Problem 30 - University Physics with Modern Physics

Question

A railroad flatcar is traveling to the right at a speed of 13.0 m/s relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (\(\textbf{Fig. E3.30}\)). 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 m/s to the right? (b) 3.0 m/s to the left? (c) zero?

Step-by-Step Solution

Answer

(a) 5.0 m/s right; (b) 16.0 m/s left; (c) 13.0 m/s left.

Step 1: Understanding the problem

We need to find out the velocity of the scooter relative to the flatcar. We have the velocity of the scooter relative to the observer on the ground and the velocity of the flatcar relative to the ground. The observer measures velocities relative to the ground, so we can express the velocities as: \( v_{sg} \) for the scooter relative to the ground, and \( v_{fg} \) for the flatcar relative to the ground. The relation we'll use is: \( v_{sf} = v_{sg} - v_{fg} \), where \( v_{sf} \) is the velocity of the scooter relative to the flatcar.

Step 2: Applying the concept for case (a)

For case (a), the scooter's velocity relative to the ground is \( v_{sg} = 18.0 \) m/s to the right, and the flatcar's velocity is \( v_{fg} = 13.0 \) m/s to the right. Using \( v_{sf} = v_{sg} - v_{fg} \), we get \( v_{sf} = 18.0 \text{ m/s} - 13.0 \text{ m/s} = 5.0 \text{ m/s} \). So, the scooter moves at 5.0 m/s to the right relative to the flatcar.

Step 3: Applying the concept for case (b)

For case (b), the scooter's velocity relative to the ground is \( v_{sg} = -3.0 \) m/s (to the left). Using the same formula \( v_{sf} = v_{sg} - v_{fg} \), we compute \( v_{sf} = (-3.0 \text{ m/s}) - (13.0 \text{ m/s}) = -16.0 \text{ m/s} \). This means the scooter is moving at 16.0 m/s to the left relative to the flatcar.

Step 4: Applying the concept for case (c)

For case (c), the scooter's velocity relative to the ground is \( v_{sg} = 0 \) m/s. Using the relation \( v_{sf} = v_{sg} - v_{fg} \) gives us \( v_{sf} = 0 \text{ m/s} - 13.0 \text{ m/s} = -13.0 \text{ m/s} \). This indicates that the scooter is moving at 13.0 m/s to the left relative to the flatcar.

Key Concepts

Velocity Addition

Velocity addition is a fundamental concept in physics that helps us understand how to calculate the velocity of an object with respect to different reference points. It's like putting pieces together to find out how fast something is truly moving from a specific perspective. For example, if we have two velocities, one of a scooter relative to the ground and another of a flatcar moving in the same or opposite direction, we can use velocity addition to find out how fast the scooter moves with respect to the flatcar.

Let's go through a quick way to add velocities:

• When both objects are moving in the same direction, you subtract the velocity of the reference (flatcar) from the observed velocity (scooter) relative to the ground.
• When they are moving in opposite directions, or when the observer needs another perspective, the same calculation is used, but considering the direction can result in positive or negative outcomes.

These calculations allow us to understand and answer questions involving relative velocities, like the exercise given about the railroad flatcar and scooter.

Reference Frames

Reference frames are perspectives that help us analyze motion from different viewpoints. Imagine standing still on the ground versus riding the flatcar; both are distinct frames of reference. The motion you perceive depends directly on your frame of reference.

Here’s how we use reference frames in motion analysis:

• A ground observer sees both the flatcar and scooter moving. The velocities here are measured relative to the stationary observer.
• A rider on the flatcar, however, views the scooter’s motion relative to their moving position. This means the flatcar's own movement becomes part of the scooter's observed velocity in that frame.

By shifting frames, we effectively change what the observer considers "at rest" or "moving," which is key to solving problems involving multiple moving objects.

Motion Analysis

Motion analysis is the detailed examination of the movement of objects, particularly focusing on speed, direction, and relative motion. When we analyze movements, we delve into how velocities change based on different reference frames and directions.

In this case:

• We utilize the given velocities of the scooter and flatcar from different frames to compute the scooter's speed relative to the flatcar.
• We consider both the speed magnitude and its direction to fully describe the motion.
• Understanding the changes in direction is crucial too, as it determines whether the velocities add up or subtract. This impacts whether the scooter appears to move faster or slower relative to the flatcar.

By meticulously breaking down these components, one can thoroughly understand how objects move relative to one another in complicated scenarios.