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Calculus 1 : Calculus

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Example Questions

Example Question #1 : Integration

The velocity of an object is given by the equation v(t) = 6t -3 . What is the position of the object at time t= 4 if the object has a position of and time t = 1 ?

Possible Answers:

Correct answer:

Explanation:

To find the position of the object we must first find the position equation of the object. The position equation can be found by integrating the velocity equation. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Using this rule we find that

$x(t) = \int (6t-3)dt = \frac{6t^{1+1}}{1+1} - \frac{3t^{0+1}}{0+1} +C = 3t^2 - 3t +C$

Using the position of the object at time t = 1 we can solve for

x(1) = 3(1)^2 - 3(1) + C = 3 - 3 + C =0+C= 6

Therefore C = 6 and x(t) = 3t^2 - 3t +6

We can now find the position at time t = 4 .

x(4) = 3(4)^2 - 3(4) +6 = 48-12 +6 = 42

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Example Question #21 : How To Find Position

The velocity of an object is given by the equation v(t) = 5t^4 + 3t^2 +5 . what is the position of the object at t= 3 , if the initial position of the object is ?

Possible Answers:

193

285

510

345

Correct answer:

285

Explanation:

The position of the object can be found by integrating the velocity of the object. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Using the power rule the position of the object is

$x(t) = \int (5t^4 + 3t^2 + 5)dt = \frac{5t^{(4+1)}}{4+1} +\frac{3t^{(2+1)}}{2+1} + \frac{5t^{(0+1)}}{0+1} +C= t^5 +t^3 +5t +C$ .

The value of can be found using the initial position of the object.

x(0) = (0)^5 +(0)^3 + 5(0) +C = 0

Therefore C = 0 and x(t) = t^5 + t^3 +5t .

The position of the object at t = 3 can now be found,

x(3) = (3)^5 + (3)^3 + 5(3) =243+ 27+ 15 = 285 .

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Example Question #882 : Spatial Calculus

The velocity of an object is given by the equation v(t) = 12t - 39 . What is the position of the object at time t = 2 , if the initial position of the object is ?

Possible Answers:

None of these.

Correct answer:

Explanation:

The position of the object can be found by integrating the velocity of the object. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Therefore the position equation of the object is

$x(t) = \int (12t-39)dt = \frac{12t^{(1+1)}}{1+1} -\frac{39t^{(0+1)}}{0+1}+C = 6t^2 - 39t +C$

We must now solve for the constant . We can do this using the initial position of the object.

x(0) = 6(0)^2 - 39(0) +C = 0 - 0 + C = 0

Therefore C = 0 and x(t) = 6t^2 - 39t

We can now find the position of the object at t = 2 .

x(2) = 6(2)^2 - 39(2) = 24 - 78 = -54

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Example Question #883 : Spatial Calculus

The velocity of an object is v(t) = 1 - 9t^2 . What is the position of the object when t = 1 , if the position of the object is at t =2 ?

Possible Answers:

Correct answer:

Explanation:

The position of the object can be found by integrating the object's velocity. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the position of the object is

$x(t) = \int (1t^0-9t^2)dt = \frac{1t^{(0+1)}}{0+1} + \frac{9t^{(2+1)}}{2+1}+C = t + 3t^3 + C$ .

We can find the value of using the position at t =2 .

x(2) = (2) - 3(2)^3 + C = 2-24+C=-22+C= 10

Therefore C= 32 and x(t) = t - 3t^3 + 32 .

x(1) = (1) - 3(1)^3 +32 = -2 + 32 = 30

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Example Question #6 : Integration

The velocity of an object is v(t) = 3t^2 . What is the position of the object if its initial position is 1000 ?

Possible Answers:

x(t) = 6t +1000

x(t) = t^3 +1000

$x(t) = \frac{t^4}{4} +1000$

x(t) = t^3 -1000

Correct answer:

x(t) = t^3 +1000

Explanation:

The position is the integral of the velocity. By integrating with the power rule we can find the object's position.

The power rule is where

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the position of the object is

$x(t) = \int (3t^2)dt = \frac{3t^{(2+1)}}{2+1} +C = t^3 +C$ .

We can solve for the constant using the object's initial position.

x(0) = (0)^3 + C = 0+C =1000

Therefore C= 1000 and x(t) = t^3 +1000 .

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Example Question #32 : How To Find Position

The velocity of an object is given by the equation v(t) = 18t^2 - 11 . What is the position of the object at time t= 3 if the initial position of the object is ?

