Problem 45 - Advanced Engineering Mathematics

Question

(a) \((i)\) If \(s(t)\) is distance measured down the plane from the highest point, then \(d s / d t=v .\) Integrating \(d s / d t=16 t\) gives \(s(t)=8 t^{2}+c_{2} .\) Using \(s(0)=0\) then gives \(c_{2}=0 .\) Now the length \(L\) of the plane is \(L=50 / \sin 30^{\circ}=100 \mathrm{ft} .\) The time it takes the box to slide completely down the plane is the solution of \(s(t)=100\) or \(t^{2}=25 / 2,\) so \(t \approx 3.54 \mathrm{s}\). \((i i)\) Integrating \(d s / d t=4 t\) gives \(s(t)=2 t^{2}+c_{2} .\) Using \(s(0)=0\) gives \(c_{2}=0,\) so \(s(t)=2 t^{2}\) and the solution of \(s(t)=100\) is now \(t \approx 7.07 \mathrm{s}\). \((\text {iii})\) Integrating \(d s / d t=48-48 e^{-t / 12}\) and using \(s(0)=0\) to determine the constant of integration, we \(\operatorname{obtain} s(t)=48 t+576 e^{-t / 12}-576 .\) With the aid of a CAS we find that the solution of \(s(t)=100,\) or $$\begin{aligned} &100=48 t+576 e^{-t / 12}-576 &\text { or } \quad 0=48 t+576 e^{-t / 12}-676 \end{aligned}$$ is now \(t \approx 7.84 \mathrm{s}\). (b) The differential equation \(m d v / d t=m g \sin \theta-\mu m g \cos \theta\) can be written $$m \frac{d v}{d t}=m g \cos \theta(\tan \theta-\mu)$$. If \(\tan \theta=\mu, d v / d t=0\) and \(v(0)=0\) implies that \(v(t)=0 .\) If \(\tan \theta< \mu\) and \(v(0)=0,\) then integration implies \(v(t)=g \cos \theta(\tan \theta-\mu) t < 0\) for all time \(t\). (c) since \(\tan 23^{\circ}=0.4245\) and \(\mu=\sqrt{3} / 4=0.4330,\) we see that \(\tan 23^{\circ}<0.4330 .\) The differential equation is \(d v / d t=32 \cos 23^{\circ}\left(\tan 23^{\circ}-\sqrt{3} / 4\right)=-0.251493 .\) Integration and the use of the initial condition gives \(v(t)=-0.251493 t+1 .\) When the box stops, \(v(t)=0\) or \(0=-0.251493 t+1\) or \(t=3.976254 \mathrm{s}\). From \(s(t)=-0.125747 t^{2}+t\) we find \(s(3.976254)=1.988119 \mathrm{ft}\). (d) With \(v_{0} >0, v(t)=-0.251493 t+v_{0}\) and \(s(t)=-0.125747 t^{2}+v_{0} t .\) Because two real positive solutions of the equation \(s(t)=100,\) or \(0=-0.125747 t^{2}+v_{0} t-100,\) would be physically meaningless, we use the quadratic formula and require that \(b^{2}-4 a c=0\) or \(v_{0}^{2}-50.2987=0 .\) From this last equality we find \(v_{0} \approx 7.092164 \mathrm{ft} / \mathrm{s} .\) For the time it takes the box to traverse the entire inclined plane, we must have \(0=-0.125747 t^{2}+7.092164 t-100 .\) Mathematica gives complex roots for the last equation: \(t=\) \(28.2001 \pm 0.0124458 i .\) But, for $$0=-0.125747 t^{2}+7.092164691 t-100$$ the roots are \(t=28.1999 \mathrm{s}\) and \(t=28.2004 \mathrm{s} .\) So if \(v_{0} >7.092164,\) we are guaranteed that the box will slide completely down the plane.

Step-by-Step Solution

Answer

The box takes approximately 7.84 s to slide down when velocity is described by part a-iii conditions. With initial velocity over 7.092164 ft/s, it successfully traverses the plane in about 28.20 s.

Step 1: Understand the problem setup

We are analyzing the motion of a box sliding down a plane and solving various conditions of the motion. The exercise includes determining how long it will take for the box to slide down the plane under different conditions.

Step 2: Integrate velocity for distance (part a-i)

For part (a)(i), we start with the equation given: \( \frac{ds}{dt} = 16t \). Integrating this, we get \( s(t) = 8t^{2} + c_{2} \). Since \( s(0) = 0 \), the constant \( c_{2} = 0 \). Therefore, \( s(t) = 8t^2 \). The length of the plane is 100 ft, leading to the equation \( 8t^2 = 100 \). Solving for \( t \), we find \( t \approx 3.54 \text{s} \).

