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Chapter 3: Problem 36

A railroad flatear 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. 3.43\()\) . What is the velocity (magnitude and direction) of the motor scooter relative to the flatcar if its 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
(a) 5.0 m/s right, (b) 16.0 m/s left, (c) 13.0 m/s left.

Step by step solution

01

Setup Problem and Understand Direction

Identify the given information: The speed of the flatcar is 13.0 m/s to the right. For each case, the speed of the motor scooter relative to the ground will be given, and you must find its speed relative to the flatcar.
02

Calculate Relative Velocity (Case a)

For case (a), the motor scooter's velocity relative to the ground is 18.0 m/s to the right. To find its velocity relative to the flatcar, subtract the velocity of the flatcar from the velocity of the motor scooter:\[ v_{s/f} = v_{s/g} - v_{f/g} = 18.0 \, \text{m/s} - 13.0 \, \text{m/s} = 5.0 \, \text{m/s} \] Thus, the motor scooter's velocity relative to the flatcar is 5.0 m/s to the right.
03

Calculate Relative Velocity (Case b)

For case (b), the motor scooter's velocity relative to the ground is 3.0 m/s to the left. Since left is the opposite of right, treat it as a negative speed with respect to right:\[ v_{s/g} = -3.0 \, \text{m/s} \] Then apply the same subtraction formula:\[ v_{s/f} = (-3.0 \, \text{m/s}) - 13.0 \, \text{m/s} = -16.0 \, \text{m/s} \] The motor scooter's velocity relative to the flatcar is 16.0 m/s to the left.
04

Calculate Relative Velocity (Case c)

For case (c), the motor scooter's velocity relative to the ground is zero.\[ v_{s/g} = 0 \, \text{m/s} \] Apply the formula:\[ v_{s/f} = 0 \, \text{m/s} - 13.0 \, \text{m/s} = -13.0 \, \text{m/s} \] The motor scooter's velocity relative to the flatcar is 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 frame of reference is essential in comparing velocities. It refers to the perspective from which we measure the motion of an object. Think of it as a stationary point or observer from which you see the world.
In this exercise, we primarily have two frames of reference:
  • The ground, where the observer is standing still.
  • The flatcar, which moves at a constant speed of 13.0 m/s to the right.
If you're observing the motor scooter's motion from the ground, you see it differently than if you were riding the flatcar. This change in observation is key to solving problems that involve relative motion. Always clearly define which frame of reference you are using for calculations or discussions.
Velocity Addition
Velocity addition involves combining speeds in the same direction. It's helpful when understanding how different movements affect the total observed motion, depending on your frame of reference.
If two objects move in the same direction, their velocities relative to one another can be found by adding their speeds. However, if they're moving towards or away from each other, it may involve subtraction. In our problem:
  • The motor scooter and flatcar moving to the right illustrate velocity addition.
By adding or subtracting these velocities, we determine how fast the scooter appears to move from the flatcar's point of view or vice versa. The direction plays a significant role here and must be consistently applied throughout your calculations.
Vector Subtraction
Vector subtraction helps when dealing with velocities in different directions. When subtracting vectors, you're essentially finding the difference in their magnitudes and directions. This is crucial when determining how one object moves with respect to another, especially when they are not aligned.
For example, calculating the velocity of the motor scooter with regard to the flatcar is a form of vector subtraction:
  • Case (a): With both moving to the right, subtract the flatcar's velocity from the scooter's.
  • Case (b): When the scooter moves left, consider its velocity as negative and subtract.
Remember, subtracting a vector that points in the opposite direction is like adding its opposite. Understanding this concept ensures accurate results in relative motion problems.
Motion Analysis
Motion analysis involves examining an object's movement to understand and break down its components. It often requires identifying direction, speed, and how these relate to other moving objects in different frames of reference.
In this exercise, motion analysis helps:
  • Determine how the observer on the ground sees the motor scooter differently from someone on the flatcar.
  • Resolve the directionality issues, such as left versus right, positive versus negative.
Through systematic analysis of each case, you see how the variables interplay to give a clear picture of motion dynamics. It's about methodically understanding and predicting an object's path based on its velocity relative to different observers.

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Most popular questions from this chapter

A canoe has a velocity of 0.40 \(\mathrm{m} / \mathrm{s}\) southeast relative to the earth. The canoe is on a river that is flowing 0.50 \(\mathrm{m} / \mathrm{s}\) east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

Tossing Your Lunch. Henrietta is going off to her physics class, jogging down the sidewalk at 3.05 \(\mathrm{m} / \mathrm{s}\) . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is 43.9 \(\mathrm{m}\) above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally 9.00 \(\mathrm{s}\) after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. (a) With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? (b) Where is Henrietta when she catches the bagels?

A rocket is launched vertically from rest with a constant upward acceleration of 1.75 \(\mathrm{m} / \mathrm{s}^{2} .\) Suddenly 22.0 \(\mathrm{s}\) after launch, an unneeded fuel tank is jettisoned by shooting it away from the rocket. Acrew member riding in the rocket measures that the initial speed of the tank is 25.0 \(\mathrm{m} / \mathrm{s}\) and that it moves perpendicular to the rocket's path. The fuel tank feels no appreciable air resistance and reels only the force of gravity once it leaves the rocket. (a) How feels only the force of gravity once it leaves the rocket. (a) How fast is the rocket moving at the instant the fuel tank is jettisoned? (b) What are the horizontal and vertical componed by (i) a crew unk's velocity just as it is jetisoned as measured by (i) a crew member in the rocket and (ii) a technician standing on the ground? (c) At what angle with respect to the horizontal does the jettisoned fuel tank initially move, as measured by a crew member in the rocket and (ii) a technician standing on the ground? (d) What maximum height above the launch pad does the jettisoned tank reach?

A maior leaguer hits a baseball so that it leaves the bat at a speed of 30.0 \(\mathrm{m} / \mathrm{s}\) and at an angle of \(36.9^{\circ}\) above the horizontal. You can ignore air resistance. (a) At what two times is the baseball at a height of 10.0 \(\mathrm{m}\) above the point at which it left the bat? (b) Calcu- late the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a). What are the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?

Leaping the River I. A car comes to a bridge during a storm and finds the bridge washed out The driver must get to the other side, so he decides to try leaping it with his car. The side of the road the car is on is 21.3 \(\mathrm{m}\) above the river, while the opposite side is a mere 1.8 \(\mathrm{m}\) above the river. The river itself is a raging torrent 61.0 \(\mathrm{m}\) wide. (a) How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands on the other side?
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