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Calculus 1 : Calculus

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Example Questions

Example Question #142 : Acceleration

The position of a particle is given by the function $x(t)=cos(3e^{5t})$ . What is the function for its acceleration?

Possible Answers:

$-75e^{5t}sin(3e^{5t})-225e^{10t}cos(3e^{5t})$

$-75e^{5t}sin(3e^{5t})-225e^{5t}cos(3e^{5t})$

$-225cos(3e^{5t})$

$-75e^{5t}sin(3e^{5t})$

$75e^{5t}sin(3e^{5t})+225e^{5t}cos(3e^{5t})$

Correct answer:

$-75e^{5t}sin(3e^{5t})-225e^{10t}cos(3e^{5t})$

Explanation:

Acceleration is given as the second time derivative of position. In this case we will use the rule of exponentials and trigonometrics to solve. Specifically we will use the following rules,

$\frac{d}{dx}e^u\rightarrow e^u\frac{du}{dx}$ ,

$\frac{d}{dx}cos(u)\rightarrow -sin(u)\frac{du}{dx}$ ,

$\frac{d}{dx}sin(u)\rightarrow cos(u)\frac{du}{dx}$ .

Begin by taking the first derivative:

$x'(t)=\frac{d}{dt}(cos(3e^{5t}))=-15e^{5t}sin(3e^{5t})$

Afterwards, acceleration may be found by deriving once more:

$a(t)=x''(t)=\frac{d}{dt}(-15e^{5t}sin(3e^{5t}))$

$a(t)=-75e^{5t}sin(3e^{5t})-225e^{10t}cos(3e^{5t})$

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Example Question #541 : Spatial Calculus

If the velocity of a particle is given by the function v(t)=ln(cos(t)) , what is the acceleration function?

Possible Answers:

tan(t)

-sin(t)cos(t)

-sin(t)ln(cos(t))

-tan(t)

sin(t)cos(t)

Correct answer:

-tan(t)

Explanation:

Acceleration is the time derivative of velocity. For this problem we will use the rules for differentiating trigonometric and logarithmic functions.

Recall that the derivative of ln(u) is $\frac{1}{u}\cdot \frac{du}{dx}$ and the derivative of cos(u) is $-sin(u)\frac{du}{dx}$ .

Applying these rules we are able to find the acceleration function.

$a(t)=v'(t)=\frac{d}{dt}(ln(cos(t)))$

$a(t)=-sin(t)\left(\frac{1}{cos(t)}\right)=-tan(t)$

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Example Question #542 : Spatial Calculus

The position of a particle is given by the function: $x(t)=tsin(e^{2t})$ . What is the acceleration function of the particle?

Possible Answers:

$-8e^{2t}cos(e^{2t})+4te^{4t}sin(e^{2t})$

$8e^{2t}cos(e^{2t})+4te^{4t}sin(e^{2t})$

$4e^{2t}cos(e^{2t})-4te^{2t}cos(e^{2t})-4te^{4t}sin(e^{2t})$

$8e^{2t}cos(e^{2t})-4te^{4t}sin(e^{2t})$

$4e^{2t}cos(e^{2t})+4te^{2t}cos(e^{2t})-4te^{4t}sin(e^{2t})$

Correct answer:

$4e^{2t}cos(e^{2t})+4te^{2t}cos(e^{2t})-4te^{4t}sin(e^{2t})$

Explanation:

Acceleration is given as the second time derivative of position:

$a(t)=x''(t)=\frac{d^2}{dt^2}(tsin(e^{2t}))$

Recall

$\frac{d}{dx}e^u\rightarrow e^u\frac{du}{dx}$ ,

$\frac{d}{dx}cos(u)\rightarrow -sin(u)\frac{du}{dx}$ ,

$\frac{d}{dx}sin(u)\rightarrow cos(u)\frac{du}{dx}$

Begin by taking the first derivative:

$\frac{d}{dt}(tsin(e^{2t}))=sin(e^{2t})+2te^{2t}cos(e^{2t})$

Then taking the second derivative gives the acceleration:

$a(t)=2e^{2t}cos(e^{2t})+2e^{2t}cos(e^{2t})+2t(2e^{2t}cos(e^{2t})-2e^{4t}sin(e^{2t})))$

$a(t)=4e^{2t}cos(e^{2t})+4te^{2t}cos(e^{2t})-4te^{4t}sin(e^{2t})$

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Example Question #544 : Calculus

Find the acceleration of a particle at time t=5 given the position function.

p(t)=4t^5-3t^4-6t^3-9t^2-15

Possible Answers:

8902

4950

None of these

8756

Correct answer:

8902

Explanation:

Velocity is the derivative of position. Acceleration is the derivative of velocity. The derivative of $x^{n}$ is $nx^{n-1}$ . The derivative of the position function must be taken twice to get the acceleration function.

v(t)=20t^4-12t^3-18t^2-18t

a(t)=80t^3-36t^2-36t-18

Plugging in t=5 gives

a(5)=80(5)^3-36(5)^2-36(5)-18

=8902

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Example Question #543 : Spatial Calculus

A ball is traveling in a path resembling the function s(x)=cos(x)-sin(x)+3x^2 .

What is the acceleration of the ball?

Possible Answers:

a(t)=sin(x)-cos(x)

a(t)=sin(x)+cos(x)+6

a(t)=sin(x)-cos(x)+6

a(t)=-sin(x)-cos(x)+6x

Correct answer:

a(t)=sin(x)-cos(x)+6

Explanation:

The acceleration is the second derivarive of the position function. For this problem we will recall the rules of differentiating trigonometric functions and the power rule.

