/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 A glider on an air track carries... [FREE SOLUTION] | 91Ó°ÊÓ

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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) Is \(v_{d}\) necessarily equal to the instantaneous velocity of the glider when it is halfway through the photogate in space? Explain. (b) Is \(v_{d}\) equal to the instantaneous velocity of the glider when it is halfway through the photogate in time? Explain.

Short Answer

Expert verified
The average velocity \(v_{d}\) is not necessarily equal to the instantaneous velocity of the glider when it is halfway through the photogate in terms of space, while it is equal to the instantaneous velocity of the glider when it is halfway through the photo gate in terms of time.

Step by step solution

01

Understanding Average and Instantaneous Velocity

Average velocity is the total displacement divided by the total time taken. An instantaneous velocity, however, is the velocity of an object at a specific point in space or a specific instance of time. Thus, when an object moves with constant acceleration, the average velocity is not necessarily equal to the instantaneous velocity.
02

Evaluating the Condition for Part (a)

Since the glider is accelerating, its velocity is not constant. So, when the glider is halfway through the photogate in terms of space, its velocity will be either less or more than the average velocity, depending on whether it is accelerating forward or backward. Thus, the average velocity \(v_{d}\) is not necessarily equal to the instantaneous velocity of the glider when it is halfway through the photogate in space.
03

Evaluating the Condition for Part (b)

When we consider time, for a body moving under uniform acceleration, the average velocity over a time interval is always equal to the instantaneous velocity at the midpoint of the time interval. So, yes, \(v_{d}\) is equal to the instantaneous velocity of the glider when it is halfway through the photogate in time.

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Key Concepts

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

Average Velocity
Average velocity is a pivotal concept in understanding how objects move. It is defined as the total displacement of an object over the total time taken to cover that distance. In mathematical terms, it is expressed as
\( v_{avg} = \frac{\Delta x}{\Delta t} \)
where \(\Delta x\) is the displacement and \(\Delta t\) is the time interval. Displacement should not be confused with distance as it is a vector quantity that considers direction. This means that if you start and end at the same point, your average velocity is zero because your overall displacement is zero. It's crucial to realize that average velocity doesn't provide insight into the variations in speed or direction that might occur during the journey.
Instantaneous Velocity
Instantaneous velocity, on the other hand, tells us the speed and direction of an object at a particular moment in time. It's akin to glancing at your speedometer while driving; the value you see represents your instantaneous velocity. In calculus terms, it's the derivative of the position with respect to time.
\( v_{inst} = \frac{d x}{d t} \)
In the context of our exercise, as the glider moves through the photogate, its speed could be increasing or decreasing due to constant acceleration. Instantaneous velocity is best visualized on a graph as the slope of the tangent to the position-time curve at any given point. This subtle distinction between average and instantaneous velocity is crucial for understanding the detailed dynamics of any motion.
Constant Acceleration
Constant acceleration is a uniformly continuous change in velocity. It means that the velocity of an object changes by an equal amount every second. When we talk about constant acceleration, the acceleration value remains the same throughout the motion. In equations of motion, it's often denoted as \( a \), and the velocity \( v \) at any time \( t \) can be found using the formula:
\( v = u + at \)
where \( u \) is the initial velocity and \( t \) is the time elapsed. Notice that even if acceleration is constant, the velocity can and does change, affecting how we compute average and instantaneous velocities.
Motion with Constant Acceleration
When an object is in motion with constant acceleration, its velocity changes linearly over time. This type of motion is described by three kinematic equations, which link displacement, initial velocity, average velocity, final velocity, acceleration, and time. One of these kinematic equations is:
\( v_{avg} = \frac{v + u}{2} \)
This implies that under constant acceleration, the average velocity over some time period is the mean of the initial and final velocities. In our exercise, when considering motion through the photogate, this equation helps us understand that average velocity is equal to the instantaneous velocity when the glider is halfway through in time, because this specific instant corresponds to the midpoint of the uniform acceleration.

Understanding through Graphs

Graphical representations like velocity-time graphs can help visualize this concept. Under constant acceleration, a velocity-time graph is a straight line, indicating that velocity increases or decreases uniformly over time.
Physics of Motion
The physics of motion, or kinematics, delves into the description of the motion of objects without considering the forces that cause this motion. It includes concepts like speed, velocity, acceleration, and displacement. In kinematic analysis, the relationships between these quantities are explored through graphical representations and mathematical equations.
A comprehensive understanding involves mastering the distinction between scalars (like speed) and vectors (like velocity and acceleration), solving problems involving multiple phases of motion, and applying the kinematic equations to predict the future position or velocity of objects in motion. Our exercise is a practical application of these principles, as it examines motion through the lens of constant acceleration and its effect on different types of velocities.

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Most popular questions from this chapter

A package is dropped from a helicopter that is descending steadily at a speed \(v_{0}\). After \(t\) seconds have elapsed, (a) what is the speed of the package in terms of \(v_{0}, g\), and \(t\) ? (b) What distance \(d\) is it from the helicopter in terms of \(g\) and \(t\) ? (c) What are the answers to parts (a) and (b) if the helicopter is rising steadily at the same speed?

Speedy Sue, driving at \(30.0 \mathrm{~m} / \mathrm{s}\), enters a one-lane tunnel. She then observes a slow-moving van \(155 \mathrm{~m}\) ahead traveling at \(5.00 \mathrm{~m} / \mathrm{s}\). Sue applies her brakes but can accelerate only at \(-2.00 \mathrm{~m} / \mathrm{s}^{2}\) because the road is wet. Will there be a collision? State how you decide. If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue's car and the van.

The average person passes out at an acceleration of \(7 g\) (that is, seven times the gravitational acceleration on Earth). Suppose a car is designed to accelerate at this rate. How much time would be required for the car to accelerate from rest to \(60.0\) miles per hour? (The car would need rocket boosters!)

A certain cable car in San Francisco can stop in \(10 \mathrm{~s}\) when traveling at maximum speed. On one occasion, the driver sees a dog a distance \(d\) in front of the car and slams on the brakes instantly. The car reaches the dog \(8.0 \mathrm{~s}\) later, and the dog jumps off the track just in time. If the car travels \(4.0 \mathrm{~m}\) beyond the position of the dog before coming to a stop, how far was the car from the dog? (Hint: You will need three equations.)

A ball is thrown upward from the ground with an initial speed of \(25 \mathrm{~m} / \mathrm{s}\); at the same instant, another ball is dropped from a building \(15 \mathrm{~m}\) high. After how long will the balls be at the same height?

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