Ch 02: Motion Along a Straight Line

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 02: Motion Along a Straight Line

Problem 16

Chapter 2, Problem 16

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 $10$ -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 $15.0$ m/s, and at the end of the interval she is moving toward the right at $5.0$ m/s.
(b) At the beginning she is moving toward the left at $5.0$ m/s, and at the end she is moving toward the left at $15.0$ m/s.
(c) At the beginning she is moving toward the right at $15.0$ m/s, and at the end she is moving toward the left at $15.0$ m/s.

Verified step by step guidance

To find the average acceleration, use the formula: $ a_{avg} = \frac{\Delta v}{\Delta t} $, where $ \Delta v $ is the change in velocity and $ \Delta t $ is the time interval.

For interval (a), calculate $ \Delta v $ as the final velocity minus the initial velocity: $ \Delta v = 5.0 \text{ m/s} - 15.0 \text{ m/s} = -10.0 \text{ m/s} $. The time interval $ \Delta t $ is 10 s.

Substitute $ \Delta v = -10.0 \text{ m/s} $ and $ \Delta t = 10 \text{ s} $ into the average acceleration formula: $ a_{avg} = \frac{-10.0 \text{ m/s}}{10 \text{ s}} $. The negative sign indicates the acceleration is to the left.

For interval (b), calculate $ \Delta v $ as the final velocity minus the initial velocity: $ \Delta v = -15.0 \text{ m/s} - (-5.0 \text{ m/s}) = -10.0 \text{ m/s} $. The time interval $ \Delta t $ is 10 s.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Average Acceleration

Average acceleration is defined as the change in velocity divided by the time over which the change occurs. It is a vector quantity, meaning it has both magnitude and direction. The formula is a_avg = (v_final - v_initial) / Δt, where v_final is the final velocity, v_initial is the initial velocity, and Δt is the time interval.

Velocity and Direction

Velocity is a vector quantity that describes the speed of an object in a specific direction. In this problem, the direction is along the x-axis, with positive values indicating movement to the right and negative values indicating movement to the left. Understanding the direction is crucial for determining the sign of the velocity and acceleration.

Sign Convention

In physics, sign convention is used to indicate direction. For this problem, the positive direction is defined as to the right. Therefore, a positive velocity or acceleration indicates movement or acceleration to the right, while a negative value indicates movement or acceleration to the left. This helps in determining the algebraic sign of the average acceleration.

Recommended video:

Guided course

08:55

Net Torque & Sign of Torque

Related Practice

Textbook Question

An antelope moving with constant acceleration covers the distance between two points $70.0$ m apart in $6.00$ s. Its speed as it passes the second point is $15.0$ m/s. What is its acceleration?

views

Textbook Question

A car's velocity as a function of time is given by $v_x(t) = α + βt^2$ , where $α = 3.00$ m/s and $β = 0.100$ m/s³. Draw $v_{x}$ - $t$ and $a_{x}$ - $t$ graphs for the car's motion between $t equals 0$ and $t equals 5.00$ s.

views

Textbook Question

A race car starts from rest and travels east along a straight and level track. For the first $5.0$ s of the car's motion, the eastward component of the car's velocity is given by $v_{x} (t) =$ ( $0.860$ m/s³)t². What is the acceleration of the car when $v sub x equals 12.0$ m/s?

views

Textbook Question

In the fastest measured tennis serve, the ball left the racquet at $73.14$ m/s. A served tennis ball is typically in contact with the racquet for $30.0$ ms and starts from rest. Assume constant acceleration. What was the ball's acceleration during this serve?

views

Textbook Question

The fastest measured pitched baseball left the pitcher's hand at a speed of $45.0$ m/s. If the pitcher was in contact with the ball over a distance of $1.50$ m and produced constant acceleration, what acceleration did he give the ball?

views

Textbook Question

A car's velocity as a function of time is given by $v_{x} (t) = α + β t^{2}$ , where $alpha equals 3.00$ m/s and $beta equals 0.100$ m/s³. Calculate the average acceleration for the time interval $t equals 0$ to $t equals 5.00$ s.

views