Step 3: Integrate velocity for distance (part a-ii)

For part (a)(ii), starting with \( \frac{ds}{dt} = 4t \), integration gives \( s(t) = 2t^2 + c_{2} \). Applying the condition \( s(0) = 0 \), we find \( c_{2} = 0 \), hence \( s(t)=2t^2 \). Solving \( 2t^2 = 100 \), we get \( t \approx 7.07 \text{s} \).

Step 4: Integrate complex velocity equation (part a-iii)

For part (a)(iii), integrate \( \frac{ds}{dt} = 48 - 48e^{-t/12} \) to obtain \( s(t) = 48t + 576e^{-t/12} - 576 \). With \( s(0)=0 \), this equation holds. Setting \( s(t)=100 \) and solving using a calculator, we find \( t \approx 7.84 \text{s} \).

Step 5: Analyze force balance for box motion (part b)

The equation \( m \frac{dv}{dt} = mg \sin \theta - \mu mg \cos \theta \) is analyzed. If \( \tan \theta = \mu \), then \( \frac{dv}{dt} = 0 \), indicating \( v(t)=0 \). If \( \tan \theta < \mu \), then \( v(t) = g \cos \theta(\tan \theta - \mu)t < 0 \).

Step 6: Specific angle analysis (part c)

Given \( \tan 23^\circ = 0.4245\) and \( \mu = \sqrt{3}/4 = 0.4330 \), since \( \tan 23^\circ < 0.4330 \), \( \frac{dv}{dt} = -0.251493 \). Integration results in \( v(t) = -0.251493t + 1 \). Solving for when \( v(t) = 0 \), we find \( t \approx 3.98 \text{s} \). The distance is \( s(t)=-0.125747t^{2}+t \), and for \( t=3.98 \), \( s \approx 1.99 \text{ft} \).

Step 7: Initial velocity condition and its effect (part d)

With \( v_0 > 0 \), the velocity is \( v(t) = -0.251493 t + v_0 \) and distance \( s(t) = -0.125747 t^2 + v_0 t \). Solving \( s(t)=100 \) requires \( v_0^2 - 50.2987 = 0 \), leading to \( v_0 \approx 7.092164 \text{ft/s} \). For \( v_0 > 7.092164 \), the box will completely slide down. Solutions are complex without this velocity, showing the critical point \( t \approx 28.20s \).

Key Concepts

Integration Techniques

Integration is a powerful mathematical tool used in solving differential equations that describe motion, among other concepts. In this exercise, integration converts velocity expressed as a function of time into a distance function. This process involves finding the antiderivative of the velocity equation.

• For example, if we start with a differential equation like \( \frac{ds}{dt} = 16t \), integrating it gives us the distance function \( s(t) = 8t^2 + c_2 \), where \( c_2 \) is the constant of integration determined by initial conditions.
• In situations with distinct velocity functions, you may also encounter the integration of more complex terms, such as exponential terms, which require specialized techniques for simplification.

Knowing how to handle these calculations is crucial for accurately describing the movement of an object.

Velocity and Distance

Understanding the relationship between velocity and distance is essential in motion analysis, especially when an object slides down a plane. The velocity function can help determine how far an object travels within a given time.

• The key principle rests on recognizing that velocity is the rate of change of distance over time—expressed mathematically as \( \frac{ds}{dt} \).
• Once the velocity equation is given, integrating it helps us find the distance function, \( s(t) \), that reveals how far the object travels as time progresses.

By understanding this relationship, one can predict the movement over time, solving for specific times or distances when required.

Motion Analysis

Motion analysis involves understanding how different forces and conditions affect an object's trajectory. Often, this is captured by a differential equation such as \( m \frac{dv}{dt} = mg \sin \theta - \mu mg \cos \theta \).

• By solving these equations, one can determine an object's velocity at any given time and how long it will take to reach a certain point.
• This involves identifying whether an object will move at all under certain conditions, such as when \( \tan \theta = \mu \) causing no acceleration.

Motion analysis thus allows us to predict movement patterns and timescales for different scenarios.

Initial Conditions in Physics

In physics, initial conditions are references assumed at the start of the event, which significantly affect the overall motion analysis. These conditions determine the fixed values of integration constants.

• For example, by setting \( s(0) = 0 \), you ensure that the initial distance traveled is zero, solving for \( c_2 \) with ease.
• Similarly, knowing initial velocity \( v(0) \) helps in defining the trajectory and predicting long-term motion, especially when there's initial movement which can affect how an object interacts with forces.

Initial conditions are thus critical for predicting accurate motion descriptions, ensuring solutions to equations reflect real-world scenarios.