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Calculus 1 : How to find acceleration

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Example Questions

Example Question #91 : Acceleration

For this function, keep in mind that acceleration is measured as $\frac{velocity}{time}$ .

A particle's position is given by the function y = 2x . Find the particle's acceleration when x=4 .

Possible Answers:

Correct answer:

Explanation:

Acceleration can be found by taking the second derivative of the position function.

The given position function is y = 2x , its first derivative using the power rule is

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

$y=2x \rightarrow y'=1\cdot 2x^{1-1}=2x^0=2$ , a constant.

Recall that the derivative of a constant always is zero we see that the second derivative of this function is .

Therefore, we have that the acceleration is always .

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Example Question #92 : Acceleration

An atom's position is defined by the equation $p(t)=12t^{3}+7t^{2}-6t+15$ . What is the acceleration of the atom at t=4 ?

Possible Answers:

300

301

302

303

299

Correct answer:

302

Explanation:

By definition, acceleration a(t) is the second derivative of a position function p(t) .

Therefore, a(t)=p''(t) .

For this particular problem we will use the power rule to find the derivative. The power rule states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Since $p(t)=12t^{3}+7t^{2}-6t+15$ , applying the power rule once we get $p'(t)=36t^{2}+14t-6$ and applying it a second time we find a(t)=p''(t)=72t+14 . Plugging in t=4 gets us a(t)=p''(t)=72(4)+14=288+14=302 .

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Example Question #481 : Spatial Calculus

A car's position is defined by the equation $p(t)=6t^{3}-3t^{2}+12t+11$ . What is its acceleration at t=3 ?

Possible Answers:

101

102

100

Correct answer:

102

Explanation:

By definition, acceleration a(t) is the second derivative of a position function p(t) .

Therefore, a(t)=p''(t) .

For this particular problem we will use the power rule to find the derivative. The power rule states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Since $p(t)=6t^{3}-3t^{2}+12t+11$ , applying the power rule once will get us $p'(t)=18t^{2}-6t+12$ .

Applying the power rule a second time we find a(t)=p''(t)=36t-6 .

Plugging in t=3 gets us a(t)=p''(t)=36(3)-6=108-6=102 .

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Example Question #481 : Spatial Calculus

A cat's position is defined by the equation $p(t)=2t^{3}+5t^{2}-11t+4$ . What is its acceleration at t=1 ?

Possible Answers:

Correct answer:

Explanation:

By definition, acceleration a(t) is the second derivative of a position function p(t) .

Therefore, a(t)=p''(t) .

For this particular problem we will use the power rule to find the derivative.

The power rule states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Since $p(t)=2t^{3}+5t^{2}-11t+4$ , applying the power rule we get $p'(t)=6t^{2}+10t-11$ .

Thus applying the power rule again we find a(t)=p''(t)=12t+10 .

Plugging in t=1 gets us a(t)=p''(t)=12(1)+10=12+10=22 .

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Example Question #482 : Spatial Calculus

What is the acceleration of an object with a position function $p(t)=5t^{3}+2t^{2}+t-12$ at t=2 ?

Possible Answers:

Correct answer:

Explanation:

Acceleration is the second derivative of position, or a(t)=p''(t) .

Given,

$p(t)=5t^{3}+2t^{2}+t-12$

we can use the power rule which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Applying this rule we get $p'(t)=15t^{2}+4t+1$ .

From here we can use the power rule again and deduce that

a(t)=p''(t)=30t+4 .

Swapping in t=2 , we get

a(2)=30(2)+4=60+4=64 .

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Example Question #96 : How To Find Acceleration

What is the acceleration of a bird with a position function $p(t)=2t^{3}-4t^{2}+7t+16$ at t=1 ?

Possible Answers:

Correct answer:

Explanation:

Acceleration is the second derivative of position, or a(t)=p''(t) .

Given,

$p(t)=2t^{3}-4t^{2}+7t+16$

we can use the power rule which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Therefore,

$p'(t)=6t^{2}-8t+7$

From here we apply the power rule a second time and deduce that,

a(t)=p''(t)=12t-8 .

Swapping in t=1 , we get

a(1)=12(1)-8=12-8=4 .

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Example Question #97 : How To Find Acceleration

What is the acceleration of a boat with a position function $p(t)=t^{3}+2t^{2}-5t+11$ at t=4 ?

Possible Answers:

Correct answer:

Explanation:

Acceleration is the second derivative of position, or a(t)=p''(t) .

