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Calculus 1 : Calculus

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Example Questions

Example Question #297 : Acceleration

The position of a particle is given by the equation p(t)=t^3+3t^2-15t+6 . What is the particle's acceleration at t=1 ?

Possible Answers:

$12\,m/s^2$

$-6\,m/s^2$

$-5\, m/s^2$

$6\,m/s^2$

Correct answer:

$12\,m/s^2$

Explanation:

To solve this, we'll need to first differentiate the position function to get the velocity function, differentiate that function to get the acceleration function, and then plug our specific time into that funtion to get the acceleration at that moment. Differentiating the position equation term-by-term, using the power rule, which states that

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

and the constant rule, which states that the derivative of a constant is 0, we get

$\begin{align*}v(t)&= p'(t) \\ p(t)&=t^3+3t^2-15t+6 \\ p'(t) &= 3t^2+(2\cdot 3)t^{2-1}-15t^{1-1}+0 \\ p'(t)&=3t^2+6t-15 \\v(t) &=3t^2+6t-15\end{align*}$

Applying the same rules to this equation gives us

$\begin{align*}a(t)&= v'(t) \\ v(t)&=3t^2+6t-15 \\ v'(t) &= (2\cdot 3)t^{2-1}+6t^{1-1}+0 \\ v'(t)&=6t+6 \\a(t) &=6t+6\end{align*}$

Evaluating at t=3 gives us

$\begin{align*} a(1)&= 6(1)+6 \\ &=6+6\\&=12\,m/s \end{align*}$

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Example Question #691 : Calculus

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

Possible Answers:

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

a(t)=1

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

a(t)=3

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

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 #692 : Calculus

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:

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

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

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

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

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

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 #301 : 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) = 4+5t, a(t)=5

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

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

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

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

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 #304 : How To Find Acceleration

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

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

Find the acceleration.

Possible Answers:

4319.4

124.7

234.2

1602.3

Answer not listed

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 #305 : 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

21.5

Answer not listed

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 #306 : 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

1.1

2.5

Answer not listed

-3.3

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 #308 : How To Find 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 #309 : How To Find Acceleration

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

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

Find the acceleration.

Possible Answers:

4.252

-3.002

Answer not listed

-1.003

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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #297 : Acceleration

Example Question #691 : Calculus

Example Question #692 : Calculus

Example Question #301 : Acceleration

Example Question #304 : How To Find Acceleration

Example Question #305 : How To Find Acceleration

Example Question #306 : How To Find Acceleration

Example Question #307 : How To Find Acceleration

Example Question #308 : How To Find Acceleration

Example Question #309 : How To Find Acceleration

All Calculus 1 Resources

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