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Calculus 1 : Velocity

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Example Questions

Example Question #301 : Velocity

The position of a point is found by the following function:

$p(t) = 3t - \cos t + \tan t \sin t$

What is the velocity at t=1 ?

Possible Answers:

$\pi$

7.6

Correct answer:

7.6

Explanation:

In order to find the velocity of any point given a position function, we must first find the derivative of the position function, p'(t)= v(t) .

This is the position function: $p(t) = 3t - \cos t + \tan t \sin t$

The derivative gives us the velocity function: $v(t) = \cos t \tan t + \sec^ 2t \sin t + \sin t +3$

You can then find the velocity of a given point by substituting it in for the variable: $v(1) = \cos (1) \tan (1) + \sec^ 2(1) \sin (1) + \sin (1) +3$

Therefore, the velocity at t= 1 is: 7.6

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Example Question #302 : Velocity

The position of a point is found by the following function:

p(t) = ln(t) + 2e^t

What is the velocity at t=0 ?

Possible Answers:

undefined

Correct answer:

undefined

Explanation:

In order to find the velocity of any point given a position function, we must first find the derivative of the position function, p'(t)= v(t) .

This is the position function: p(t) = ln(t) + 2e^t

The derivative gives us the velocity function: $v(t) = \frac{1}{t} + 2e^t$

You can then find the velocity of a given point by substituting it in for the variable: $v(0) = \frac{1}{(0)} + 2e^{(0)}$

Therefore, the velocity at t= 0 is: undefined

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Example Question #303 : Velocity

The position of a point is found by the following function:

$p(t) = e^t + 3t^{(\frac{1}{3})} +4t^6$

What is the velocity at t=2 ?

Possible Answers:

234.7

730

776.02

Correct answer:

776.02

Explanation:

In order to find the velocity of any point given a position function, we must first find the derivative of the position function, p'(t)= v(t) .

This is the position function: $p(t) = e^t + 3t^{(\frac{1}{3})} +4t^6$

The derivative gives us the velocity function: $v(t) = e^t + t^{-\frac{2}{3}} + 24t^5$

You can then find the velocity of a given point by substituting it in for the variable: $v(2) = e^{(2)} + (2)^{-\frac{2}{3}} + 24(2)^5$

Therefore, the velocity at t=2 is: 776.02

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Example Question #304 : Velocity

The position of a point is found by the following function:

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

What is the velocity at t=3 ?

Possible Answers:

137,592

10,457

2,176

2,345

8,990

Correct answer:

137,592

Explanation:

In order to find the velocity of any point given a position function, we must first find the derivative of the position function, p'(t)= v(t) .

This is the position function: p(t) = (4t^2+2t)^3 - 5

The derivative gives us the velocity function: v(t) = 3(4t^2+2t)^2 (8t+2)

You can then find the velocity of a given point by substituting it in for the variable: v(3) = 3(4(3)^2+2(3))^2 (8(3)+2)

Therefore, the velocity at t= 3 is: 137,592

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Example Question #305 : Velocity

The position of a certain point is given by the following function:

$p(t)= 2t^{\frac{1}{2}} + 3e^t +4$

What is the velocity at t=4 ?

Possible Answers:

23.2

76.3

100.4

165.8

Correct answer:

165.8

Explanation:

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: p'(t)= v(t)

In this case, the position function is: $p(t)= 2t^{\frac{1}{2}} + 3e^t +4$

The velocity function is found by taking the derivative of the position function: $v(t)= t^{-\frac{1}{2}} + 3e^t$

Finally, to find the velocity, substitute t=4 into the velocity function: $v(4)= (4)^{-\frac{1}{2}} + 3e^{(4)}$

Therefore, the answer is: 165.8

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Example Question #306 : Velocity

The position of a certain point is given by the following function:

$p(t)= \frac{1}{3}t^3 + ln(t) - 23$

What is the velocity at t=3 ?