Given,

$p(t)=t^{3}+2t^{2}-5t+11$

we can use the power rule which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Applying the power rule we find,

$p'(t)=3t^{2}+4t-5$ .

From here we apply the power rule a second time and deduce that,

a(t)=p''(t)=6t+4 .

Swapping in t=4 , we get

a(4)=6(4)+4=24+4=28 .

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Example Question #94 : Acceleration

Satoshi is pacing back and forth in an erratic way. His position at any given time is defined by the following function:

x(t) = 5 sin(t) - cos (t^2)

Find a function describing his acceleration at any point in time.

Possible Answers:

2sin(t^2) - 4t^2sin(t^2) - 5sin(t)

2sin(t^2) - 4t^2cos(t^2) - 5sin(t)

2sin(t^2) + 4t^2sin(t^2) - 5sin(t)

2sin(t^2) + 4t^2cos(t^2) - 5sin(t)

Correct answer:

2sin(t^2) + 4t^2cos(t^2) - 5sin(t)

Explanation:

We are given the position function of Satoshi; we can find his acceleration function by taking the second derivative of his position with respect to time.

The first step will be to find the velocity function by taking the first derivative of the position function.

The position function is:

x(t) = 5 sin(t) - cos (t^2)

To take the derivative of this function, let's handle each term separately.

The derivative of

5 sin(t) is 5 cos(t)

To find the derivative of the

-cos(t^2)

term, remember to find the derivative of the term inside of the parenthesis, and multiply it by the derivative of the outside term:

2tsin(t^2)

In this manner we find our velocity function:

v(t) = 5 cos(t) + 2tsin(t^2)

To find acceleration, we then take the derivative of velocity with respect to time.

Since the second term, 2tsin(t²), is a product, we must use the product rule for derivatives:

Derivative of f(t)g(t) = f'(t)g(t) + f(t)g'(t) ):

2sin(t^2) + 4t^2cos(t^2)

Adding this to the derivative of

5 cos(t)

which is

-5 sin (t)

we can define our acceleration function as:

2sin(t^2) + 4t^2cos(t^2) - 5sin(t)

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Example Question #95 : Acceleration

The position of a very erratic particle can be described with the following equation:

$x(t)=5t^2 + 2sin(3t) + 4e^{-t^3}$

Write an equation that describes its acceleration at any point in time.

Possible Answers:

$10-18sin(3t)-24te^{-t^3}+36t^4e^{-t^3}$

$10+18sin(3t)+24te^{-t^3}-36t^4e^{-t^3}$

$10+18sin(3t)+24te^{-t^3}+36t^4e^{-t^3}$

$10+18sin(3t)-24te^{-t^3}-36t^4e^{-t^3}$

Correct answer:

$10-18sin(3t)-24te^{-t^3}+36t^4e^{-t^3}$

Explanation:

Acceleration is defined as the second derivative of the position function with respect to time. Taking the first derivative will give us a velocity (remember that for trigonometric functions and exponentials, take the derivative of the inside terms as well and multiply them with the derivative of the outside terms):

$v(t)=10t+6cos(3t)-12t^2e^{-t^3}$

Taking the derivative of this, remembering to use the product rule for derivatives, in turn gives us the acceleration:

$a(t) =10-18sin(3t)-24te^{-t^3}+36t^4e^{-t^3}$

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Example Question #491 : Calculus

The position of a given chicken is defined by the equation $p(t)=2t^{3}-12t^{2}+5t+9$ . What is the acceleration of the chicken at t=3 ?

Possible Answers:

Correct answer:

Explanation:

Acceleration a(t) is defined as the second derivative of position p(t) , or a(t)=p''(t) .

Therefore, given

$p(t)=2t^{3}-12t^{2}+5t+9$ and using the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$

where $n\neq0$ ,

$p'(t)=6t^{2}-24t+5$ and

a(t)=p''(t)=12t-24 .

Consequently, at t=3 ,

a(3)=p''(3)=12(3)+24=36-24=12 .

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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 #91 : Acceleration

Example Question #92 : Acceleration

Example Question #481 : Spatial Calculus

Example Question #481 : Spatial Calculus

Example Question #482 : Spatial Calculus

Example Question #96 : How To Find Acceleration

Example Question #97 : How To Find Acceleration

Example Question #94 : Acceleration

Example Question #95 : Acceleration

Example Question #491 : Calculus

All Calculus 1 Resources

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