/*! 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 62 A boat is cruising in a straight... [FREE SOLUTION] | 91Ó°ÊÓ

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A boat is cruising in a straight line at a constant speed of \(2.6 \mathrm{m} / \mathrm{s}\) when it is shifted into neutral. After coasting \(12 \mathrm{m}\) the engine is engaged again, and the boat resumes cruising at the reduced constant speed of \(1.6 \mathrm{m} / \mathrm{s}\). Assuming constant acceleration while coasting. (a) how long did it take for the boat to coast the \(12 \mathrm{m} ?\) (b) What was the boat's acceleration while it was coasting? \((c)\) When the boat had coasted for \(6.0 \mathrm{m},\) was its speed \(2.1 \mathrm{m} / \mathrm{s},\) more than \(2.1 \mathrm{m} / \mathrm{s},\) or less than \(2.1 \mathrm{m} / \mathrm{s}\) ? Explain.

Short Answer

Expert verified
(a) 5.71 seconds, (b) -0.175 m/s², (c) Speed is more than 2.1 m/s.

Step by step solution

01

Understanding the problem

We need to calculate the time taken to coast 12 m, the acceleration during this coast, and determine the speed at 6 m under constant acceleration.
02

Identifying known values and equations

The initial speed of the boat is 2.6 m/s, the final speed after coasting is 1.6 m/s, and the distance coasted is 12 m. The equation for motion under constant acceleration is: \( v^2 = u^2 + 2a s \), where \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration, and \( s \) is distance.
03

Calculate the acceleration

Using the equation \( v^2 = u^2 + 2a s \), we solve for \( a \):\[ (1.6)^2 = (2.6)^2 + 2a(12) \]\[ 2.56 = 6.76 + 24a \]\[ 24a = 2.56 - 6.76 \]\[ 24a = -4.2 \]\[ a = \frac{-4.2}{24} \approx -0.175 \text{ m/s}^2 \]
04

Calculate the time to coast 12m

Using the equation \( v = u + at \), solving for time \( t \):\[ 1.6 = 2.6 + (-0.175)t \]\[ -1.0 = -0.175t \]\[ t = \frac{-1.0}{-0.175} \approx 5.71 \text{ seconds} \]
05

Determine speed at 6 meters

Now using \( v^2 = u^2 + 2a s \) with \( s = 6 \) m:\[ v^2 = 2.6^2 + 2(-0.175)(6) \]\[ v^2 = 6.76 - 2.1 \]\[ v^2 = 4.66 \]\[ v = \sqrt{4.66} \approx 2.16 \text{ m/s} \]
06

Conclusion

(a) It took approximately 5.71 seconds to coast 12 meters. (b) The boat's acceleration was approximately -0.175 m/s² while coasting. (c) After coasting for 6 meters, the speed was about 2.16 m/s, which is more than 2.1 m/s.

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that focuses on the movement of objects without analyzing the forces causing the motion. In kinematics, we frequently deal with quantities such as velocity, acceleration, displacement, and time.
This concept helps us understand how an object moves from one point to another under specific conditions, such as constant acceleration or uniform velocity.
Kinematics concentrates on describing the motion and utilizing equations that relate these quantities.
Kinematic equations assume a simple format:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
In these equations:
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is acceleration,
  • \( s \) is displacement, and
  • \( t \) is time.
By understanding these relationships, one can predict how an object will move over time, which is crucial for solving problems related to motion, like the one involving the boat in our exercise.
Displacement
Displacement is a crucial concept to grasp when discussing motion. It's different from distance, as displacement is a vector quantity that represents the change in position of an object.
This means it has both magnitude and direction, focusing on the shortest path between the starting point and the ending point of the object's movement.
In the boat exercise, the 12 m distance the boat coasted is actually its displacement.
This means the boat shifted 12 m along a straight line - no more, no less. Displacement focuses on this change in position without regard for the path taken or the initial and final speeds.
Why is displacement important? Understanding displacement allows us to compute other quantities, like acceleration or time, using kinematic equations. It acts as a fundamental piece of the puzzle in mastering motion-related problems.
Final Velocity
Final velocity refers to the speed with which an object ends its motion after undergoing a phase of acceleration or deceleration. It's a key output in the kinematic equations.
In our exercise, the boat began at a velocity of 2.6 m/s and ended coasting at a reduced speed of 1.6 m/s.
This is the boat's final velocity - it shows us how much the speed has changed after coasted under constant acceleration.
Finding the final velocity can involve kinematic equations. In this specific scenario, the equation used was:
  • \( v^2 = u^2 + 2as \)
Solving this equation helps us understand the boat's behavior after it coasts to a stop, assisting in determining how much time it took to reach this final state or the possible speed at an intermediate point.
Constant Speed
Constant speed means that an object moves at an unchanging rate over a period of time. This implies zero acceleration, as the speed remains steady.
In the given scenario, after completing its coasting phase, the boat resumes at a new constant speed of 1.6 m/s. Before hitting "neutral," it cruised with a constant speed of 2.6 m/s.
Constant speed is easier to analyze than accelerating motion. Without acceleration, you avoid the complexity of changing velocities.
Using the formula:
  • \( s = vt \)
Where
  • \( s \) is displacement,
  • \( v \) is a constant speed, and
  • \( t \) is time,
  • you can calculate distance-wise motion effortlessly.
    For students, recognizing when an object moves at constant speed or undergoes constant acceleration is fundamental for solving motion problems effectively.

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