/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 A speedboat moving at \(30.0 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A speedboat moving at \(30.0 \mathrm{~m} / \mathrm{s}\) approaches a no-wake buoy marker \(100 \mathrm{~m}\) ahead. The pilot slows the boat with a constant acceleration of \(-3.50 \mathrm{~m} / \mathrm{s}^{2}\) by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy?

Short Answer

Expert verified
The time taken by the boat to reach the buoy is approximately 8.57 s. At this time, the velocity of the boat when it reaches the buoy is approximately 0 m/s.

Step by step solution

01

Analyzing given data

First, list the known quantities: initial velocity \(v_i = 30.0 \mathrm{~m/s}\), final point (buoy) is \(100 \mathrm{~m}\) away (distance \(d = 100 \mathrm{~m}\)), and acceleration \(a = -3.50 \mathrm{~m/s^2}\). The negative sign of the acceleration indicates it opposes the motion, hence is also called deceleration.
02

Calculate Time

Use the equation for distances that have constant acceleration: \[d = v_it + 1/2 a t^2\]. Substituting the known values, we get \(100 = 30t + 1/2*-3.5t^2\). This is a quadratic equation in the form \(at^2 + bt + c = 0\). Solving this equation will yield two solutions, but only one is physically meaningful (time cannot be negative).
03

Calculate Final Velocity

Use the value of time from Step 2 and acceleration to find the final velocity with: \(v_f = v_i + at\).

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 91Ó°ÊÓ!

Key Concepts

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

Kinematics
In the realm of physics, kinematics is the branch that deals with the motion of objects without considering the causes of this motion, such as forces. It's all about understanding how things move. Kinematics focuses on several key quantities:
  • Displacement is the change in position of an object. In this scenario, the speedboat is initially moving toward a buoy marker 100 meters ahead, making this the displacement.

  • Velocity describes how fast something is moving in a given direction. Initially, the speedboat has a velocity of 30.0 meters per second.

  • Acceleration defines how quickly the velocity changes. In our exercise, the boat's acceleration is -3.50 meters per second squared, indicating it is slowing down.
Understanding these quantities is crucial, as they are foundational to solving problems related to motion. The target is usually to find unknown variables like time, distance, or final speed, using kinematic equations.
Constant acceleration
Constant acceleration implies that an object's velocity changes by the same amount in every equal time period. For instance, in our exercise, the speedboat reduces its velocity at a constant rate of -3.50 meters per second squared. This means each second, the speedboat slows down by 3.50 meters every second.
When dealing with constant acceleration, we can rely on specific formulas for calculations. One commonly used equation is:
  • For distance: \[d = v_it + \frac{1}{2} a t^2\]
  • For final velocity: \[v_f = v_i + at\]
These equations help determine the time it takes for an object to cover a certain distance or its final velocity after a period of time under constant acceleration. Utilizing these kinematic equations allows us to predict and understand the object's motion without needing detailed knowledge of the forces at play.
Quadratic equations
Quadratic equations often appear in kinematics, especially when dealing with constant acceleration. A quadratic equation has the general form:\[ax^2 + bx + c = 0\]In the context of our problem, solving for time, we derived an equation:\[100 = 30t + \frac{1}{2}(-3.5)t^2\]This equation is quadratic in nature because it includes a term with \(t^2\). To find the time taken for the boat to reach the buoy, we solve this quadratic equation.
Quadratic equations may yield two potential solutions. However, in physical contexts, not all solutions are viable. For instance, in calculating time, negative values aren't meaningful. Thus, we focus on the positive value, which represents actual time.
Deceleration
Deceleration occurs when an object slows down, meaning its velocity decreases over time. In physics, deceleration is just a type of negative acceleration.
Considering the exercise, the boat experiences a deceleration of -3.50 meters per second squared. This negative sign indicates that the acceleration is opposite to the direction of motion.
  • A deceleration value indicates a reduction in speed.
  • It affects the time it takes for an object to stop or reach a predetermined position.
  • Deceleration is crucial in safety calculations, like determining stopping distances for vehicles.
Through deceleration, we can understand how external factors like friction or applied forces affect an object's motion, slowing it down effectively.

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

A ranger in a national park is driving at \(35.0 \mathrm{mi} / \mathrm{h}\) when a deer jumps into the road \(200 \mathrm{ft}\) ahead of the vehicle. After a reaction time \(t\), the ranger applies the brakes to produce an acceleration \(a=-9.00 \mathrm{ft} / \mathrm{s}^{2}\). What is the maximum reaction time allowed if she is to avoid hitting the deer?

A model rocket is launched straight upward with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\), It accelerates with a constant upward acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) until its engines stop at an altitude of \(150 \mathrm{~m}\). (a) What can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

A parachutist with a camera descends in free fall at a speed of \(10 \mathrm{~m} / \mathrm{s}\). The parachutist releases the camera at an altitude of \(50 \mathrm{~m}\). (a) How long does it take the camera to reach the ground? (b) What is the velocity of the camcra just before it hits the ground?

A truck on a straight road starts from rest and accelerates at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) until it reaches a speed of \(20 \mathrm{~m} / \mathrm{s}\). Then the truck travels for \(20 \mathrm{~s}\) at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional \(5.0 \mathrm{~s}\). (a) How long is the truck in motion? (b) What is the average velocity of the truck during the motion described?

Two cars travel in the same direction along a straight highway, one at a constant speed of \(55 \mathrm{mi} / \mathrm{h}\) and the other at \(70 \mathrm{mi} / \mathrm{h}\). (a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination \(10 \mathrm{mi}\) away? (b) How far must the faster car travel before it has a 15 -min lead on the slower car?

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.