/*! 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 33 A speedboat starts from rest and... [FREE SOLUTION] | 91Ó°ÊÓ

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A speedboat starts from rest and accelerates at \(+2.01 \mathrm{~m} / \mathrm{s}^{2}\) for \(7.00 \mathrm{~s}\). At the end of this time, the boat continues for an additional \(6.00 \mathrm{~s}\) with an acceleration of \(+0.518 \mathrm{~m} / \mathrm{s}^{2}\) Following this, the boat accelerates at \(-1.49 \mathrm{~m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{~s}\). (a) What is the velocity of the boat at \(t=21.0 \mathrm{~s} ?\) (b) Find the total displacement of the boat.

Short Answer

Expert verified
The velocity at \( t = 21.0 \, \mathrm{s} \) is \( 5.198 \, \mathrm{m/s} \) and total displacement is \( 211.395 \) meters.

Step by step solution

01

Calculate Velocity After First Acceleration

The boat starts from rest, with its initial velocity, \( u_1 = 0 \). It accelerates at \( a_1 = 2.01 \, \mathrm{m/s^2} \) for \( t_1 = 7.00 \, \mathrm{s} \). Use the formula \( v = u + at \) to find the velocity at the end of this interval. \[ v_1 = 0 + (2.01) \times (7.00) = 14.07 \, \mathrm{m/s} \]
02

Calculate Velocity After Second Acceleration

Now the boat accelerates at \( a_2 = 0.518 \, \mathrm{m/s^2} \) for \( t_2 = 6.00 \, \mathrm{s} \). Use the same velocity formula with \( u = 14.07 \, \mathrm{m/s} \). \[ v_2 = 14.07 + (0.518)(6.00) = 17.178 \, \mathrm{m/s} \]
03

Calculate Final Velocity

The boat then accelerates at \( a_3 = -1.49 \, \mathrm{m/s^2} \) for \( t_3 = 8.00 \, \mathrm{s} \). Plug in the values where \( u = 17.178 \, \mathrm{m/s} \).\[ v_3 = 17.178 + (-1.49)(8.00) = 5.198 \, \mathrm{m/s} \]
04

Calculate Displacement During First Interval

Use the formula \( s = ut + \frac{1}{2} a t^2 \) to find the displacement during the first interval with \( u = 0 \), \( t = 7.00 \, \mathrm{s} \), and \( a = 2.01 \, \mathrm{m/s^2} \).\[ s_1 = 0 + \frac{1}{2}(2.01)(7.00)^2 = 49.245 \, \mathrm{m} \]
05

Calculate Displacement During Second Interval

Now calculate the displacement with \( u = 14.07 \, \mathrm{m/s} \), \( t = 6.00 \, \mathrm{s} \), and \( a = 0.518 \, \mathrm{m/s^2} \).\[ s_2 = (14.07)(6.00) + \frac{1}{2}(0.518)(6.00)^2 = 90.726 \, \mathrm{m} \]
06

Calculate Displacement During Third Interval

Calculate the displacement with \( u = 17.178 \, \mathrm{m/s} \), \( t = 8.00 \, \mathrm{s} \), and \( a = -1.49 \, \mathrm{m/s^2} \).\[ s_3 = (17.178)(8.00) + \frac{1}{2}(-1.49)(8.00)^2 = 71.424 \, \mathrm{m} \]
07

Find Total Displacement

Add the displacements from all intervals to find the total displacement.\[ s_{\text{total}} = s_1 + s_2 + s_3 = 49.245 + 90.726 + 71.424 = 211.395 \, \mathrm{m} \]

