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Chapter 1: Problem 2

Repeat Example 1.2. Compute the velocity to \(t=10 \mathrm{s}\), with a step size of (a) 1 and (b) 0.5 s. Can you make any statement regarding the errors of the calculation based on the results?

Short Answer

Expert verified
The velocity formula is given by \(v(t) = gt\), where g 鈮� -9.8 m/s虏. For both step sizes (1s and 0.5s), the velocity at \(t = 10s\) is -98 m/s. No difference is observed in the results because the velocity function is linear with respect to time, and using a smaller step size does not lead to a noticeable improvement in accuracy in this case.

Step by step solution

01

Write down the formula for velocity

The formula for velocity as a function of time is given by: \(v(t) = gt\), where \(g\) is the acceleration due to gravity (approximately -9.8 m/s虏) and \(t\) is the time.
02

Calculate velocity with a step size of 1s

We will calculate the velocity for each time step from \(t=0\) to \(t=10s\) using the step size of 1s. To do this, substitute the time steps into the formula and calculate the corresponding velocities: - \(v(0) = -9.8 \cdot 0 = 0\) - \(v(1) = -9.8 \cdot 1 = -9.8\) - \(v(2) = -9.8 \cdot 2 = -19.6\) - \(\dots\) - \(v(10) = -9.8 \cdot 10 = -98\) The velocity at \(t = 10s\) with a step size of 1s is \(v = -98 m/s\).
03

Calculate velocity with a step size of 0.5s

Next, we will calculate the velocity for each time step from \(t=0\) to \(t=10s\) using the step size of 0.5s: - \(v(0) = -9.8 \cdot 0 = 0\) - \(v(0.5) = -9.8 \cdot 0.5 = -4.9\) - \(v(1) = -9.8 \cdot 1 = -9.8\) - \(v(1.5) = -9.8 \cdot 1.5 = -14.7\) - \(\dots\) - \(v(10) = -9.8 \cdot 10 = -98\) The velocity at \(t = 10s\) with a step size of 0.5s is \(v = -98 m/s\).
04

Compare the results and discuss errors

For both step sizes (1s and 0.5s), the computed velocity at \(t = 10s\) is -98 m/s. As a result, there doesn't seem to be any difference between the calculations using the given step sizes. However, by using smaller step sizes, we generally expect to have increased accuracy in our calculation, especially for functions that have more complex or non-linear behavior. In this case, since the velocity function is linear with respect to time, using a smaller step size does not lead to any noticeable improvement in accuracy, and hence, both calculations yield the same value.

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Most popular questions from this chapter

In our example of the free-falling parachutist, we assumed that the acceleration due to gravity was a constant value of \(9.8 \mathrm{m} / \mathrm{s}^{2}\) Although this is a decent approximation when we are examining falling objects near the surface of the earth, the gravitational force decreases as we move above sea level. A more general representation based on Newton's inverse square law of gravitational attraction can be written as $$g(x)=g(0) \frac{R^{2}}{(R+x)^{2}}$$ where \(g(x)=\) gravitational acceleration at altitude \(x\) (in \(\mathrm{m}\) ) measured upwards from the earth's surface \(\left(\mathrm{m} / \mathrm{s}^{2}\right), g(0)=\) gravitational acceleration at the earth's surface \(\left(\cong 9.8 \mathrm{m} / \mathrm{s}^{2}\right),\) and \(R=\) the earth's radius \(\left(\cong 6.37 \times 10^{6} \mathrm{m}\right)\) (a) In a fashion similar to the derivation of Eq. (1.9) use a force balance to derive a differential equation for velocity as a function of time that utilizes this more complete representation of gravitation. However, for this derivation, assume that upward velocity is positive. (b) For the case where drag is negligible, use the chain rule to express the differential equation as a function of altitude rather than time. Recall that the chain rule is $$\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}$$ (c) Use calculus to obtain the closed form solution where \(v=v_{0}\) at \(x=0\) (d) Use Euler's method to obtain a numerical solution from \(x=0\) to \(100,000 \mathrm{m}\) using a step of \(10,000 \mathrm{m}\) where the initial velocity is \(1400 \mathrm{m} / \mathrm{s}\) upwards. Compare your result with the analytical solution.

Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature), $$\frac{d T}{d t}=-k\left(T-T_{a}\right)$$ where \(T=\) the temperature of the body \(\left(^{\circ} \mathrm{C}\right), t=\operatorname{time}(\min ), k=\) the proportionality constant (per minute), and \(T_{a}=\) the ambient temperature \(\left(^{\circ} \mathrm{C}\right)\). Suppose that a cup of coffee originally has a temperature of \(68^{\circ} \mathrm{C}\). Use Euler's method to compute the temperature from \(t=0\) to 10 min using a step size of 1 min if \(T_{a}=21^{\circ} \mathrm{C}\) and \(k=\) \(0.1 / \mathrm{min}\).

Water accounts for roughly \(60 \%\) of total body weight. Assuming it can be categorized into six regions, the percentages go as follows. Plasma claims \(4.5 \%\) of the body weight and is \(7.5 \%\) of the total body water. Dense connective tissue and cartilage occupies \(4.5 \%\) of the total body weight and \(7.5 \%\) of the total body water. Interstitial lymph is \(12 \%\) of the body weight, which is \(20 \%\) of the total body water. Inaccessible bone water is roughly \(7.5 \%\) of the total body water and \(4.5 \%\) total body weight. If intracellular water is \(33 \%\) of the total body weight and transcellular water is \(2.5 \%\) of the total body water, what percent of total body weight must the transcellular water be and what percent of total body water must the intracellular water be?

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area. $$\frac{d V}{d t}=-k A$$ where \(V=\) volume \(\left(\mathrm{mm}^{3}\right), t=\) time \((\min ), k=\) the evaporation rate \((\mathrm{mm} / \mathrm{min}),\) and \(A=\) surface area \(\left(\mathrm{mm}^{2}\right) .\) Use Euler's method to compute the volume of the droplet from \(t=0\) to 10 min using a step size of 0.25 min. Assume that \(k=0.1 \mathrm{mm} / \mathrm{min}\) and that the droplet initially has a radius of \(3 \mathrm{mm}\). Assess the validity of your results by determining the radius of your final computed volume and verifying that it is consistent with the evaporation rate.

A storage tank contains a liquid at depth \(y\) where \(y=0\) when the tank is half full. Liquid is withdrawn at a constant flow rate \(Q\) to meet demands. The contents are resupplied at a sinusoidal rate \(3 Q \sin ^{2}(t)\) Equation (1.13) can be written for this system as $$\begin{aligned} &\frac{d(A y)}{d x}=3 Q \sin ^{2}(t)-Q\\\ &\left(\begin{array}{c} \text { change in } \\ \text { volume } \end{array}\right)=(\text { inflow })-(\text { outflow }) \end{aligned}$$ or, since the surface area \(A\) is constant $$\frac{d y}{d x}=3 \frac{Q}{A} \sin ^{2}(t)-\frac{Q}{A}$$ Use Euler's method to solve for the depth \(y\) from \(t=0\) to 10 d with a step size of \(0.5 \mathrm{d} .\) The parameter values are \(A=1200 \mathrm{m}^{2}\) and \(Q=500 \mathrm{m}^{3} / \mathrm{d} .\) Assume that the initial condition is \(y=0\)
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