Possible Answers:

139

Correct answer:

139

Explanation:

The position of the object can be found by integrating the velocity. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Using this rule we find that

$x(t) = \int (18t^2-11)dt = \frac{18t^{(2+1)}}{2+1} - \frac{11t^{(0+1)}}{0+1} +C = 6t^3 - 11t +C$ .

We can find the value of using the initial position of the object.

x(0) = 6(0)^3 - 11(0) +C = 0 - 0 +C =10

Therefore C = 10 and x(t) = 6t^3 - 11t +10 .

x(3) = 6(3)^3 - 11(3) + 10 = 162 - 11(3) + 10 = 139

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Example Question #33 : How To Find Position

The velocity of an object is given by the equation v(t) = 44t^3 + t^2 +2 . What is the position of the object at time t =1 , if the initial position of the object is ?

Possible Answers:

$9 \frac{1}{3}$

$15 \frac{1}{3}$

$11 \frac{1}{3}$

Correct answer:

$15 \frac{1}{3}$

Explanation:

The position of the object can be found by integrating the velocity. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Using this rule to integrate the velocity gives us

$x(t) = \int (44t^3 + t^2 + 2)dt = \frac{44t^{(3+1)}}{3+1} + \frac{t^{(2+1)}}{2+1} +\frac{2t^{(0+1)}}{0+1} +C$

$x(t) = 11t^4 + \frac{t^3}{3} + 2t +C$ .

The value of can be found by using the initial position of the object.

$x(0)= 11(0)^4 +\frac{(0)^3}{3} + 2(0) +C = 0 + 0+0+C = 2$

Therefore C = 2 and $x(t) = 11t^4 + \frac{t^3}{3} + 2t + 2$ .

Using the position equation we find the position at t = 1 .

$x(1) = 11(1)^4 + \frac{(1)^3}{3} + 2(1) +2 = 11 + \frac{1}{3} + 2 +2 = 15 \frac{1}{3}$

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Example Question #882 : Calculus

Find the position of an object at $t=2\pi$ if the velocity function is v(t)= cos(t) .

Possible Answers:

$\frac{\sqrt2}{2}$

$\frac{1}{2}$

Correct answer:

Explanation:

To determine the position of an object given the velocity function, integrate once to obtain the position function.

$x(t)=\int v(t)\: dt=\int cos(t)\:dt= sin(t)$

Substitute the value $t=2\pi$ into the position function.

$sin(2\pi)=0$

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Example Question #35 : How To Find Position

Suppose the acceleration function is described by a(t)=sin(2t)+cos(t) . What is the position when $t=\frac{\pi}{2}$ ?

Possible Answers:

$\frac{\sqrt2}{2}$

$-\frac{\sqrt2}{2}$

Correct answer:

Explanation:

Obtain the position function by integrating the acceleration twice.

a(t)=sin(2t)+cos(t)

$v(t)= \int a(t)\:dt = -\frac{1}{2}cos(2t)+sin(t)$

Integrate again to obtain the position function.

$x(t)= \int v(t)\:dt= \int -\frac{1}{2}cos(2t)+sin(t)\:dt$

$x(t)=-\frac{1}{4}sin(2t)-cos(t)$

Substitute $t=\frac{\pi}{2}$ .

$x(t=\frac{\pi}{2})=-\frac{1}{4}sin\left(2\left(\frac{\pi}{2}\right)\right)-cos\left(\frac{\pi}{2}\right)=0-0=0$

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Example Question #36 : How To Find Position

Find a vector perpendicular to (6,5) .

Possible Answers:

(-5,6)

(6,-5)

(-6,5)

(5,-6)

(-6,-5)

Correct answer:

(5,-6)

Explanation:

By definition, any vector (a,b) has a perpendicular vector (b,-a) . Given a vector (6,5) , the perpendicular vector is (5,-6) .

We can verify this further by noting that the product of any vector and its perpendicular vector is equal to , or $x\times y=0$ . Taking the product of (6,5) and (5,-6) , we get:

$(6,5)\times (5,-6)=(6\times 5)+(5\times -6)=30+(-30)=0$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : Integration

Example Question #21 : How To Find Position

Example Question #882 : Spatial Calculus

Example Question #883 : Spatial Calculus

Example Question #6 : Integration

Example Question #32 : How To Find Position

Example Question #33 : How To Find Position

Example Question #882 : Calculus

Example Question #35 : How To Find Position

Example Question #36 : How To Find Position

All Calculus 1 Resources

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