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(I) A rolling ball moves from \(x_{1}=3.4 \mathrm{~cm}\) to \(x_{2}=-4.2 \mathrm{~cm}\) during the time from \(t_{1}=3.0 \mathrm{~s}\) to \(t_{2}=5.1 \mathrm{~s} .\) What is its average velocity?

Short Answer

Expert verified
The average velocity is approximately \(-3.62 \, \text{cm/s}\).

Step by step solution

01

Understanding Average Velocity

Average velocity is defined as the change in position divided by the change in time. The formula to calculate the average velocity is: \\[ v_{avg} = \frac{\Delta x}{\Delta t} \] \where \\( \Delta x = x_2 - x_1 \) and \\( \Delta t = t_2 - t_1 \).
02

Calculating Change in Position

First, we need to find the change in position. We have \\( x_1 = 3.4 \, \text{cm} \) and \\( x_2 = -4.2 \, \text{cm} \). Thus, the change in position \( \Delta x \) is: \\[ \Delta x = x_2 - x_1 = -4.2 \, \text{cm} - 3.4 \, \text{cm} = -7.6 \, \text{cm} \]
03

Calculating Change in Time

Next, we determine the change in time. We have \\( t_1 = 3.0 \, \text{s} \) and \\( t_2 = 5.1 \, \text{s} \). Hence, the change in time \( \Delta t \) is: \\[ \Delta t = t_2 - t_1 = 5.1 \, \text{s} - 3.0 \, \text{s} = 2.1 \, \text{s} \]
04

Calculating Average Velocity

We substitute \( \Delta x = -7.6 \, \text{cm} \) and \( \Delta t = 2.1 \, \text{s} \) into the average velocity formula: \\[ v_{avg} = \frac{-7.6 \, \text{cm}}{2.1 \, \text{s}} \approx -3.62 \, \text{cm/s} \]
05

Final Answer

The average velocity of the rolling ball is approximately \\( -3.62 \, \text{cm/s} \). The negative sign indicates that the ball is moving in the opposite direction of the positive axis.

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Key Concepts

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

Change in Position
When discussing the concept of change in position, it refers to how far an object has moved from its initial point to its final one. It is an essential part of understanding motion. Change in position is denoted by \( \Delta x \), and it is calculated as the difference between two positions: the final position \( x_2 \) and the initial position \( x_1 \). 1. Formula: - \( \Delta x = x_2 - x_1 \) - - This formula tells us the net change in where the object is located.In the given exercise, the position of the ball changes from \( x_1 = 3.4 \, \text{cm} \) to \( x_2 = -4.2 \, \text{cm} \). Calculating this gives us \( \Delta x = -4.2 \, \text{cm} - 3.4 \, \text{cm} = -7.6 \, \text{cm} \).The negative sign indicates that the object moved in the opposite direction on the axis from the positive starting point. Understanding this directionality in motion is crucial, especially when interpreting real-world movements.
Change in Time
The change in time, often symbolized as \( \Delta t \), measures the duration over which the change in position occurs. This is a key element in many physics calculations, especially those involving motion.1. Understanding change in time:- It allows us to quantify how long an event lasts, providing a sense of speed and movement over periods.2. Formula: - \( \Delta t = t_2 - t_1 \)To calculate it, we subtract the start time \( t_1 \) from the end time \( t_2 \). In the exercise example, the time changes from \( t_1 = 3.0 \, \text{s} \) to \( t_2 = 5.1 \, \text{s} \), resulting in \( \Delta t = 5.1 \, \text{s} - 3.0 \, \text{s} = 2.1 \, \text{s} \).This calculated duration helps us to compare how fast different objects move, given the same or different durations. Time change helps identify not only the efficiency of a motion but also its nature across varying lengths of cycles.
Velocity Calculation
Velocity calculation is central to understanding how fast something moves in a particular direction. Average velocity is a straightforward way to express this. We compute it by dividing the change in position by the change in time.1. Formula:- \( v_{avg} = \frac{\Delta x}{\Delta t} \) This ratio gives us a clear view of how rapid the movement was, with direction included due to the sign (positive or negative) of \( \Delta x \).2. Example Calculation:- Using the previous calculations: - The change in position \( \Delta x = -7.6 \, \text{cm} \) - The change in time \( \Delta t = 2.1 \, \text{s} \)Using these values in the velocity formula:\[ v_{avg} = \frac{-7.6 \, \text{cm}}{2.1 \, \text{s}} \approx -3.62 \, \text{cm/s} \] This result shows us that the object moved at an average speed of \(-3.62 \, \text{cm/s}\), and the negative value indicates its direction. The concept of velocity changes our understanding of motion by linking time, distance, and direction together, providing a comprehensive picture of an object's movement over time.

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Most popular questions from this chapter

(1I) The position of a small object is given by \(x=34+10 t-2 t^{3},\) where \(t\) is in seconds and \(x\) in meters. (a) Plot \(x\) as a function of \(t\) from \(t=0\) to \(t=3.0 \mathrm{s}\) (b) Find the average velocity of the object between 0 and 3.0 \(\mathrm{s}\) (c) At what time between 0 and 3.0 \(\mathrm{s}\) is the instantaneous velocity zero?

A fast-food restaurant uses a conveyor belt to send the burgers through a grilling machine. If the grilling machine is \(1.1 \mathrm{~m}\) long and the burgers require \(2.5 \mathrm{~min}\) to cook, how fast must the conveyor belt travel? If the burgers are spaced \(15 \mathrm{~cm}\) apart, what is the rate of burger production (in burgers/min)?

In putting, the force with which a golfer strikes a ball is planned so that the ball will stop within some small distance of the cup, say \(1.0 \mathrm{~m}\) long or short, in case the putt is missed. Accomplishing this from an uphill lie (that is, putting the ball downhill, see Fig. \(2-48\) ) is more difficult than from a downhill lie. To see why, assume that on a particular green the ball decelerates constantly at \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) going downhill, and constantly at \(2.8 \mathrm{~m} / \mathrm{s}^{2}\) going uphill. Suppose we have an uphill lie \(7.0 \mathrm{~m}\) from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range \(1.0 \mathrm{~m}\) short to \(1.0 \mathrm{~m}\) long of the cup. Do the same for a downhill lie \(7.0 \mathrm{~m}\) from the cup. What in your results suggests that the downhill putt is more difficult?

(I) A sprinter accelerates from rest to \(9.00 \mathrm{~m} / \mathrm{s}\) in \(1.28 \mathrm{~s}\). What is her acceleration in (a) \(\mathrm{m} / \mathrm{s}^{2} ;\) (b) \(\mathrm{km} / \mathrm{h}^{2}\) ?

(III) The acceleration of a particle is given by \(a=A \sqrt{t}\) where \(A=2.0 \mathrm{~m} / \mathrm{s}^{5 / 2} .\) At \(t=0, v=7.5 \mathrm{~m} / \mathrm{s}\) and \(x=0\) (a) What is the speed as a function of time? (b) What is the displacement as a function of time? (c) What are the acceleration, speed and displacement at \(t=5.0 \mathrm{~s} ?\)

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