Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Starting from rest, a boat increases its speed to \(5.00 \mathrm{~m} / \mathrm{s}\) with constant acceleration. a) What is the boat's average speed? b) If it takes the boat 4.00 s to reach this speed, how far has it traveled?

Short Answer

Expert verified
Question: A boat accelerates from rest to a speed of 5.00 m/s with constant acceleration. If it takes 4.00 seconds to reach this speed, find a) The boat's average speed, and b) The distance the boat travels. Answer: a) The boat's average speed is 2.50 m/s. b) The boat travels a distance of 10.00 meters.

Step by step solution

01

Calculate the average speed

The initial speed of the boat is 0 m/s (as it starts from rest), and the final speed is given as 5.00 m/s. We can calculate the average speed using the formula: Average speed = (Initial speed + Final speed) / 2 Average speed = (0 + 5.00) / 2 Average speed = 2.50 m/s The boat's average speed is 2.50 m/s.
02

Calculate the acceleration

We can calculate the acceleration using the formula: Acceleration = (Final speed - Initial speed) / time We are given that it takes the boat 4.00 seconds to reach the final speed of 5.00 m/s, so: Acceleration = (5.00 - 0) / 4.00 Acceleration = 5.00 / 4.00 Acceleration = 1.25 m/s² The boat's constant acceleration is 1.25 m/s².
03

Calculate the distance traveled

Now, we can find the distance traveled, using the formula: Distance = Initial speed × time + 0.5 × acceleration × time² Here, Initial speed = 0 m/s, acceleration = 1.25 m/s², and time = 4.00 s. Substitute the values in the formula: Distance = 0 × 4.00 + 0.5 × 1.25 × (4.00)² Distance = 0 + 0.5 × 1.25 × 16.00 Distance = 0.5 × 1.25 × 16.00 Distance = 10.00 m The boat travels a distance of 10.00 meters.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Average Speed Calculation
Understanding how to calculate average speed is essential in kinematics to describe how fast an object moves over a given period. The average speed gives us a simple way to quantify movement by considering the total distance traveled and the total time taken, without the complexities of accounting for variations in speed during the trip.

For an object starting from rest, like the boat in our example, the initial speed is zero. To get the average speed, we simply add the initial speed to the final speed and divide by two. Mathematically represented as:
\[\begin{equation}\text{Average speed} = \frac{\text{Initial speed} + \text{Final speed}}{2}\end{equation}\]
When the final speed is given, and the initial speed is zero, the calculation becomes even simpler, illustrating a foundational concept in kinematics: the mean value of two numbers.
Constant Acceleration
The concept of constant acceleration is a cornerstone in kinematics, especially when dealing with problems involving objects starting from rest or in uniform motion. Acceleration is defined as the rate at which the velocity of an object changes with time.

In cases where acceleration is constant, the formula we use is:
\[\begin{equation}\text{Acceleration} = \frac{\text{Final speed} - \text{Initial speed}}{\text{Time}}\end{equation}\]
For objects starting from a standstill, with an initial speed of zero, the formula simplifies considerably. With constant acceleration, the motion of an object can be easily predicted, emphasizing a predictable linear relationship between velocity and time, which forms a basis for more complex motion equations.
Distance Traveled
Calculating the distance traveled by an object under the influence of constant acceleration involves another core principle in kinematics. This value is essential for understanding the space an object covers as it moves. The general equation for this distance is sized to account for the initial velocity, time of travel, and acceleration:
\[\begin{equation}\text{Distance} = \text{Initial speed} \times \text{time} + 0.5 \times \text{acceleration} \times \text{time}^2\end{equation}\]
When starting from rest, the initial speed is zero, meaning the first term of the equation drops away, leaving us with the product of half the acceleration and the square of the time. This simplification highlights the quadratic relationship between distance and time under constant acceleration.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You drop a water balloon straight down from your dormitory window \(80.0 \mathrm{~m}\) above your friend's head. At \(2.00 \mathrm{~s}\) after you drop the balloon, not realizing it has water in it your friend fires a dart from a gun, which is at the same height as his head, directly upward toward the balloon with an initial velocity of \(20.0 \mathrm{~m} / \mathrm{s}\). a) How long after you drop the balloon will the dart burst the balloon? b) How long after the dart hits the balloon will your friend have to move out of the way of the falling water? Assume the balloon breaks instantaneously at the touch of the dart.

A car is traveling due west at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the car after \(3.00 \mathrm{~s}\) if its acceleration is \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) due west. Assume the acceleration remains constant. a) \(17.0 \mathrm{~m} / \mathrm{s}\) west b) \(17.0 \mathrm{~m} / \mathrm{s}\) east c) \(23.0 \mathrm{~m} / \mathrm{s}\) west d) \(23.0 \mathrm{~m} / \mathrm{s}\) east e) \(11.0 \mathrm{~m} / \mathrm{s}\) south

An object starts from rest and has an acceleration given by \(a=B t^{2}-\frac{1}{2} C t,\) where \(B=2.0 \mathrm{~m} / \mathrm{s}^{4}\) and \(C=-4.0 \mathrm{~m} / \mathrm{s}^{3}\). a) What is the object's velocity after 5.0 s? b) How far has the object moved after \(t=5.0\) s?

The vertical position of a ball suspended by a rubber band is given by the equation $$ y(t)=(3.8 \mathrm{~m}) \sin (0.46 t / \mathrm{s}-0.31)-(0.2 \mathrm{~m} / \mathrm{s}) t+5.0 \mathrm{~m} $$ a) What are the equations for velocity and acceleration for this ball? b) For what times between 0 and \(30 \mathrm{~s}\) is the acceleration zero?

Which of these statement(s) is (are) true? 1\. An object can have zero acceleration and be at rest. 2\. An object can have nonzero acceleration and be at rest. 3\. An object can have zero acceleration and be in motion. a) 1 only b) 1 and 3 c) 1 and 2 d) \(1,2,\) and 3

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free