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Chapter 2: Problem 37

Suppose that the velocity function of a particle moving along a line is \(v(t)=3 t^{3}+2\). Find the average velocity of the particle over the time interval \(1 \leq \mathrm{t} \leq 4\) by integrating.

Short Answer

Expert verified
Answer: The average velocity is \(\frac{789}{12}\).

Step by step solution

01

Integrate the velocity function

First, we need to integrate the given velocity function \(v(t)=3t^3+2\) with respect to \(t\) to obtain the displacement function. To integrate, we apply the power rule for integration: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where \(n\) is a constant and \(C\) is the constant of integration. Using the power rule, the integrated velocity function (displacement function) is: \( s(t) = \int v(t)dt = \int(3t^3 + 2)dt = \frac{3t^4}{4} + 2t + C\).
02

Determine the limits of integration

The given time interval is \(1 \leq t \leq 4\). These will be our limits of integration when integrating the velocity function to find the total displacement over the time interval.
03

Calculate the total displacement

To find the total displacement, we will evaluate the displacement function between the limits of integration. That means, we will subtract the displacement at the initial time (\(t=1\)) from the displacement at the final time (\(t=4\)): Total Displacement = \(s(4) - s(1) = \left(\frac{3(4)^4}{4} + 2(4)\right) - \left(\frac{3(1)^4}{4} + 2(1)\right)\). Let's evaluate this expression: Total Displacement = \((192 + 8) - (\frac{3}{4} + 2) = 200 - \frac{11}{4} = \frac{789}{4}\)
04

Calculate the average velocity

The average velocity is the total displacement divided by the length of the time interval. The length of the time interval is \((4 - 1)=3\). So, we will divide the total displacement by 3: Average Velocity = \(\frac{\text{Total Displacement}}{\text{Length of Time Interval}} = \frac{789/4}{3} = \frac{789}{12}\). Hence, the average velocity of the particle over the time interval \(1 \leq t \leq 4\) is \(\frac{789}{12}\).

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Most popular questions from this chapter

Starting from \(\frac{1}{1+x}-1+x-x^{2}+\ldots+x^{2 n-1}=\frac{x^{2 n}}{1+x}\) show that \(t-\frac{t^{2}}{2}+\frac{t^{2}}{3}-\ldots-\frac{t^{2 n}}{2 n} \leq \ln (1+t) \leq t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-+\frac{t^{2 n+1}}{2 n+1}\) for \(\mathrm{t} \geq 0\).

Evaluate the following integrals : (i) \(\int_{1}^{\infty} \frac{d x}{x^{2}(x+1)}\) (ii) \(\int_{0}^{\infty} x^{3} e^{-x^{2}} d x\) (iii) \(\int_{0}^{\frac{1}{6}} \frac{\mathrm{dx}}{\mathrm{x} \ln ^{2} \mathrm{x}}\) (iv) \(\int_{-\infty}^{\infty} \frac{d x}{x^{2}+2 x+2}\)

Let \(\mathrm{P}_{\mathrm{n}}\) denote the polynomial of degree \(\mathrm{n}\) given by \(\mathrm{P}_{\mathrm{n}}(\mathrm{x})=\mathrm{x}+\frac{\mathrm{x}^{2}}{2}+\frac{\mathrm{x}^{3}}{3}+\ldots .+\frac{\mathrm{x}^{\mathrm{n}}}{\mathrm{n}}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{x}^{\mathrm{k}}}{\mathrm{k}}\). Then, for every \(x<1\) and every \(n \geq 1\), prove that \(-\ln (1-x)=P_{n}(x)+\int_{0}^{x} \frac{u^{n}}{1-u} d u\)

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Showthat \(\int_{0}^{\pi} \frac{\ell \mathrm{n}(1+\mathrm{a} \cos \mathrm{x})}{\cos \mathrm{x}} \mathrm{dx}=\pi \sin ^{-1} \mathrm{a},(|\mathrm{a}|<1)\)
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