$\frac{d}{dx}cos(u)=-sin(u)\frac{du}{dx}$

$\frac{d}{dx}sin(u)=cos(u)\frac{du}{dx}$

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

Having been provided with the position function, we first need to find the first derivative.

s'(x)=-sin(x)-cos(x)+6x

From here we take the derivative again to find the acceleration of the ball.

$s^{''}(x)=-cos(x)+sin(x)+6$ .

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Example Question #544 : Spatial Calculus

If the position function of an object is x(t)=16t^2+4t+5 , what is its acceleration function?

Possible Answers:

a(t)=0

a(t)=32t+4

a(t)=32

a(t)=32t

a(t)=1

Correct answer:

a(t)=32

Explanation:

For this problem, you must remember that the acceleration function is the second derivative of the position function.

Therefore, you must take the derivative twice. The first derivative (also the velocity function) is v(t)=32t+4 .

To find acceleration, you must take the derivative of velocity.

That becomes a(t)=32 .

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Example Question #545 : Spatial Calculus

Given that the position function of a paper airplane is S(t)=-t^3+4t^2+6t-8 and its initial veloctiy is $6\frac{ft}{sec}$ find its acceleration at seconds.

Possible Answers:

None of the above.

$-5 \frac{ft}{s^2}$

$5 \frac{ft}{s^2}$

$4 \frac{ft}{s^2}$

$-4 \frac{ft}{s^2}$

Correct answer:

$-4 \frac{ft}{s^2}$

Explanation:

In order to solve this proble, it must first be realized that the derivative of a position function is the velocity function and the derivative of the velocity function becomes the acceleration function. In this problem we are given the position function and asked to find the acceleration after a certain amount of time. By taking the double derivative of the function and plugging in the time, we will be able to find the acceleration of the paper plane at that time. In order to take the derivative of the position function, we must first know the power rule, $\frac{d}{dx}x^n=nx^{n-1}$ .

Applying the power rule to the position function, we find the first derivative to be v(t)=-3t^2+8t+6 .

Applying the power rule a second time to the veloctiy function, we find the second derivative to be a(t)=-6t+8 .

Now, by simply plugging t=2 seconds, we find that the acceleration of the paper airplane at that time is $-4\frac{ft}{s^2}$ .

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Example Question #546 : Spatial Calculus

Find the acceleration function a(t) if the velocity function is v(t)=3ln(t) - t .

Possible Answers:

$a(t) = \frac{t}{3} - 1$

$a(t) = \frac{3}{t} - 1$

$a(t) = \frac{3}{t^2} - 1$

$a(t) = \frac{3}{t} - t$

$a(t) = \frac{t^3}{3} - t$

Correct answer:

$a(t) = \frac{3}{t} - 1$

Explanation:

Acceleration is equal to the derivative of velocity with respect to time.

Recall the following rules to help solve this problem.

Power Rule: $\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$

Natural Log Rule: $\frac{\mathrm{d} }{\mathrm{d} t}(aln(bt)) = \frac{a}{t}$

Therefore, the acceleration function, by the power rule and natural log rule, is

$a(t) = v'(t) = \frac{3}{t} - 1$

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Example Question #541 : Spatial Calculus

Find the acceleration function a(t) if the position function is s(t) = cos(t)+3t^2 - 2 .

Possible Answers:

a(t) = 6 - cos(t)

a(t) = 6 + cos(t)

a(t) = 6 - sin(t)

a(t) = 6t + sin(t)

a(t) = 6t - cos(t)

Correct answer:

a(t) = 6 - cos(t)

Explanation:

Acceleration is equal to the second derivative of position with respect to time.

Recall the following rules to help solve the problem.

Power Rule:
$\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$

Differentiation rules for sine and cosine:

$\\\frac{\mathrm{d} }{\mathrm{d} t}(asin(bt)) = abcos(bt)\\ \frac{\mathrm{d} }{\mathrm{d} t}(acos(bt)) = -absin(bt)\\$

Therefore the acceleration function, by the differentiation rule for cosine and the power rule, is

$\\a(t) = \frac{\mathrm{d} }{\mathrm{d} t}\left(\frac{\mathrm{d} }{\mathrm{d} t}(cos(t)+3t^2 - 2)\right)\\ \\ a(t)= \frac{\mathrm{d} }{\mathrm{d} t}(-sin(t)+6t)\\ \\ a(t)= 6 - cos(t)$

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Example Question #548 : Spatial Calculus

Find the average acceleration on the interval $0\leq t \leq 7$ if the velocity function is v(t) = 3t^2 - 8t - 35 .

Possible Answers:

Correct answer:

Explanation:

Average acceleration is defined as $a_{avg} = \frac{\Delta v}{\Delta t}$ .

Therefore, on the interval $0 \leq t \leq 7$ ,

$\\a_{avg} = \frac{v(7)-v(0)}{7-0}\\ a_{avg} = \frac{56 - (-35)}{7}\\ a_{avg} = 13$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #142 : Acceleration

Example Question #541 : Spatial Calculus

Example Question #542 : Spatial Calculus

Example Question #544 : Calculus

Example Question #543 : Spatial Calculus

Example Question #544 : Spatial Calculus

Example Question #545 : Spatial Calculus

Example Question #546 : Spatial Calculus

Example Question #541 : Spatial Calculus

Example Question #548 : Spatial Calculus

All Calculus 1 Resources

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