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Calculus 1 : Velocity

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Example Questions

Example Question #341 : Spatial Calculus

A particle's position is described by $s(t)=3t^{2}+2t+5$ , where is in feet, and is in minutes. What is the velocity of the particle after $60\; seconds?$

Possible Answers:

362 feet/second

362 feet/minute

8 feet/minute

10 feet/minute

8 feet/second

Correct answer:

8 feet/minute

Explanation:

The velocity of the particle can be found by taking the derivative of the position function. Then, we simply need to plug in the time to get the final answer. However, the time must be converted to minutes, so that the units match the problem statement.

v(t)=s'(t)=6t+2.

v(t)=6(1minute)+2=8feet/minute .

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Example Question #342 : Spatial Calculus

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

$p(t)= 4t^3 - \cos(t^2) + e^{2t}$

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: $p(t)= 4t^3 - \cos(t^2) + e^{2t}$

Then take the derivative of the position function to get the velocity function: $v(t)= 12t^2 - 2t\cos(t^2) + 2e^{2t}$

Then, plug t= 0 into the velocity function: $v(0)= 12({\color{Blue} 0})^2 + 2({\color{Blue} 0})\sin({\color{Blue} 0}^2) + 2e^{2({\color{Blue} 0})}$

Therefore, the answer is:

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Example Question #343 : Spatial Calculus

Find the velocity of the particle at t=1, its position function given by the following:

$s(t)=t^5+t^3+\cos(t)$

Possible Answers:

$8+\sin(1)$

$8-\sin(1)$

Correct answer:

$8-\sin(1)$

Explanation:

To find the velocity of the particle, we must find the first derivative of the position function:

$v(t)=s'(t)=5t^4+3t^2-\sin(t)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Now, to find the velocity at t=1 simply plug this into the velocity function:

$v(1)=5+3-\sin(1)=8-\sin(1)$

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Example Question #341 : Velocity

The position of a particle is represented by f(t) = 4t³ – 3t + 15

What is its instantaneous velocity at time t = 3?

Possible Answers:

275

284

105

324

None of the other answers

Correct answer:

105

Explanation:

First we must find the simple derivative:

f'(t) = 12t² – 3

The instantaneous velocity is f'(3) or 12 * 3 * 3 – 3 = 105

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Example Question #342 : Velocity

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

$p(t)= \sin(4t + e^t)$

Find the velocity.

Possible Answers:

7.4

3.1

Answer not listed

2.7

6.9

Correct answer:

2.7

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: $p(t)= \sin(4t + e^t)$

Then take the derivative of the position function to get the velocity function: $v(t)= \cos(4t + e^t) (4+e^t)$

Then, plug t=0 into the velocity function: $v(0)= \cos(4({\color{Blue} 0}) + e^{\color{Blue} 0}) (4+e^{\color{Blue} 0})$

Therefore, the answer is: 2.7

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Example Question #346 : Spatial Calculus

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

p(t)= ln(2t+3)

Find the velocity.

Possible Answers:

$\frac{3}{5}$

$\frac{2}{11}$

$\frac{4}{7}$

Answer not listed

$\frac{1}{2}$

Correct answer:

$\frac{2}{11}$

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(2t+3)

Then take the derivative of the position function to get the velocity function: $v(t)= \frac{2}{2t+3}$

Then, plug t=4 into the velocity function: $v(4)= \frac{2}{2({\color{Blue} 4})+3}$

Therefore, the answer is: $\frac{2}{11}$

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Example Question #343 : Calculus

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

$p(t)= -\cos^2 (3t + 5t^4)$

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: $p(t)= -\cos^2 (3t + 5t^4)$

Then take the derivative of the position function to get the velocity function: $v(t)= 2 \sin (3t + 5t^4) (3+20t^3)$

Then, plug t=0 into the velocity function: $v(0)= 2 \sin (3({\color{Blue} 0}) + 5({\color{Blue} 0})^4) (3+20({\color{Blue} 0})^3)$

Therefore, the answer is:

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Example Question #344 : Calculus

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

$p(t)= (6t^3 - 7t^2 + e^{2t})^2$

Find the velocity.

Possible Answers:

32.5

Answer not listed

528.3

126.9

239.9

Correct answer:

239.9

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: $p(t)= (6t^3 - 7t^2 + e^{2t})^2$

Then take the derivative of the position function to get the velocity function: $v(t)= 2(6t^3 - 7t^2 + e^{2t})(18t^2 - 14t + 2e^{2t})$

Then, plug t=1 into the velocity function: $v(1)= 2(6({\color{Blue} 1})^3 - 7({\color{Blue} 1})^2 + e^{2({\color{Blue} 1})})(18({\color{Blue} 1})^2 - 14({\color{Blue} 1}) + 2e^{2({\color{Blue} 1})})$

Therefore, the answer is: 239.9

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Example Question #349 : Spatial Calculus

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

$p(t)= ln(2t) + \cos(2t) - e^{2t}$

Find the velocity.

Possible Answers:

-26.8

98.2

Answer not listed

-15.6

12.4

Correct answer:

-15.6

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: $p(t)= ln(2t) + \cos(2t) - e^{2t}$

Then take the derivative of the position function to get the velocity function: $v(t)= \frac{2}{2t} - 2\sin(2t) - 2e^{2t}$

Then, plug t=1 into the velocity function: $v(1)= \frac{2}{2({\color{Blue}1 })} - 2\sin(2({\color{Blue} 1})) - 2e^{2({\color{Blue} 1})}$

Therefore, the answer is: -15.6

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Example Question #350 : Spatial Calculus

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

p(t)= (6t^2+5t)^2

Find the velocity.

Possible Answers:

1972

2145

7821

Answer not listed

3961

Correct answer:

1972

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)= (6t^2+5t)^2

Then take the derivative of the position function to get the velocity function: v(t)= 2(6t^2+5t) (12t +5)

Then, plug t=2 into the velocity function: $v(2)= 2(6({\color{Blue} 2})^2+5({\color{Blue} 2})) (12({\color{Blue} 2}) +5)$

Therefore, the answer is: 1972

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Calculus 1 : Velocity

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #341 : Spatial Calculus

Example Question #342 : Spatial Calculus

Example Question #343 : Spatial Calculus

Example Question #341 : Velocity

Example Question #342 : Velocity

Example Question #346 : Spatial Calculus

Example Question #343 : Calculus

Example Question #344 : Calculus

Example Question #349 : Spatial Calculus

Example Question #350 : Spatial Calculus

All Calculus 1 Resources

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