along the line from \(B\) to \(A\) at a constant speed of \(3.00 \mathrm{m} / \mathrm{s}\) What is (a) her average speed over the entire trip? (b) her average velocity over the entire trip?

Short Answer

Expert verified
The average speed for the entire trip will be derived from the total distance and time, while the average velocity will be zero, since the start and end points are the same, leaving a total displacement of zero.

Step by step solution

01

Determine Total Distance and Total Time

The total distance for the whole trip is twice the distance from A to B. Let's denote this distance as \(d\). The total time can be calculated as the sum of the time taken to travel from A to B and from B to A, which in turn is the distance divided by speed for each part of the trip.
02

Calculate Average Speed

The average speed \(V_{av}\) is the total distance divided by total time. Plug in the calculated total distance and total time in this formula to find the average speed.
03

Calculate Average Velocity

As the start and end points are the same, the displacement for the whole trip is zero. As the formula for average velocity is displacement divided by time elapsed, the average velocity will also be zero.

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Key Concepts

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

Constant Speed
When we talk about constant speed, it means traveling at the same pace over a specific distance without changing. In this exercise, our person travels first from point A to point B at a constant speed of \(5.00 \, \text{m/s}\). Constant speed is a straightforward concept.
  • The speed remains the same throughout the entire journey.
  • No accelerations or decelerations occur; it's a steady flow.
  • This allows us to calculate the time easily using the formula \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\).

The constant speed simplifies calculations, as we only focus on how long it takes to cover the distance rather than any variations in speed.
Average Velocity
Average velocity is slightly different than average speed because it considers the direction of travel. It's calculated by dividing the total displacement by the total time.
In this problem:
  • The person goes from A to B and back to A.
  • The starting and ending points are the same, making the total displacement zero.
Hence, the formula for average velocity is:\[ V_{\text{avg}} = \frac{\text{Total Displacement}}{\text{Total Time}} \].
Since the displacement is zero, the average velocity is also zero. It's important to note that average velocity highlights directional changes, whereas average speed focuses only on how fast an object is moving.
Displacement
Displacement is a vector quantity that refers to how far out of place an object is, considering its starting and final positions. Unlike total distance, which measures the entire path traveled, displacement evaluates the change in position from the start to the endpoint. Consider our exercise:
  • The person starts at point A, travels to B, and returns to A.
  • The initial and final positions are identical, setting displacement to zero.
This effectively demonstrates why average velocity, calculated using displacement, also results in zero. Displacement is crucial in physics as it defines not just how far but in what direction an object moves. Simplicity arises here, as the distance traveled does not affect the displacement if the starting and ending points coincide.

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Most popular questions from this chapter

A glider on an air track carries a flag of length \(\ell\) through a stationary photogate, which measures the time interval \(\Delta t_{d}\) during which the flag blocks a beam of infrared light passing across the photogate. The ratio \(v_{d}=\ell / \Delta t_{d}\) is the average velocity of the glider over this part of its motion. Suppose the glider moves with constant acceleration. (a) Argue for or against the idea that \(v_{d}\) is equal to the instantaneous velocity of the glider when it is halfway through the photogate in space. (b) Argue for or against the idea that \(v_{d}\) is equal to the instantaneous velocity of the glider when it is halfway through the photogate in time.

An object moving with uniform acceleration has a velocity of \(12.0 \mathrm{cm} / \mathrm{s}\) in the positive \(x\) direction when its \(x\) coordinate is \(3.00 \mathrm{cm} .\) If its \(x\) coordinate \(2.00 \mathrm{s}\) later is \(-5.00 \mathrm{cm},\) what is its acceleration?

A ball starts from rest and accelerates at \(0.500 \mathrm{m} / \mathrm{s}^{2}\) while moving down an inclined plane \(9.00 \mathrm{m}\) long. When it reaches the bottom, the ball rolls up another plane, where, after moving \(15.0 \mathrm{m},\) it comes to rest. (a) What is the speed of the ball at the bottom of the first plane? (b) How long does it take to roll down the first plane? (c) What is the acceleration along the second plane? (d) What is the ball's speed \(8.00 \mathrm{m}\) along the second plane?

A particle moves along the \(x\) axis. Its position is given by the equation \(x=2+3 t-4 t^{2}\) with \(x\) in meters and \(t\) in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at \(t=0\).

A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is \(10.0 \mathrm{m} / \mathrm{s}\), and the distance from the limb to the level of the saddle is \(3.00 \mathrm{m}\). (a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move? (b) How long is he in the air?
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