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Calculus 1 : Velocity

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Example Question #381 : Velocity

The velocity function of a particle and a position of this particle at a known time are given by v(t)=27sin^3(t^3) and p(0)=3 . Approximate p(3) using Euler's Method and three steps.

Possible Answers:

-33.08

-72.71

45.24

19.09

-22.91

Correct answer:

45.24

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

In the case of this problem, this can be rewritten as:

$p(t_n)=p(t_{n+1}) +\Delta t v(t_n)$

To calculate the step size find the difference between the final and initial value of and divide by the number of steps to be used:

$\Delta t = \frac{t_f-t_i}{Steps}$

For this problem, we are told v(t)=27sin^3(t^3) and p(0)=3

Knowing this, we may take the steps to estimate our function value at our desired value:

$\Delta t = \frac{3-0}{3}=1$

p_0=3;t_0=0

p_1=3+(1)27sin^3(0^3)=3

p_2=3+(1)27sin^3(1^3)=19.09

p_3=19.09+(1)27sin^3(2^3)=45.24

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Example Question #382 : Calculus

The velocity function of a particle and a position of this particle at a known time are given by $v(t)=\frac{sin(t^3)}{6}$ and $p(\pi)=0$ . Approximate $p(4\pi)$ using Euler's Method and three steps.

Possible Answers:

-0.660

-0.138

0.384

-0.209

Correct answer:

0.384

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

In the case of this problem, this can be rewritten as:

$p(t_n)=p(t_{n+1}) +\Delta t v(t_n)$

To calculate the step size find the difference between the final and initial value of and divide by the number of steps to be used:

$\Delta t = \frac{t_f-t_i}{Steps}$

For this problem, we are told $v(t)=\frac{sin(t^3)}{6}$ and $p(\pi)=0$

Knowing this, we may take the steps to estimate our function value at our desired value:

$\Delta t = \frac{4\pi-\pi}{3}=\pi$

$p_0=0;t_0=\pi$

$p_1=0+(\pi)\frac{sin(\pi^3)}{6}=-0.209$

$p_2=-0.209+(\pi)\frac{sin((2\pi)^3)}{6}=-0.138$

$p_3=-0.138+(\pi)\frac{sin((3\pi)^3)}{6}=0.384$

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Example Question #381 : Velocity

The position of a t=4 is given by the following functions:

p(t) = 2t + 3

Find the velocity.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: p'(t) = v(t)

In this case, the position function is: p(t) = 2t + 3

Then take the derivative of the position function to get the velocity function: v(t) = 2

Then, plug t=4 into the velocity function: v(4) = 2

Therefore, the answer is:

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Example Question #382 : Velocity

The position of a t=2 is given by the following functions:

$p(t) = - \cos(3t - 6)$

Find the velocity.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: p'(t) = v(t)

In this case, the position function is: $p(t) = - \cos(3t - 6)$

Then take the derivative of the position function to get the velocity function: $v(t) = 3\sin(3t - 6)$

Then, plug t=2 into the velocity function: $v(2) = 3\sin(3({\color{Blue} 2}) - 6)$

Therefore, the answer is:

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Example Question #383 : Velocity

The position of a t=1 is given by the following functions:

p(t) = ln(4t^2 + 6t)

Find the velocity.

Possible Answers:

$\frac{7}{5}$

$\frac{5}{7}$

Answer not listed

$\frac{3}{4}$

$\frac{2}{5}$

Correct answer:

$\frac{7}{5}$

Explanation:

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: p'(t) = v(t)

In this case, the position function is: p(t) = ln(4t^2 + 6t)

Then take the derivative of the position function to get the velocity function: $v(t) = \frac{8t + 6}{4t^2+6t}$

Then, plug t=1 into the velocity function: $v(1) = \frac{8({\color{Blue} 1})+ 6}{4({\color{Blue} 1})^2+6({\color{Blue} 1})}$

Therefore, the answer is: $\frac{7}{5}$

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Example Question #382 : Velocity

The position of a t=0 is given by the following function:

$p(t) = e^{3t-ln(7)}$

Find the velocity.

