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Calculus 1 : How to find acceleration

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Example Questions

Example Question #251 : Acceleration

$f(x)=\sin(x) +x^{2$

Using the function above, what is the acceleration of an object moving in plane at x=1 ? Round to the nearest hundredth.

Possible Answers:

-2.54

1.16

2.54

4.54

-1.16

Correct answer:

1.16

Explanation:

In order to find the acceration of an object in a plane at any give point , you must first differentiate the first derivative.

In this case, the first derivative is: $f'(x)=\cos (x)+2x$

When you differentiate the first derivative, you get the second derivative: $f''(x)= 2-\sin (x)$

This is the function that gives you the acceleration of an object vector at point .

You plug x=1 into the second derivative and therefore, the correct answer is 1.16 .

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Example Question #252 : Acceleration

$f(x)= e^{x}+x^{2}-3$

Using the function above, what is the acceleration of an object moving in plane at x=5 ? Round to the nearest hundredth.

Possible Answers:

-150.41

100.34

158.41

150.41

-158.41

Correct answer:

150.41

Explanation:

In order to find the acceration of an object in a plane at any give point , you must first differentiate the first derivative.

In this case, the first derivative is: $f'(x)=e^{x}+2x$

When you differentiate the first derivative, you get the second derivative: $f''(x)= e^{x}+2$

This is the function that gives you the acceleration of an object vector at point .

You plug x=5 into the second derivative and therefore, the correct answer is 150.41 .

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Example Question #253 : Acceleration

$f(x)= ln(x)-3x^{2}+2$

Using the function above, what is the acceleration of an object moving in plane at x=4 ? Round to the nearest hundredth.

Possible Answers:

-6.06

23.75

-23.75

6.06

-0.06

Correct answer:

-6.06

Explanation:

In order to find the acceration of an object in a plane at any give point , you must first differentiate the first derivative.

In this case, the first derivative is: $f'(x)=\frac{1}{x}-6x$

When you differentiate the first derivative, you get the second derivative: $f''(x)= -\frac{1}{x^{2}}-6$

This is the function that gives you the acceleration of an object vector at point .

You plug x=4 into the second derivative and therefore, the correct answer is -6.06 .

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Example Question #254 : Acceleration

$f(x)= \cos (x) +e^{x}$

Using the function above, what is the acceleration of an object moving in plane at x=0 ? Round to the nearest hundredth.

Possible Answers:

Undefined

Correct answer:

Explanation:

In order to find the acceration of an object in a plane at any give point , you must first differentiate the first derivative.

In this case, the first derivative is: $f'(x)=e^{x}-\sin (x)$

When you differentiate the first derivative, you get the second derivative: $f''(x)= e^{x}-\cos(x)$

This is the function that gives you the acceleration of an object vector at point .

You plug x=0 into the second derivative and therefore, the correct answer is .

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Example Question #255 : Acceleration

$f(x)= \csc (x)+x^{3}$

What is the slope of this curve at x=6 ? Round to two decimal points.

Possible Answers:

16.98

12.05

-15.93

Correct answer:

16.98

Explanation:

To find the slope of a curve at a certain point, you must first find the derivative of that curve.

The derivative of this curve is $f'(x)=3x^{2}-\cot (x)\csc (x)$ .

Then, plug x=6 into the derivative and you will get 16.98 , which is the slope.

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Example Question #256 : Acceleration

f(x)= ln(x)-2x^3

What is the slope of this curve at x=0 ? Round to two decimal points.

Possible Answers:

-2.34

2.34

Undefined

Correct answer:

Undefined

Explanation:

To find the slope of a curve at a certain point, you must first find the derivative of that curve.

The derivative of this curve is $f'(x)=\frac{1}{x}-6x^{2}$ .

Then, plug x=0 into the derivative. The slope of this curve at point x=0 is undefined.

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Example Question #257 : Acceleration

The position at a certain point is given by:

$f(x)=\frac{1}{2}x^{2} - 4x + 5$

What is the acceleration at x=5 ?

Possible Answers:

Correct answer:

Explanation:

In order to find the acceleration of a given point, you must first find the derivative of the position function which gives you the velocity function:

f'(x)=x-4

Then, you differentiate the velocity function to get the acceleration function:

f''(x)=1

Then, find f''(x) when x=5 .

The answer is .

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Example Question #258 : Acceleration

The position at a certain point is given by:

$f(x)=\frac{3}{5}x^{5} - 6x + 5$

What is the acceleration at x=2 ?

Possible Answers:

198

Correct answer:

Explanation:

In order to find the acceleration of a given point, you must first find the derivative of the position function which gives you the velocity function:

$f'(x)=3x^{4}-6$

Then, you differentiate the velocity function to get the acceleration function:

f''(x)=12x^3

Then, find f''(x) when x=2 .

The answer is:

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Example Question #258 : Acceleration

The position at a certain point is given by:

$f(x)=(\sin x)^{2} +2x^{3}$

What is the acceleration at $x=\pi$ ?

Possible Answers:

2.98

39.7

34.2

213.7

Correct answer:

39.7

Explanation:

In order to find the acceleration of a given point, you must first find the derivative of the position function which gives you the velocity function:

$f'(x)=2 \cos x \sin x +6x^{2}$

Then, you differentiate the velocity function to get the acceleration function:

$f''(x)= -2\sin ^{2}x + 2\cos^2x + 12x$

Then, find f''(x) when $x=\pi$ .

The answer is: 39.7

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Example Question #259 : Acceleration

The position at a certain point is given by:

$f(x)= x^{x-3}$

What is the acceleration at x=4 ?

Possible Answers:

23.86

54.09

12.46

31.35

41.84

Correct answer:

12.46

Explanation:

In order to find the acceleration of a given point, you must first find the derivative of the position function which gives you the velocity function:

$f'(x)= x^{x-3}(ln(x)+\frac{x-3}{x})$

Then, you differentiate the velocity function to get the acceleration function:

$f''(x)= (x^{x-3})(ln(x)+\frac{x-3}{x})^{2} + (\frac{2}{x} - \frac{x-3}{x^{2}}) (x^{x-3})$

Then, find f''(x) when x=4 .

The answer is: 12.46

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Calculus 1 : How to find acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #251 : Acceleration

Example Question #252 : Acceleration

Example Question #253 : Acceleration

Example Question #254 : Acceleration

Example Question #255 : Acceleration

Example Question #256 : Acceleration

Example Question #257 : Acceleration

Example Question #258 : Acceleration

Example Question #258 : Acceleration

Example Question #259 : Acceleration

All Calculus 1 Resources

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