/*! 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 36 A railroad flatcar is traveling ... [FREE SOLUTION] | 91Ó°ÊÓ

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A railroad flatcar is traveling to the right at a speed of \(13.0 \mathrm{~m} / \mathrm{s}\) relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (Fig. E3.36). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter's velocity relative to the observer on the ground is (a) \(18.0 \mathrm{~m} / \mathrm{s}\) to the right? (b) \(3.0 \mathrm{~m} / \mathrm{s}\) to the left? (c) zero?

Short Answer

Expert verified
The velocity of the scooter relative to the flatcar is (a) $5.0 m/s$ to the right, (b) $16.0 m/s$ to the left, (c) $13.0 m/s$ to the left.

Step by step solution

01

Identify given data

The initial data given consists of the speed of the railroad flatcar relative to the observer on the ground, this being $13.0 m/s$ and towards the right. In addition, there are three scenarios for the velocity of the scooter relative to the same observer and we should calculate the speed of the scooter relative to the flatcar for each such scenario.
02

Calculate velocity for scenario (a)

In scenario (a), the scooter moves to the right (the same direction as the flatcar) at a velocity of $18.0 m/s$. Since they are moving in the same direction, we subtract the flatcar's velocity from the scooter's velocity, so \(V_{scooter-flatcar} = V_{scooter-ground} - V_{flatcar-ground} = 18.0 m/s - 13.0 m/s = 5.0 m/s\) to the right.
03

Calculate velocity for scenario (b)

In scenario (b), the scooter moves to the left (opposite direction to the flatcar's motion) at $3.0 m/s$. In this case, we add the magnitudes of scooter's and flatcar's velocities (since they are moving in opposite directions). So \(V_{scooter-flatcar} = V_{scooter-ground} + V_{flatcar-ground} = 3.0 m/s + 13.0 m/s = 16.0 m/s\) to the left.
04

Calculate velocity for scenario (c)

In scenario (c), the scooter is at rest relative to the ground ($0 m/s$ velocity). Hence, for the person in the flatcar, the scooter will appear to be moving to the right with a velocity equal to the flatcar's. This means that \(V_{scooter-flatcar} = - V_{flatcar-ground} = -13.0 m/s = 13.0 m/s\) to the left.

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Key Concepts

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

Frame of Reference
Understanding the concept of a frame of reference is crucial when analyzing motion in physics. Simply put, a frame of reference is the viewpoint from which an observer measures the position and velocity of an object. Imagine you're standing still on a platform, watching a train pass by—that's one frame of reference. Now, consider if you were sitting inside the train; your observations about other objects' motion would be completely different. In the given exercise, there are two frames of reference: the observer standing on the ground and the moving flatcar. The velocities calculated depend significantly on which frame is chosen as the point of reference.

Why does this matter? When solving problems in physics, specifying the frame of reference helps to avoid confusion and enables us to correctly understand and predict the behavior of objects in motion. It is the foundational step that dictates how we apply other concepts, such as vector subtraction or addition, as we'll see in the next sections.
Vector Subtraction
Now, let's dive into vector subtraction, a mathematical tool used for finding the relative velocity between two objects moving in the same or opposite directions. Vectors are quantities having both magnitude and direction, which is why when we talk about velocity, we deal with vectors. When two objects are moving along a straight line in the same direction and we want to find the velocity of one relative to the other, we subtract their velocities—as in scenario (a) of our exercise.

Remember, vector subtraction is not the same as subtracting simple numbers because direction plays a vital role. If the velocities are in the same direction, as in the scooter and flatcar, the vector representing the relative velocity points in the direction of the larger velocity vector, and its magnitude is the difference between the two velocities. However, when the velocities are in opposite directions, the magnitudes add up—but that's a story for vector addition that we encounter in scenario (b).
Motion in One Dimension

Understanding Motion Along a Line

Motion in one dimension occurs when an object is moving along a straight line, such as a car on a straight road or a train on its tracks—like the flatcar in our exercise. This type of motion is the simplest form of motion to analyze because it only involves one spatial dimension, which greatly simplifies calculations of velocity and acceleration. You can think of it as motion along a number line, where you only have two directions: forward (right) or backward (left).

When solving problems involving motion in one dimension, we apply vector arithmetic like subtraction or addition, depending on whether the objects are moving toward each other or away from each other. In the exercise, the different scenarios illustrate perfectly how one-dimensional motion can be assessed with these concepts: moving together in the same direction (scenario a), moving opposite to each other (scenario b), and one object being at rest relative to another frame of reference (scenario c).

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Most popular questions from this chapter

An object is projected with initial speed \(v_{0}\) from the edge of the roof of a building that has height \(H .\) The initial velocity of the object makes an angle \(\alpha_{0}\) with the horizontal. Neglect air resistance. (a) If \(\alpha_{0}\) is \(90^{\circ}\), so that the object is thrown straight up (but misses the roof on the way down), what is the speed \(v\) of the object just before it strikes the ground? (b) If \(\alpha_{0}=-90^{\circ}\), so that the object is thrown straight down, what is its speed just before it strikes the ground? (c) Derive an expression for the speed \(v\) of the object just before it strikes the ground for general \(\alpha_{0}\). (d) The final speed \(v\) equals \(v_{1}\) when \(\alpha_{0}\) equals \(\alpha_{1}\). If \(\alpha_{0}\) is increased, does \(v\) increase, decrease, or stay the same?

A "moving sidewalk" in an airport terminal moves at \(1.0 \mathrm{~m} / \mathrm{s}\) and is \(35.0 \mathrm{~m}\) long. If a woman steps on at one end and walks at \(1.5 \mathrm{~m} / \mathrm{s}\) relative to the moving sidewalk, how much time does it take her to reach the opposite end if she walks (a) in the same direction the sidewalk is moving? (b) In the opposite direction?

A model of a helicopter rotor has four blades, each \(3.40 \mathrm{~m}\) long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev \(/\) min. (a) What is the linear speed of the blade tip, in \(\mathrm{m} / \mathrm{s} ?\) (b) What is the radial acceleration of the blade tip expressed as a multiple of \(g ?\)

For this equipment to land at the front of the ship, at what a ship, which is moving at \(45.0 \mathrm{~cm} / \mathrm{s}\), before the ship can dock. This equipment is thrown at \(15.0 \mathrm{~m} / \mathrm{s}\) at \(60.0^{\circ}\) above the horizontal from the top of a tower at the edge of the water, \(8.75 \mathrm{~m}\) above the ship's deck (Fig. \(\mathbf{P 3 . 5 4}\) ). For this equipment to land at the front of the ship, at what distance \(D\) from the dock should the ship be when the equipment is thrown? Ignore air resistance.

Firefighters use a high-pressure hose to shoot a stream of water at a burning building. The water has a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) as it leaves the end of the hose and then exhibits projectile motion. The firefighters adjust the angle of elevation \(\alpha\) of the hose until the water takes \(3.00 \mathrm{~s}\) to reach a building \(45.0 \mathrm{~m}\) away. Ignore air resistance; assume that the end of the hose is at ground level. (a) Find \(\alpha\). (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

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