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Calculus 1 : Acceleration

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Example Questions

Example Question #301 : How To Find Acceleration

If $x(t)=\frac{1}{4}x^2-2x+1$ , what is the acceleration function?

Possible Answers:

$a(t)=\frac{1}{2}x$

a(t)=3

$a(t)=-\frac{1}{2}$

$a(t)=\frac{1}{2}$

a(t)=1

Correct answer:

$a(t)=\frac{1}{2}$

Explanation:

To find the acceleration, you must first find the velocity function, which is the derivative of position. To take the derivative, multiply the exponent by the coefficient in front of the x and then decrease the exponent by 1. Therefore, the velocity function is: $v(t)=\frac{1}{2}x-2$ . To find acceleration now, take the derivative of the velocity function, which is $a(t)=\frac{1}{2}$ .

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Example Question #302 : How To Find Acceleration

Find the acceleration of a particle whose position (meters) is described by s(t)=6t^2+4t+1. , and time is in seconds.

Possible Answers:

$24 \frac{m}{s^{2}}$

$30\frac{m}{s^{2}}$

$(12t+4) \frac{m}{s^{2}}$

$12 \frac{m}{s^{2}}$

$16 \frac{m}{s^{2}}$

Correct answer:

$12 \frac{m}{s^{2}}$

Explanation:

The acceleration is the second derivative of the position. The derivative of a number (without any variables) is zero. Therefore, all of the terms drop out except for the leading term.

$s'(t)=12t+4. \;s''(t)=12\frac{m}{s^{2}}.$

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Example Question #303 : How To Find Acceleration

The position of a particle is described by s(t)=(4+5t)^2 . Find the velocity and the acceleration as a function of time.

Possible Answers:

v(t)=30t+16, a(t)=30

v(t) = 4+5t, a(t)=5

v(t)=25t+16, a(t)=25

v(t)=50t, a(t)=0

v(t)=50t+40, a(t)=50

Correct answer:

v(t)=50t+40, a(t)=50

Explanation:

The velocity can be found by taking the first derivative of the position function. The second derivative determines the acceleration. This derivative involves the chain rule. s'(t)=2(4+5t)*5=10(4+5t)=40+50t .

s''(t)=a(t)=50.

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Example Question #301 : Acceleration

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

$p(t)= (3x^2 + e^{2x})^2$

Find the acceleration.

Possible Answers:

1602.3

124.7

4319.4

Answer not listed

234.2

Correct answer:

1602.3

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; and then the derivative of the velocity function to find the acceleration function: p''(t)=v'(t)=a(t)

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

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

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

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

Therefore, the answer is: 1602.3

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Example Question #302 : How To Find Acceleration

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

$p(t)= \sin(t^2+3t^4)$

Find the acceleration.

Possible Answers:

1602.3

Answer not listed

21.5

Correct answer:

Explanation:

In this case, the position function is: $p(t)= \sin(t^2+3t^4)$

Then take the derivative of the position function to get the velocity function:

$v(t)= \cos(t^2+3t^4) (2t+12t^3)$

Then take the derivative of the velocity function to get the acceleration function:

$a(t)= (36t^2+2) \cos(3t^4+t^2) - (12t^3+2t)^2 \sin(3t^4+t^2)$

Then, plug t= 0 into the acceleration function: $a(0)= (36({\color{Blue} 0})^2+2) \cos(3({\color{Blue} 0})^4+ {\color{Blue} 0}^2) - (12({\color{Blue} 0})^3+2({\color{Blue} 0}))^2 \sin(3({\color{Blue} 0})^4+{\color{Blue} 0}^2)$

Therefore, the answer is:

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Example Question #303 : How To Find Acceleration

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

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

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

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

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

Then take the derivative of the velocity function to get the acceleration function: a(t)= 1

Then, plug t= 0 into the acceleration function: a(0)= 1

Therefore, the answer is:

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Example Question #307 : How To Find Acceleration

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

p(t)= -ln(2t^2 + 5t)

Find the acceleration.

Possible Answers:

7.4

Answer not listed

-3.3

1.1

2.5

Correct answer:

1.1

Explanation:

In this case, the position function is: p(t)= -ln(2t^2 + 5t)

Then take the derivative of the position function to get the velocity function: $v(t)= - \frac{4t+5}{2t^2 + 5t}$

Then take the derivative of the velocity function to get the acceleration function: $a(t)= \frac{(4t+5)^2}{(2t^2 + 5t)^2} - \frac{4}{2t^2+5t}$

Then, plug t= 1 into the acceleration function: $a(1)= \frac{(4({\color{Blue} 1})+5)^2}{(2({\color{Blue} 1})^2 + 5({\color{Blue} 1}))^2} - \frac{4}{2({\color{Blue} 1})^2+5({\color{Blue} 1})}$

Therefore, the answer is: 1.1

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Example Question #302 : Acceleration

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

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

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

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 take the derivative of the velocity function to get the acceleration function: $a(t)= 4t^2\sin(t^2) - 2\cos(t^2) + 4e^{2t} + 24t$

Then, plug t= 0 into the acceleration function: $a(0)= 4({\color{Blue} 0})^2\sin({\color{Blue} 0}^2) - 2\cos({\color{Blue} 0}^2) + 4e^{2({\color{Blue} 0})} + 24({\color{Blue} 0})$

Therefore, the answer is:

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Example Question #692 : Spatial Calculus

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

$p(t)= \cos(2t +e^t)$

Find the acceleration.

Possible Answers:

-3.002

4.252

-1.003

Answer not listed

Correct answer:

-3.002

Explanation:

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

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

Then take the derivative of the velocity function to get the acceleration function: $a(t)= -e^t \sin(e^t+2t) - (e^t+2)^2 \cos(e^t+2t)$

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

Therefore, the answer is: -3.002

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Example Question #310 : How To Find Acceleration

Find the acceleration function of the falling rock if its position is given by the following function:

s(t)=5-20t-8t^2

Possible Answers:

-8t

-20-16t

-16t

Correct answer:

Explanation:

To find the acceleration function of the falling rock, we must take the second derivative of the position function for the rock:

v(t)=s'(t)=-20-16t

a(t)=v'(t)=-16

The derivatives were found using the following rule:

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

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #301 : How To Find Acceleration

Example Question #302 : How To Find Acceleration

Example Question #303 : How To Find Acceleration

Example Question #301 : Acceleration

Example Question #302 : How To Find Acceleration

Example Question #303 : How To Find Acceleration

Example Question #307 : How To Find Acceleration

Example Question #302 : Acceleration

Example Question #692 : Spatial Calculus

Example Question #310 : How To Find Acceleration

All Calculus 1 Resources

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