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Chapter 2: Problem 53

The acceleration of a motorcycle is given by \(a_{x}(t)=A t-B t^{2},\) where \(A=1.50 \mathrm{m} / \mathrm{s}^{3}\) and \(B=0.120 \mathrm{m} / \mathrm{s}^{4}\). The motorcycle is at rest at the origin at time \(t=0 .\) (a) Find its position and velocity as functions of time. (b) Calculate the maximum velocity it attains.

Short Answer

Expert verified
Position: \(x(t)=\frac{A}{6}t^3-\frac{B}{12}t^4\); Velocity: \(v(t)=\frac{A}{2}t^2-\frac{B}{3}t^3\); Max Velocity: \(\frac{A^3}{6B^2}\).

Step by step solution

01

Find the velocity function

To find the velocity function, we need to integrate the acceleration function with respect to time. The acceleration function is given as \(a_x(t) = A t - B t^2\). Integrating it, we get the velocity function:\[ v_x(t) = \int (A t - B t^2) \, dt = \frac{A}{2} t^2 - \frac{B}{3} t^3 + C \]Since the motorcycle is 'at rest' at \(t=0\), \(v_x(0) = 0\). This implies that \(C = 0\). Thus, the velocity function is:\[ v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3 \].
02

Find the position function

To find the position function, integrate the velocity function with respect to time. We know:\[ v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3 \].Integrating this, we obtain:\[ x(t) = \int \left( \frac{A}{2} t^2 - \frac{B}{3} t^3 \right) dt = \frac{A}{6} t^3 - \frac{B}{12} t^4 + D \].Since the motorcycle is 'at the origin' at \(t=0\), \(x(0) = 0\). This implies \(D = 0\). Thus, the position function is:\[ x(t) = \frac{A}{6} t^3 - \frac{B}{12} t^4 \].
03

Find the time when the velocity is maximum

The velocity will be maximum when its derivative (acceleration) is zero. We have:\[ v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3 \].Taking the derivative:\[ a_x(t) = \frac{d}{dt} \left( \frac{A}{2} t^2 - \frac{B}{3} t^3 \right) = A t - B t^2 \].Set the acceleration equal to zero and solve for \(t\):\[ A t - B t^2 = 0 \].This factors to:\[ t(A - B t) = 0 \].Thus, \(t = 0\) or \(t = \frac{A}{B}\). We disregard \(t = 0\) because we are looking for a maximum after the initial start. Hence, \(t = \frac{A}{B}\).
04

Calculate the maximum velocity

Substitute \(t = \frac{A}{B}\) into the velocity function to calculate its value at that time:\[ v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3 \].Thus,\[ v_x\left(\frac{A}{B}\right) = \frac{A}{2} \left(\frac{A}{B}\right)^2 - \frac{B}{3} \left(\frac{A}{B}\right)^3 \].Simplifying,\[ v_x\left(\frac{A}{B}\right) = \frac{A^3}{2B^2} - \frac{A^3}{3B^2} = \frac{3A^3}{6B^2} - \frac{2A^3}{6B^2} = \frac{A^3}{6B^2} \].

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Key Concepts

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

Acceleration Function
Acceleration is the rate at which the velocity of an object changes with time. In this exercise, the acceleration function of the motorcycle is given by \( a_x(t) = A t - B t^2 \). This represents the acceleration as a function of time where \( A \) and \( B \) are constants.

