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Calculus 1 : Calculus

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Example Questions

Example Question #211 : How To Find Acceleration

A driver is speeding through the interstate. The distance traveled by the driver over a certain amount of time can be calculated by the equation

$x(t) = 20\frac{m}{s^{2}} (t^{2}) + 10m$

where is the distance traveled in meters and is time in seconds .

What is the equation for the acceleration of the driver?

Possible Answers:

$a(t) = 20\tfrac{m}{s} (t^{2})$

$a(t) = 2\tfrac{m}{s^{2}}$

$a(t) = 40\tfrac{m}{s} (t)$

$a(t) = 20\tfrac{m}{s} (t)$

$a(t) = 20\tfrac{m}{s^{2}}$

Correct answer:

$a(t) = 20\tfrac{m}{s^{2}}$

Explanation:

To find the velocity of the driver, we take the derivative of the position equation. The velocity equation is

$v(t) = 20\tfrac{m}{s} (t)$

Then to find the acceleration of the toy car, we take the derivative again. The acceleration equation is

$a(t) = 20\tfrac{m}{s}$

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Example Question #611 : Spatial Calculus

A driver is speeding through the interstate. The distance traveled by the driver over a certain amount of time can be calculated by the equation

$x(t) = 20\frac{m}{s^{2}} (t^{2}) + 10m$

where is the distance traveled in meters and is time in seconds .

What is the instantaneous acceleration of the driver at any time?

Possible Answers:

$10\tfrac{m}{s^{2}}$

$2\tfrac{m}{s^{2}}$

$20\tfrac{m}{s^{2}}$

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

$40\tfrac{m}{s^{2}}$

Correct answer:

$20\tfrac{m}{s^{2}}$

Explanation:

To find the velocity of the driver, we take the derivative of the position equation. The velocity equation is

$v(t) = 20\tfrac{m}{s} (t)$

Then to find the acceleration of the toy car, we take the derivative again. The acceleration equation is

$a(t) = 20\tfrac{m}{s}$

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Example Question #612 : Spatial Calculus

What is the acceleration of an object following the path of y=e^tsin(t) at $t=2\pi$ ?

Possible Answers:

$-2e^{2\pi}$

$e^{2\pi}$

$2e^{2\pi}$

Correct answer:

$2e^{2\pi}$

Explanation:

Remember that acceleration is the second derivative of position with respect to time.

By the product rule, we know that

$\frac{d}{dt}(f(t)*g(t))=f'(t)*g(t)+g'(t)f(t)$ , where and are functions that have first derivatives and .

Keeping this in mind:

y(t)=e^tsin(t)

y'(t)=e^tsin(t)+e^tcos(t)

To solve for the second derivative, we have to take the product rule for each term:

y''(t)=e^tsin(t)+e^tcos(t)+e^tcos(t)-e^tsin(t)

=2e^tcos(t)

Solving for $y''(2\pi)$ ,

$y''(2\pi)=2e^{2\pi}cos({2\pi})=2e^{2\pi}$

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Example Question #613 : Spatial Calculus

You are given the position function:

X(t) = 6t^2 + 2t -5

Find the acceleration a(t) at time .

Possible Answers:

a(2) = 12

a(2) = 0

a(2) = -5

a(2) = 24

a(2) = 14

Correct answer:

a(2) = 12

Explanation:

The acceleration function is given by the second derivative of the position function. Take the first derivative of X(t) using the power rule:

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

Applying the power rule to this equation, noting the constant of in the first term:

$\frac{dX}{dt} X(t) = 12t + 2$

This is the velocity function. Take the derivative once again to find the acceleration.

$\frac{d^2X}{dt^2} X(t) = 12$

Because there is no factor of t in the acceleration equation, you do not need to plug in to solve for a(2) . The acceleration is a constant .

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Example Question #614 : Spatial Calculus

Given the position function:

What is the instantaneous acceleration at time (t=2 seconds)?