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Key Concepts

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

Velocity Calculation
Velocity is a measure of how fast an object is moving and in what direction. It's calculated using the formula:
  • \( v = u + at \)
where:
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is the acceleration,
  • \( t \) is the time over which the acceleration occurs.
In our speedboat problem, the boat starts from rest, meaning \( u = 0 \) for the first interval. The boat's velocity increases due to acceleration. After multiple intervals of acceleration, both positive and negative, the final velocity can be calculated after each stage by adding the previous velocity to the product of acceleration and time for that interval.
For instance, after the first phase, using \( u = 0 \), \( a = 2.01 \, \mathrm{m/s^2} \), and \( t = 7.00 \, \mathrm{s} \), the velocity becomes \( 14.07 \, \mathrm{m/s} \). This process repeats for subsequent stages of acceleration.
By understanding and applying this formula, we can calculate how the velocity changes over different phases of motion.
Acceleration
Acceleration is how quickly an object changes its velocity. It can be a result of speeding up, slowing down, or changing direction. It's crucial to differentiate between positive acceleration (speeding up) and negative acceleration (slowing down), often called deceleration.
The formula to determine the final velocity given an initial velocity and acceleration is:
  • \( v = u + at \)
In our scenario:
  • First, the boat accelerates positively at \( 2.01 \, \mathrm{m/s^2} \).
  • Then, it continues with a reduced positive acceleration of \( 0.518 \, \mathrm{m/s^2} \).
  • Finally, the boat experiences a negative acceleration or deceleration of \( -1.49 \, \mathrm{m/s^2} \).
The overall behavior of the speedboat is shaped by these different accelerations. Acceleration affects not just the speed but also the displacement and how far the boat travels during each phase. Keeping track of the sign of the acceleration is crucial, as it informs whether the velocity increases or decreases with time.
Displacement
Displacement refers to the change in position of an object. It's not just about how far an object traveled but also its direction. The formula used to find the displacement during an interval of accelerated motion is:
  • \( s = ut + \frac{1}{2}at^2 \)
where:
  • \( s \) is the displacement,
  • \( u \) is the initial velocity,
  • \( a \) is the acceleration,
  • \( t \) is the time duration of motion.
For our speedboat, this equation is applied three times, each for different phases with varying accelerations. By summing these calculated displacements:
  • First part: \( 49.245 \, \mathrm{m} \)
  • Second part: \( 90.726 \, \mathrm{m} \)
  • Third part: \( 71.424 \, \mathrm{m} \)
we find the total displacement of the boat. Thus, the total displacement, which represents how far from its start the boat finally is, calculates as \( 211.395 \, \mathrm{m} \). Displacement is essential for understanding the actual journey and final position of an object in motion.

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Most popular questions from this chapter

Consult Interactive Solution \(2.17\) at before beginning this problem. A car is traveling along a straight road at a velocity of \(+36.0 \mathrm{~m} / \mathrm{s}\) when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is \(\bar{a}_{1}\). For the next six seconds the car slows down further, and its average acceleration is \(\bar{a}_{2}\). The velocity of the car at the end of the eighteen-second period is \(+28.0 \mathrm{~m} / \mathrm{s}\). The ratio of the average acceleration values is \(\bar{a}_{1} / \bar{a}_{2}=1.50\). Find the velocity of the car at the end of the initial twelve-second interval.

ssm The greatest height reported for a jump into an airbag is \(99.4 \mathrm{~m}\) by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of \(39 \mathrm{~m} / \mathrm{s}(88 \mathrm{mi} / \mathrm{h})\). To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.

A train has a length of \(92 \mathrm{~m}\) and starts from rest with a constant acceleration at time \(t=0 \mathrm{~s} .\) At this instant, a car just reaches the end of the train. The car is moving with a constant velocity. At a time \(t=14 \mathrm{~s},\) the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time \(t=28 \mathrm{~s},\) the car is again at the rear of the train. Find the magnitudes of (a) the car's velocity and (b) the train's acceleration.

Interactive Solution \(\underline{2.31}\) at offers help in modeling this problem. A car is traveling at a constant speed of \(33 \mathrm{~m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is \(2.5 \mathrm{~km}\) away?

Two cars cover the same distance in a straight line. Car \(\mathrm{A}\) covers the distance at a constant velocity. Car B starts from rest, maintains a constant acceleration, and covers the distance in the same time as car A. (a) Is car B's final velocity smaller than, equal to, or greater than car A's constant velocity? Explain. (b) Is car B's average velocity smaller than, equal to, or greater than car A's constant velocity? Why? (c) Given a value for the time to cover the distance and the fact that car B starts from rest, what other variable must be known before car B's acceleration can be determined? Provide a reason for your answer. Problem Both cars cover a distance of \(460 \mathrm{~m}\) in \(210 \mathrm{~s}\). Assume that they are moving in the \(+x\) direction. Determine (a) the constant velocity of car \(\mathrm{A},\) (b) the final velocity of car \(\mathrm{B},\) and \((\mathrm{c})\) the acceleration of car \(\mathrm{B}\). Verify that your answers are consistent with your answers to the Concept Questions.

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