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Chapter 2: Problem 40

A motorboat traveling on a straight course slows uniformly from \(60 \mathrm{~km} / \mathrm{h}\) to \(40 \mathrm{~km} / \mathrm{h}\) in a distance of \(50 \mathrm{~m}\) What is the boat's acceleration?

Short Answer

Expert verified
The boat's acceleration is approximately -1.55 m/s虏.

Step by step solution

01

Understand the Problem

We are given an initial speed, a final speed, and the distance over which the speed changes. We need to find the acceleration. This is a physics problem involving motion with uniform acceleration.
02

List the Known Values

Initial speed ( \( v_i \)): 60 km/h.Final speed ( \( v_f \)): 40 km/h.Distance ( \( d \)): 50 m.We need to convert the speeds to meters per second, since the distance is given in meters.
03

Convert Units

Convert the speeds from km/h to m/s:\[ v_i = 60 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 16.67 \text{ m/s} \]\[ v_f = 40 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 11.11 \text{ m/s} \]
04

Use the Kinematic Equation

We need to use the equation \[ v_f^2 = v_i^2 + 2ad \]where \( a \) is acceleration, \( v_i \) is initial velocity, \( v_f \) is final velocity, and \( d \) is distance.Plug in the known values:\[ (11.11 \text{ m/s})^2 = (16.67 \text{ m/s})^2 + 2a(50 \text{ m}) \]
05

Solve for Acceleration

Rearrange the equation from step 4 to solve for \( a \):\[ 123.43 = 278.89 + 100a \]\[ 100a = 123.43 - 278.89 \]\[ 100a = -155.46 \]\[ a = \frac{-155.46}{100} \]\[ a = -1.5546 \text{ m/s}^2 \]

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Key Concepts

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

Uniform Acceleration
Uniform acceleration refers to a steady change in velocity over time.
In situations where acceleration is uniform, the object's velocity changes at a constant rate.
This means that if a car or boat is speeding up or slowing down uniformly, it gains or loses the same amount of speed each second. In the example provided with the motorboat:
  • We start with an initial speed of 60 km/h.
  • The boat slows down to a final speed of 40 km/h over a set distance.
This change in speed over a specific distance is an example of uniform acceleration.
It's important to note that deceleration, a type of uniform acceleration, occurs when the speed decreases. In our problem, this is exactly what happens, as the boat is slowing down over a uniform distance and time frame.
Velocity Conversion
Velocity conversion is crucial when solving kinematics problems, especially when the units for velocity and distance differ.
In our problem, the velocities are initially given in km/h.
To solve the problem, we need to convert velocities to m/s since the distance is in meters. Here's a straightforward way to convert km/h to m/s:
  • Multiply the velocity in km/h by 1000 to change kilometers to meters.
  • Divide the result by 3600 to change hours to seconds.
For our motorboat:
  • The initial velocity changes from 60 km/h to 16.67 m/s.
  • The final velocity changes from 40 km/h to 11.11 m/s.
By converting these units, we ensure consistency in the units, which is essential for accurate calculations regarding motion and acceleration.
Kinematic Equations
Kinematic equations enable us to solve problems related to motion, particularly when dealing with constant acceleration.
One of the key equations involves the final and initial velocities, the acceleration, and the distance:\[ v_f^2 = v_i^2 + 2ad \]In this equation:
  • \( v_f \) is the final velocity,
  • \( v_i \) is the initial velocity,
  • \( a \) is the acceleration, and
  • \( d \) is the distance traveled.
For the motorboat slowing down, we substitute:
  • The final velocity: 11.11 m/s
  • The initial velocity: 16.67 m/s
  • The distance: 50 m
Rearranging and solving the equation allows us to find the boat鈥檚 acceleration, computed as \(-1.5546\ m/s^2\).
Mastering these equations and knowing how to rearrange them is key to tackling any kinematics-related challenge effectively.

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Most popular questions from this chapter

A pollution-sampling rocket is launched straight upward with rockets providing a constant acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2}\) for the first \(1000 \mathrm{~m}\) of flight. At that point the rocket motors cut off and the rocket itself is in free fall. Ignore air resistance. (a) What is the rocket's speed when the engines cut off? (b) What is the maximum altitude reached by this rocket? (c) What is the time it takes to get to its maximum altitude?

A student driving home for the holidays starts at 8: 00 AM to make the \(675-\mathrm{km}\) trip, practically all of which is on nonurban interstate highways. If she wants to arrive home no later than \(3: 00 \mathrm{PM},\) what must be her minimum average speed? Will she have to exceed the \(65-\mathrm{mi} / \mathrm{h}\) speed limit?

A hospital patient is given 500 cc of saline by IV. If the saline is received at a rate of \(4.0 \mathrm{~mL} / \mathrm{min},\) how long will it take for the half liter to run out?

A train on a straight, level track has an initial speed of \(35.0 \mathrm{~km} / \mathrm{h}\). A uniform acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) is applied while the train travels \(200 \mathrm{~m}\). (a) What is the speed of the train at the end of this distance? (b) How long did it take for the train to travel the \(200 \mathrm{~m} ?\)

A car accelerates from rest at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) for \(5.0 \mathrm{~s}\). (a) What is the speed of the car at the end of that time? (b) How far does the car travel in this time?
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