Possible Answers:

488

472

224

Correct answer:

472

Explanation:

The acceleration is simply the second derivative of the position function:

a(t)=s''(t)

To find the derivatives in this particular problem use the power rule which states,

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

Applying the power rule once, the first derivative of the given function gives us the velocity:

v(t)=s'(t)=15t^4-8t

Applying the power rule a second time, the derivative of this will give us our acceleration function:

a(t)=60t^4-8

At time (t=2 seconds):

a(2)=60(2)^3-8=60(8)-8=480-8=472

This is one of the answer choices.

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Example Question #611 : Spatial Calculus

Given the position function:

s(t)=4t^3-5t^2+5t-100

What is the instantaneous acceleration at time (t=2 seconds)?

Possible Answers:

Correct answer:

Explanation:

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

To find the derivatives in this particular problem use the power rule which states,

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

$a(t)=s''(t)=\frac{d^2}{dt^2}[4t^3-5t^2+5t-100]=\frac{d}{dt}[12t^2-10t+5]=24t-10$

At time (t=2 seconds):

a(2)=24(2)-10=48-10=38

This is one of the answer choices.

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Example Question #616 : Spatial Calculus

Given the velocity function:

v(t)=3t^2-9t+14

Find the instantaneous acceleration at time (t=5 seconds)

Possible Answers:

Correct answer:

Explanation:

The instantaneous acceleration can be found by taking the derivative of the velocity function, which is given:

To find the derivatives in this particular problem use the power rule which states,

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

$a(t)=v'(t)=\frac{d}{dt}[3t^2-9t+14]=6t-9$

At time (t=5 seconds):

a(5)=6(5)-9=30-9=21

This is one of the answer choices.

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Example Question #611 : Spatial Calculus

Find the acceleration function if the velocity function is

v(t)=et .

Possible Answers:

a(t)=e+t

a(t)=e+1

a(t)=t

a(t)=e

Correct answer:

a(t)=e

Explanation:

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

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

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

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

and by applying this to the given function

$a(t)=\frac{d}{dt}[et]=1*et^{1-1}=e$ .

Hence

a(t)=e .

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Example Question #228 : How To Find Acceleration

Find the acceleration function if the velocity function is

v(t)=e^2 .

Possible Answers:

a(t)=0

a(t)=1

a(t)=e

a(t)=e^2

Correct answer:

a(t)=0

Explanation:

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

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

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

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

and by applying this to the given function

$a(t)=\frac{d}{dt}[e^2]=0$ .

Hence

a(t)=0 .

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Example Question #617 : Spatial Calculus

If the velocity of an asteroid is given by g(x) , find the function which models the acceleration of the asteroid.

g(x)=5sin(x)

Possible Answers:

cos(5x)

-5sin(x)

-5cosx

5cosx

Correct answer:

5cosx

Explanation:

If the velocity of an asteroid is given by g(x), find the function which models the acceleration of the asteroid.

g(x)=5sin(x)

Recall that a velocity function is the first derivative of a position function, and that an acceleration function is the first derivative of a velocity function.

So, to find the acceleration function, we need to find the derivative of g(x)

Begin by recalling or looking up the rule for derivative of sine:

$\frac{d}{dx}sinx=cosx$

So the answer to our question can be found via:

$\frac{d}{dx}5sinx=5cosx$

To be clear, our answer is:

g'(x)=5cosx

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #211 : How To Find Acceleration

Example Question #611 : Spatial Calculus

Example Question #612 : Spatial Calculus

Example Question #613 : Spatial Calculus

Example Question #614 : Spatial Calculus

Given the position function:

What is the instantaneous acceleration at time (t=2 seconds)?

Example Question #611 : Spatial Calculus

Given the position function:

Example Question #616 : Spatial Calculus

Given the velocity function:

Example Question #611 : Spatial Calculus

Example Question #228 : How To Find Acceleration

Example Question #617 : Spatial Calculus

All Calculus 1 Resources

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