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exis [7]
3 years ago
9

If a ball is thrown straight up into the air with an initial velocity of 100 ft/s, it height in feet after t second is given by

y = 100t – 16.
a.) Find the average velocity for the time period begining when t = 2 and lasting for the given amount of time.

0.5 seconds


(ii) 0.1 seconds.


(iii) 0.01 seconds.


b.) Based on the above results, guess what the instantaneous velocity of the ball is when t = 2 seconds.
Answer:
Mathematics
1 answer:
Sergeu [11.5K]3 years ago
4 0

Answer:

11 0.1 seconds

Step-by-step explanation:

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Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.
andriy [413]

Answer:

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Step-by-step explanation:

We have been given a indefinite integral \int \frac{2}{3x\left(3x-5\right)}dx. We are asked to find the indefinite integral.

We will use partial fraction formula to solve our given problem.

\frac{2}{3x\left(3x-5\right)}=\frac{3}{5(3x-5)}-\frac{1}{5x}

\int \frac{2}{3x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx

\frac{2}{3}\int \frac{1}{x\left(3x-5\right)}dx=\frac{2}{3}\int \frac{3}{5(3x-5)}-\frac{1}{5x}dx

Using difference rule of integrals, we will get:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)

Now, we need to use u-substitution as:

Let u=3x-5.

\frac{du}{dx}=3

dx=\frac{1}{3}du

\int \frac{3}{5(3x-5)}dx= \frac{3}{5}\int \frac{1}{(u)}*\frac{1}{3}du=\frac{3}{5}*\frac{1}{3}\int \frac{1}{(u)}du=\frac{1}{5}ln|u|=\frac{1}{5}ln|3x-5|

\int \frac{1}{5x}dx=\frac{1}{5}\int \frac{1}{x}dx=\frac{1}{5}ln|x|

Substitute back these values:

\frac{2}{3}(\int \frac{3}{5(3x-5)}dx-\int \frac{1}{5x}dx)=\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)

Let us add a constant C.

\frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C

Therefore, our required integral would be \frac{2}{3}(\frac{1}{5}ln|3x-5|-\frac{1}{5}ln|x|)+C.

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3 years ago
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kondaur [170]

Answer:

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The correct answer is the second one.

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