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Calculus 1 : Acceleration

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Example Questions

Example Question #311 : How To Find Acceleration

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

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

Find the acceleration.

Possible Answers:

-20.5

12.8

51.3

-65.4

Answer not listed

Correct answer:

-20.5

Explanation:

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: p''(t)= v'(t)=a(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 take the derivative of the velocity function to get the acceleration function: $a(t)= e^t \cos(4t + e^t) -(4+e^t)^2 \sin(4t+e^t)$

Then, plug t=0 into the acceleration function: $a(0)= e^{\color{Blue} 0} \cos(4({\color{Blue} 0}) + e^{\color{Blue} 0}) -(4+e^{\color{Blue} 0})^2 \sin(4({\color{Blue} 0})+e^{\color{Blue} 0})$

Therefore, the answer is: -20.5

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Example Question #312 : How To Find Acceleration

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

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

Find the acceleration.

Possible Answers:

842

Answer not listed

123

126

543

Correct answer:

842

Explanation:

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 take the derivative of the velocity function to get the acceleration function: a(t)= 432t^2 + 360t + 50

Then, plug t=1 into the acceleration function: $a(1)= 432({\color{Blue} 1})^2 + 360({\color{Blue} 1}) + 50$

Therefore, the answer is: 842

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Example Question #313 : How To Find Acceleration

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

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

Find the acceleration.

Possible Answers:

0.07

-0.12

0.78

Answer not listed

-0.03

Correct answer:

-0.03

Explanation:

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 take the derivative of the velocity function to get the acceleration function: $a(t)= -\frac{4}{(2t+3)^2}$

Then, plug t=4 into the acceleration function: $a(t)= -\frac{4}{(2({\color{Blue} 4})+3)^2}$

Therefore, the answer is: -0.03

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Example Question #314 : How To Find Acceleration

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

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

Find the acceleration.

Possible Answers:

2354.8

123.9

23.4

Anwer not listed

Correct answer:

Anwer not listed

Explanation:

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 take the derivative of the velocity function to get the acceleration function: $a(t)=2(18t^2 - 14t + 2e^{2t})^2 + 2(6t^3 - 7t^2 + e^{2t})(36t - 14 + 4e^{2t})$

Then, plug t=1 into the acceleration function: $a(1)=2(18({\color{Blue} 1})^2 - 14({\color{Blue} 1}) + 2e^{2({\color{Blue} 1})})^2 + 2(6({\color{Blue} 1})^3 - 7({\color{Blue} 1})^2 + e^{2({\color{Blue} 1})})(36({\color{Blue} 1}) - 14 + 4e^{2({\color{Blue} 1})})$

Therefore, the answer is: 1364.03

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Example Question #315 : How To Find Acceleration

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

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

Find the velocity.

Possible Answers:

Answer not listed

-28.9

7.6

-34.6

23.8

Correct answer:

-28.9

Explanation:

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 take the derivative of the velocity function to get the acceleration function: $a(t)= - \frac{1}{t^2} - 4\cos(2t) - 4e^{2t}$

Then, plug t=1 into the acceleration function: $a(1)= - \frac{1}{{\color{Blue} 1}^2} - 4\cos(2({\color{Blue} 1})) - 4e^{2({\color{Blue} 1})}$

Therefore, the answer is: -28.9

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Example Question #311 : Acceleration

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

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

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

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 take the derivative of the velocity function to get the acceleration function: $a(t)= 120 \sin (3t + 5t^4)+ 2(3+20t^3)^2 \cos(3t+5t^4)$

Then, plug t=0 into the acceleration function: $a(0)= 120 \sin (3({\color{Blue} 0})+ 5({\color{Blue} 0})^4)+ 2(3+20({\color{Blue} 0})^3)^2 \cos(3({\color{Blue} 0})+5({\color{Blue} 0})^4)$

Therefore, the answer is:

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Example Question #317 : How To Find Acceleration

Find the acceleration function of the spaceship given its position function:

$s(t)=t^4+e^{2t}$

Possible Answers:

$12t^2+4e^{2t}$

$12t^2+4e^{4t}$

$12t^2+e^{2t}$

$8t^3+4e^{2t}$

Correct answer:

$12t^2+4e^{2t}$

Explanation:

To find the acceleration function of the spaceship, we must find the second derivative of the position function of the spaceship:

$v(t)=s'(t)=4t^3+2e^{2t}$

$a(t)=s''(t)=12t^2+4e^{2t}$

The derivatives were found using the following rule:

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

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Example Question #318 : How To Find Acceleration

Find the acceleration function if the velocity function is

v(t)=23.4t .

Possible Answers:

a(t)=23.4

a(t)=0

a(t)=11.7t^2

a(t)=2.34x

Correct answer:

a(t)=23.4

Explanation:

In order to find the acceleration function from the velocity function we need to take the derivative of the velocity function

$\frac{d}{dt}[v(t)]=a(t)$

When taking the derivative, we use the power rule which states

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

Applying this rule we get

$\frac{d}{dt}[23.4t]=1*23.4t^{1-1}=23.4t^0=23.4*1$

As such

a(t)=23.4

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Example Question #319 : How To Find Acceleration

Find the acceleration function a(t) if the velocity function is

v(t)=-32t+40

Possible Answers:

$a(t)=-32+\frac{40}{t}$

a(t)=8

a(t)=-32

a(t)=0

Correct answer:

a(t)=-32

Explanation:

In order to find the acceleration function from the velocity function we need to take the derivative of the velocity function

$\frac{d}{dt}[v(t)]=a(t)$

When taking the derivative, we use the power rule which states

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

Applying this rule we get

$\frac{d}{dt}[-32t+40]=-32*1t^{1-1}+0*40t^{0-1}=-32t^0$

As such

a(t)=-32

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Example Question #320 : How To Find Acceleration

Find the acceleration function if the velocity function is

v(t)=-32t

Possible Answers:

a(t)=0

a(t)=-16t^2

$a(t)\, DNE$

a(t)=-32

Correct answer:

a(t)=-32

Explanation:

In order to find the acceleration function from the velocity function we need to take the derivative of the velocity function

$\frac{d}{dt}[v(t)]=a(t)$

When taking the derivative, we use the power rule which states

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

Applying this rule we get

$\frac{d}{dt}[-32t]=-32*1t^{1-1}=-32t^0$

As such

a(t)=-32

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #311 : How To Find Acceleration

Example Question #312 : How To Find Acceleration

Example Question #313 : How To Find Acceleration

Example Question #314 : How To Find Acceleration

Example Question #315 : How To Find Acceleration

Example Question #311 : Acceleration

Example Question #317 : How To Find Acceleration

Example Question #318 : How To Find Acceleration

Example Question #319 : How To Find Acceleration

Example Question #320 : How To Find Acceleration

All Calculus 1 Resources

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