91Ó°ÊÓ

An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a \(10 \mathrm{~s}\) interval. What are the magnitude, the algebraic sign, and the direction of the average acceleration in each interval? Assume that the positive direction is to the right. (a) At the beginning of the interval, the astronaut is moving toward the right along the \(x\) -axis at \(15.0 \mathrm{~m} / \mathrm{s}\), and at the end of the interval she is moving toward the right at \(5.0 \mathrm{~m} / \mathrm{s}\). (b) At the beginning she is moving toward the left at \(5.0 \mathrm{~m} / \mathrm{s},\) and at the end she is moving toward the left at \(15.0 \mathrm{~m} / \mathrm{s}\). (c) At the beginning she is moving toward the right at \(15.0 \mathrm{~m} / \mathrm{s}\), and at the end she is moving toward the left at \(15.0 \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Determine the change in velocity for interval (a)

The change in velocity (\(\Delta v\)) is final velocity minus initial velocity. Since both velocity values are towards the right and positive, \(\Delta v = v_f - v_i = 5.0 m/s - 15.0 m/s = -10.0 m/s\).

02

Calculate average acceleration for interval (a)

The average acceleration (\(a\)) is \(\Delta v / \Delta t\). We know \(\Delta t = 10s\), so \(a = \Delta v / \Delta t = -10.0 m/s / 10 s = -1.0 m/s²\). Since the value is negative, the acceleration is towards the left.

03

Determine the change in velocity for interval (b)

\(v_i\) for interval b is towards the left and hence negative. So \(\Delta v = v_f - v_i = -15.0 m/s - (-5.0 m/s) = -10.0 m/s\).

04

Calculate average acceleration for interval (b)

Using the same formula as in Step 2, \(a = \Delta v / \Delta t = -10.0 m/s / 10 s = -1.0 m/s²\). The acceleration is the same magnitude as in interval (a), but also towards the left.

05

Determine the change in velocity for interval (c)

Here's where things change. The final velocity (\(v_f\)) is towards the left, so it will be negative, while the initial velocity (\(v_i\)) is towards the right so it will be positive. Thus, \(\Delta v = v_f - v_i = -15.0 m/s - 15.0 m/s = -30.0 m/s\).

06

Calculate average acceleration for interval (c)

Again using our formula from Step 2, \(a = \Delta v / \Delta t = -30.0 m/s / 10 s = -3.0 m/s²\). The acceleration is three times more than in intervals (a) and (b) and it is also towards the left.

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Key Concepts

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

Average Acceleration

Average acceleration is the rate at which velocity changes over a specific period of time. It's a little like how your speedometer might tell you how fast you're increasing your speed when you step on a car's accelerator.

To find average acceleration, you need two things:

• The change in velocity, represented by \( \Delta v \).
• The time interval over which the change occurs, denoted as \( \Delta t \).

Put together, the formula for average acceleration \( a \) is:\[a = \frac{\Delta v}{\Delta t}\]In this exercise, each interval is \( 10 \text{ seconds} \), which makes calculating \( a \) a bit simpler.

The unit of acceleration is \( \text{m/s}^2 \), which tells us how much the velocity changes each second. If the acceleration is negative, it simply means the object is slowing down in the direction of its current motion, or it is accelerating in the opposite direction.

Velocity Change

Velocity change is all about how fast something is moving in a specific direction, and whether that speed or direction is changing.

Consider interval (a) from our exercise. Initially, the astronaut moves to the right at \( 15.0 \text{ m/s} \), but by the end of the time period, her speed is \( 5.0 \text{ m/s} \) in the same direction. The change in velocity is calculated as:

• \( \Delta v = v_f - v_i = 5.0 \text{ m/s} - 15.0 \text{ m/s} = -10.0 \text{ m/s} \)

This tells us that the astronaut's velocity decreased by \( 10.0 \text{ m/s} \) to the right, or effectively she decelerated.

Different intervals in the exercise may result in different magnitudes of change, and the direction (positive or negative) indicates whether it moves to the left or right based on the initial reference direction. This is important as it helps us determine the resulting motion after the time interval.

Directional Motion Analysis

Directional motion analysis is crucial when you want to know not just how fast something is moving, but in what direction.

Let’s look at interval (c) in our exercise. Here, the astronaut moves initially to the right at \( 15.0 \text{ m/s} \) and eventually switches to the left at the same speed, \( -15.0 \text{ m/s} \) in our defined positive direction to the right. This tells a lot about the motion:

• The astronaut turned around, going from a positive to a negative velocity.
• The total change in velocity \( \Delta v \) is \( -15.0 \text{ m/s} - 15.0 \text{ m/s} = -30.0 \text{ m/s} \).
This change is twice as dramatic as the other intervals mentioned because not only did the speed decrease, it also changed direction. Now, the acceleration, calculated using the average acceleration formula, \[ a = \frac{-30.0 \space \text{m/s}}{10 \space \text{s}} = -3.0 \space \text{m/s}^2 \], signifies a strong movement to the left.

Understanding directional changes in velocity helps in predicting and illustrating trajectories and outcomes in dynamics, offering insight into the comprehensive movement of the object.