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Chapter 2: Problem 3

(1) A particle at \(t_{1}=-2.0 \mathrm{s}\) is at \(x_{1}=4.3 \mathrm{cm}\) and at \(t_{2}=4.5 \mathrm{s}\) is at \(x_{2}=8.5 \mathrm{cm} .\) What is its average velocity? Can you calculate its average speed from these data?

Short Answer

Expert verified
Average velocity is approximately 0.646 cm/s. Average speed cannot be determined without complete path information.

Step by step solution

01

Identify the time interval

The time interval is the difference between the two time points given: \( t_2 = 4.5 \text{ s} \) and \( t_1 = -2.0 \text{ s} \). To find the time interval, calculate \( t_2 - t_1 \).
02

Calculate the time interval

Subtract the initial time from the final time: \( \Delta t = t_2 - t_1 = 4.5 \text{ s} - (-2.0 \text{ s}) = 4.5 \text{ s} + 2.0 \text{ s} = 6.5 \text{ s} \).
03

Identify the displacement of the particle

The displacement is the change in position from \( x_1 \) to \( x_2 \). This is given by \( x_2 = 8.5 \text{ cm} \) and \( x_1 = 4.3 \text{ cm} \). We calculate \( x_2 - x_1 \).
04

Calculate the displacement

Subtract the initial position from the final position: \( \Delta x = x_2 - x_1 = 8.5 \text{ cm} - 4.3 \text{ cm} = 4.2 \text{ cm} \).
05

Calculate average velocity

Average velocity is the displacement divided by the time interval: \( v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{4.2 \text{ cm}}{6.5 \text{ s}} \approx 0.646 \text{ cm/s} \).
06

Determine the possibility to calculate average speed

Average speed requires total distance traveled divided by the time interval. With only two positions provided, assuming straight-line motion, average speed equals average velocity, but true average speed needs complete path information. With only start and end points, average speed cannot be accurately determined.

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Key Concepts

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

Displacement Calculation
In physics, understanding displacement helps in knowing how far out of the starting point an object has moved. It is a vector quantity, meaning it has both magnitude and direction. In our exercise, to calculate displacement, we need the initial position, \(x_1 = 4.3 \, \text{cm}\), and the final position, \(x_2 = 8.5 \, \text{cm}\). Displacement is the difference between these two positions. The formula for calculating displacement is:
  • \( \Delta x = x_2 - x_1 \)

Using the given values:
  • \( \Delta x = 8.5 \, \text{cm} - 4.3 \, \text{cm} = 4.2 \, \text{cm} \)
It's important to note that displacement can be different from distance if the path taken is not a straight line, but in this case, we assume straight-line motion.
Time Interval
The time interval is a crucial aspect when calculating the rate of motion, like speed or velocity. It represents the duration during which the motion occurs. For this exercise, the time interval is determined by the subtraction of the initial time \(t_1 = -2.0 \, \text{s}\) from the final time \(t_2 = 4.5 \, \text{s}\). Here's how you find the time interval:
  • Calculate \( \Delta t = t_2 - t_1 \)

Plugging in the values:
  • \( \Delta t = 4.5 \, \text{s} - (-2.0 \, \text{s}) = 4.5 \, \text{s} + 2.0 \, \text{s} = 6.5 \, \text{s} \)
Time intervals should always be expressed in positive numbers, as they represent the elapsed time.
Average Speed Determination
Average speed is a scalar quantity that measures how fast an object moves, regardless of direction. It's different from velocity, which is vectorial. The definition of average speed is the total distance traveled divided by the total time taken.In this situation, we have simplified the path as a straight line, which means average speed equals average velocity. If more complex paths were involved, we would need additional data on the total distance traveled. However, here we use the points provided:
  • Average Velocity (which in this case equals average speed with the given data): \( v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{4.2 \, \text{cm}}{6.5 \, \text{s}} \approx 0.646 \, \text{cm/s} \)
Remember, if there's no clear path information, true average speed needs full route data beyond just start and end points.

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Most popular questions from this chapter

A police car at rest, passed by a speeder traveling at a constant 130 \(\mathrm{km} / \mathrm{h}\) , takes off in hot pursuit. The police officer catches up to the speeder in 750 \(\mathrm{m}\) , maintaining a constant acceleration. (a) Qualitatively plot the position vs. time graph for both cars from the police car's start to the catch-up point. Calculate \((b)\) how long it took the police officer to overtake the speeder, (c) the required police car acceleration, and \((d)\) the speed of the police car at the overtaking point.

(II) A car traveling at \(105 \mathrm{~km} / \mathrm{h}\) strikes a tree. The front end of the car compresses and the driver comes to rest after traveling \(0.80 \mathrm{~m} .\) What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of " \(g\) 's," where \(1.00 g=9.80 \mathrm{~m} / \mathrm{s}^{2}\).

Suppose a car manufacturer tested its cars for front-end collisions by hauling them up on a crane and dropping them from a certain height. \((a)\) Show that the speed just before a car hits the ground, after falling from rest a vertical distance \(H,\) is given by \(\sqrt{2 g H}\). What height corresponds to a collision at \((b) 50 \mathrm{~km} / \mathrm{h} ?\) (c) \(100 \mathrm{~km} / \mathrm{h} ?\)

(III) A bowling ball traveling with constant speed hits the pins at the end of a bowling lane \(16.5 \mathrm{~m}\) long. The bowler hears the sound of the ball hitting the pins \(2.50 \mathrm{~s}\) after the ball is released from his hands. What is the speed of the ball, assuming the speed of sound is \(340 \mathrm{~m} / \mathrm{s} ?\)

(II) A stone is thrown vertically upward with a speed of \(24.0 \mathrm{~m} / \mathrm{s}\). \((a)\) How fast is it moving when it reaches a height of \(13.0 \mathrm{~m} ?(b)\) How much time is required to reach this height? (c) Why are there two answers to (b)?
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