Possible Answers:

Answer not listed

$\frac{6}{7}$

$\frac{4}{7}$

$\frac{3}{7}$

$\frac{1}{7}$

Correct answer:

$\frac{3}{7}$

Explanation:

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: p'(t) = v(t)

In this case, the position function is: $p(t) = e^{3t-ln(7)}$ = $p(t) = \frac{e^{3t}}{7}$

Then take the derivative of the position function to get the velocity function: $v(t) = \frac{3e^{3t}}{7}$

Then, plug t=0 into the velocity function: $v(0) = \frac{3e^{3({\color{Blue} 0})}}{7}$

Therefore, the answer is: $\frac{3}{7}$

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Example Question #385 : Velocity

The position of a t=2 is given by the following function:

p(t) = 4t^2 - e^t

Find the velocity.

Possible Answers:

8.4

8.6

6.1

Answer not listed

5.3

Correct answer:

8.6

Explanation:

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: p'(t) = v(t)

In this case, the position function is: p(t) = 4t^2 - e^t

Then take the derivative of the position function to get the velocity function: v(t) = 8t - e^t

Then, plug t=2 into the velocity function: $v(2) = 8({\color{Blue} 2}) - e^{{\color{Blue} 2}}$

Therefore, the answer is: 8.6

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Example Question #386 : Velocity

The position of a t=0 is given by the following function:

$p(t) = ln(3t + \sin(3t))$

Find the velocity.

Possible Answers:

Undefined

Answer not listed

Correct answer:

Undefined

Explanation:

In order to find the velocity of a certain point, you first find the derivative of the position function to get the velocity function: p'(t) = v(t)

In this case, the position function is: $p(t) = ln(3t + \sin(3t))$

Then take the derivative of the position function to get the velocity function: $v(t) = \frac{3+ 3\cos(3t)}{3t + \sin(3t)}$

Then, plug t=0 into the velocity function: $v(0) = \frac{3+ 3\cos(3({\color{Blue} 0}))}{3({\color{Blue} 0}) + \sin(3({\color{Blue} 0}))}$

Therefore, the answer is: Undefined

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Example Question #383 : Velocity

Given the position function of a projectile in meters, find the velocity of the object at t=3 seconds.

s(t)=4t^2-8t+3

Possible Answers:

15 m/s

20 m/s

16 m/s

4 m/s

None of the other answers.

Correct answer:

16 m/s

Explanation:

The velocity of the projectile at any given point in time is modeled by the first derivative of the position function.

s'(t)=8t-8=v(t)=velocity

The velocity of the projectile at t=3 is then

v(3)=s'(3)=8(3)-8= 24-8=16 m/s

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Example Question #388 : Velocity

Given the following position function, determine the velocity, , when t=2 :

S(t)= t^3 + 5t

Possible Answers:

V(2)= 21

V(2)=16

V(2)= 15

V(2)=17

Correct answer:

V(2)=17

Explanation:

The velocity can be determined at any given time by taking the first derivative of the position function.

In this case, the derivative of

S(t)=t^3+5t

is the velocity function

V(t)= 3t^2+5 ,

using the power rule

$\frac{\mathrm{d} }{\mathrm{d} x}x^n= nx^{n-1}$ .

The next step is to substitute 2 into the velocity equation for t, and solving to obtain,

V(t)= 3t^2+5

$\\V(2)= 3(2)^2+5 \\V(2)=3(4)+5 \\V(2)=12+5$

V(2)= 17

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Calculus 1 : Velocity

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #381 : Velocity

Example Question #382 : Calculus

Example Question #381 : Velocity

Example Question #382 : Velocity

Example Question #383 : Velocity

Example Question #382 : Velocity

Example Question #385 : Velocity

Example Question #386 : Velocity

Example Question #383 : Velocity

Example Question #388 : Velocity

All Calculus 1 Resources

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