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Chapter 2: Problem 44

A bowling ball moves from \(x_{1}=3.5 \mathrm{~cm}\) to \(x_{2}=-4.7 \mathrm{~cm}\) during the time interval from \(t_{1}=3.0 \mathrm{~s}\) to \(t_{2}=5.5 \mathrm{~s}\). What is the ball's average velocity?

Short Answer

Expert verified
The average velocity of the ball is -3.28 cm/s.

Step by step solution

01

Determining Displacement and Time Interval

First, we need to determine the displacement of the ball, which is the change in position. The displacement is achieved by subtracting the initial position from the final position and is given by the equation \(d = x_2 - x_1 = -4.7cm - 3.5cm = -8.2cm\). Next, we calculate the time interval. The elapsed time for the ball's displacement is determined by subtracting the initial time from the final time, and is given by the equation \(\Delta t = t_2 - t_1 = 5.5s - 3.0s = 2.5s\)
02

Calculating Average Velocity

Now we can find the average velocity using the formula: Velocity = Displacement/Time. Substituting the calculated values into the formula, we get the average velocity as \(v = d / \Delta t = -8.2cm / 2.5s = -3.28 cm/s.\)

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Key Concepts

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

Understanding Displacement
Displacement is the change in position of an object. Unlike distance, which is the total path covered, displacement considers only the start and endpoint, providing the shortest path between them.
In the exercise, displacement is calculated by subtracting the initial position from the final position using the formula:
  • \(d = x_2 - x_1\)
  • Where \(x_1 = 3.5 \text{ cm}\) and \(x_2 = -4.7 \text{ cm}\)
  • So, \(d = -4.7 \text{ cm} - 3.5 \text{ cm} = -8.2 \text{ cm}\)
This negative value indicates that the ball moved towards the negative direction on the coordinate axis.
Calculating Time Interval
A time interval is the period over which the displacement occurs. It is vital to determine the duration or time taken for any motion. In our scenario, we calculate it using:
  • \(\Delta t = t_2 - t_1\)
  • Where \(t_1 = 3.0 \text{ s}\) and \(t_2 = 5.5 \text{ s}\)
  • \(\Delta t = 5.5 \text{ s} - 3.0 \text{ s} = 2.5 \text{ s}\)
This time interval is crucial for calculating velocity as it represents how long it took for the displacement to occur.
Average Velocity Calculation
Average velocity gives us the rate at which an object changes its position over a time period. It is calculated by dividing displacement by the time interval:
  • Formula: \(v = \frac{d}{\Delta t}\)
  • Substitute the known values: \(v = \frac{-8.2 \text{ cm}}{2.5 \text{ s}}\)
  • Result: \(v = -3.28 \text{ cm/s}\)
The negative sign of velocity indicates motion in the opposite direction to the chosen positive axis. Understanding this calculation provides insight into how fast and in what direction the object is moving.

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Most popular questions from this chapter

Biology A sperm whale can accelerate at about \(0.1\) \(\mathrm{m} / \mathrm{s}^{2}\) when swimming on the surface of the ocean. How far will a whale travel if it starts at a speed of \(1 \mathrm{~m} / \mathrm{s}\) and accelerates to a speed of \(2.25 \mathrm{~m} / \mathrm{s}\) ? Assume the whale travels in a straight line. SSM

鈥 two-stage rocket blasts off vertically from rest on a launchpad. During the first stage, which lasts for \(15.0 \mathrm{~s}\), the acceleration is a constant \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) upward. After \(15.0 \mathrm{~s}\), the first engine stops and the second stage engine fires, producing an upward acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\) that lasts for \(12.0 \mathrm{~s}\). At the end of the second stage, the engines no longer fire and therefore cause no acceleration, so the rocket coasts to its maximum altitude. (a) What is the maximum altitude of the rocket? (b) Over the time interval from blastoff at the launchpad to the instant that the rocket falls back to the launchpad, what are its (i) average speed and (ii) average velocity? Ignore the effects of air resistance.

Derive the equation that relates position to speed and acceleration but in which the time variable does not appear. Start with the basic equation for the definition of acceleration, \(a=\left(v-v_{0}\right) / t\), solve for \(t\), and substitute the resulting expression into the position versus time equation, \(x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}\).

Alex climbs to the top of a tall tree while his friend Gary waits on the ground below. Alex throws down a ball at \(4 \mathrm{~m} / \mathrm{s}\) from \(50 \mathrm{~m}\) above the ground at the same time Gary throws a ball up. At what speed must Gary throw a ball up in order for the two balls to cross paths \(25 \mathrm{~m}\) above the ground? The starting height of the ball thrown upward is \(1.5 \mathrm{~m}\) above the ground. Ignore the effects of air resistance. SSM

Under what circumstances will the displacement and the distance traveled be the same? When will they be different?
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