Possible Answers:

ln(t)

-13.67

13.67

4.98

Correct answer:

-13.67

Explanation:

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: p'(t)= v(t)

In this case, the position function is: $p(t)= \frac{1}{3}t^3 + ln(t) - 23t$

The velocity function is found by taking the derivative of the position function: $v(t)= t^2 + \frac{1}{t} - 23$

Finally, to find the velocity, substitute t=3 into the velocity function: $v(3)= (3)^2 + \frac{1}{(3)} - 23$

Therefore, the answer is: -13.67

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Example Question #307 : Velocity

The position of a certain point is given by the following function:

$p(t)= 6t^{-\frac{1}{6}} + 4t^5 - 4t^2$

What is the velocity at t=1 ?

Possible Answers:

9.6

Correct answer:

Explanation:

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: p'(t)= v(t)

In this case, the position function is: $p(t)= 6t^{-\frac{1}{6}} + 4t^5 - 4t^2$

The velocity function is found by taking the derivative of the position function: $v(t)= -t^{-\frac{7}{6}} + 20t^4 - 8t$

Finally, to find the velocity, substitute t=1 into the velocity function: $v(1)= -(1)^{-\frac{7}{6}} + 20(1)^4 - 8(1)$

Therefore, the answer is:

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Example Question #308 : Velocity

The position of a certain point is given by the following function:

$p(t)= -\cos t + 3t^2 + 6ln(t)$

What is the velocity at $t=\pi$ ?

Possible Answers:

20.8

4.3

$\pi$

1.2

Correct answer:

20.8

Explanation:

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: p'(t)= v(t)

In this case, the position function is: $p(t)= -\cos t + 3t^2 + 6ln(t)$

The velocity function is found by taking the derivative of the position function: $v(t)= \sin t + 6t + \frac{6}{t}$

Finally, to find the velocity, substitute $t=\pi$ into the velocity function: $v(\pi)= \sin (\pi) + 6(\pi) + \frac{6}{(\pi)}$

Therefore, the answer is: 20.8

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Example Question #309 : Velocity

The position of a certain point is given by the following function:

$p(t)= -\cos t + \sin(t) - 45e^t + 5t^2$

What is the velocity at t=0 ?

Possible Answers:

$\pi$

Correct answer:

Explanation:

In order to find the velociy of a certain point, you must find the derivative of the position function which gives us the velocity function: p'(t)= v(t)

In this case, the position function is: $p(t)= -\cos t + \sin t - 45e^t + 5t^2$

The velocity function is found by taking the derivative of the position function: $v(t)= \sin t + \cos t - 45e^t + 10t$

Finally, to find the velocity, substitute t=0 into the velocity function: $v(0)= \sin (0) + \cos(0) - 45e^{(0)} + 10(0)$

Therefore, the answer is:

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Example Question #310 : Velocity

The velocity function of a particle and a position of this particle at a known time are given by $v(t)=\frac{tan(t)}{t^3}$ p(1)=0 . Approximate p(1.6) using Euler's Method and three steps.

Possible Answers:

0.697

0.423

0.119

1.445

1.032

Correct answer:

1.032

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 $v(t)=\frac{tan(t)}{t^3}$ p(1)=0

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

$\Delta t = \frac{1.6-1}{3}=0.2$

p_0=0;t_0=1

$p_1=0+(0.2)(\frac{tan(1)}{1^3})=0.311$

$p_2=0.311+(0.2)(\frac{tan(1.2)}{1.2^3})=0.609$

$p_3=0.609+(0.2)(\frac{tan(1.4)}{1.4^3})=1.032$

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Calculus 1 : Velocity

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #301 : Velocity

Example Question #302 : Velocity

Example Question #303 : Velocity

Example Question #304 : Velocity

Example Question #305 : Velocity

Example Question #306 : Velocity

Example Question #307 : Velocity

Example Question #308 : Velocity

Example Question #309 : Velocity

Example Question #310 : Velocity

All Calculus 1 Resources

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