Key points to remember about acceleration functions:
  • The function can show how acceleration varies over time.
  • A positive acceleration means an increase in velocity, whereas a negative value (often referred to as deceleration) indicates a decrease.
  • The terms \( At \) and \(-Bt^2 \) represent linear and quadratic components, respectively. As time increases, the squared term can eventually dominate, leading to negative acceleration, or deceleration.
  • Most importantly in this exercise, it affects how we calculate subsequent functions like velocity and position.
Velocity Function
Velocity is the rate of change of position. In order to find the velocity function, we integrate the acceleration function with respect to time.
The given function for velocity \( v_x(t) \) is derived from integrating \( a_x(t) = A t - B t^2 \). The integration leads to:
\[ v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3 + C \]
Where \( C \) is the constant of integration.
  • To solve for \( C \), we use the initial condition. Since the motorcycle is at rest at \( t = 0 \), we know \( v_x(0) = 0 \). Therefore, \( C = 0 \).
  • This results in the simplified function: \( v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3 \).
  • The velocity function reveals how an object's velocity changes over time, considering the initial acceleration.
Understanding velocity functions is crucial for predicting the future position of the object, which is the next step.
Position Function
The position function indicates the location of an object over time. To derive it, we integrate the velocity function \( v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3 \) with respect to time. This results in:
\[ x(t) = \frac{A}{6} t^3 - \frac{B}{12} t^4 + D \]
  • The constant \( D \) is determined using the initial condition \( x(0) = 0 \), giving us \( D = 0 \).
  • Thus, the position function becomes: \[ x(t) = \frac{A}{6} t^3 - \frac{B}{12} t^4 \]
  • This function helps in knowing where the motorcycle is at any given time \( t \).
Position functions provide insight into the path and distance travelled by an object, based on its velocity over time.
Integration in Physics
Integration is a fundamental tool in physics, used to find quantities like velocity or position when given acceleration. It involves the reverse process of differentiation and allows us to accumulate values over time.
  • In this context, it's used first to find the velocity by integrating the acceleration function.
  • Subsequently, the position function is obtained from the velocity through another integration.
  • It requires considering initial conditions (such as starting at rest or the origin) to solve for constants of integration.
  • In physics, integration helps in constructing a thorough understanding of motion and dynamics by linking various motion parameters such as speed, displacement, and time.
This exercise showcases how integration connects each of these aspects and is crucial in solving motion-related physics problems.
Maximum Velocity Calculation
Calculating the maximum velocity involves determining when the velocity function reaches its highest point. This happens when the acceleration is zero.
  • We find when \( a_x(t) = A t - B t^2 = 0 \). Solving, we find \( t = \frac{A}{B} \).
  • The time \( t = 0 \) is ignored, as it refers to the starting point where the motorcycle was at rest.
  • By substituting \( t = \frac{A}{B} \) into the velocity function \( v_x(t) = \frac{A}{2} t^2 - \frac{B}{3} t^3 \), we find the maximum velocity:
    \[ v_x\left(\frac{A}{B}\right) = \frac{A^3}{6B^2} \]
Understanding this calculation is crucial as it shows not just when but how fast at maximum speed the motorcycle reaches. The result of this calculation provides crucial insights into the dynamics of motion entailed.

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Most popular questions from this chapter

Launch of the Space Shuttle. At launch the space shuttle weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 \(\mathrm{km} / \mathrm{h}\) , and at the end of the first 1.00 \(\mathrm{min}\) its speed is 1610 \(\mathrm{km} / \mathrm{h}\) . (a) What is the average acceleration (\(\operatorname{in}\) \(\mathrm{m} / \mathrm{s}^{2}\) ) of the shuttle (i) during the first \(8.00 \mathrm{s},\) and (ii) between 8.00 \(\mathrm{s}\) and the end of the first 1.00 \(\mathrm{min}\) ? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals, what distance does the shuttle travel (i) during the first \(8.00 \mathrm{s},\) and ( ii) during the interval from 8.00 s to 1.00 \(\mathrm{min}\)?

A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time sled is 14.4 \(\mathrm{m}\) from the top, 2.00 s after that it is 25.6 \(\mathrm{m}\) from the top, 2.00 s later 40.0 \(\mathrm{m}\) from the top, and 2.00 s later it is 57.6 \(\mathrm{m}\) from the top. (a) What is the magnitude of the average velocity of the sled during each of the 2.00 -s intervals after passing the \(14.4-\mathrm{m}\) point? (b) What is the acceleration of the sled? (c) What is the speed of the sled when it passes the \(14.4-\mathrm{m}\) point? (d) How much time did it take to go from the top to the \(14.4-\mathrm{m}\) point? (e) How far did the sled go during the first second after passing the \(14.4-\mathrm{m}\) point?

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.420 s to pass from the top to the bottom of this window, which is 1.90 \(\mathrm{m}\) high. How far is the top of the window below the windowsill from which the flowerpot fell?

Determined to test the law of gravity for himself, a student walks off a skyscraper 180 \(\mathrm{m}\) high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student. Superman leaves the roof with an initial speed \(v_{0}\) that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. (a) What must the value of \(v_{0}\) be so that Superman catches the student just before they reach the ground? (b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a). (c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?

A ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 \(\mathrm{m}\) during the second 5.0 \(\mathrm{s}\) of its motion. How far did it roll during the first 5.0 s of motion?
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