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Calculus 1 : Calculus

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Example Questions

Example Question #15 : Position

The velocity of an object is given by the equation f'(t) = 10x - 6x^2 . What is the position of the object at t=2 , if the object has an initial position of ?

Possible Answers:

Correct answer:

Explanation:

The position of the object can be found by integrating the velocity equation f'(t) = 10x -6t^2 given.

The position equation is

$f(x)=\int f'(t) = \int 10x -6t^2$

Increase the exponent of each term and then divide that term by the number number that is in the exponent.

f(t) = 5t^2 - 2t^3 + C

We can solve for the constant using the initial position,

f(0) = 0 - 0 +C = 0

Therefore C = 0

Now we can solve for the position at t=2

f(2) = 5(2)^2 - 2(2)^3 = 20 - 16 = 4

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Example Question #16 : Position

If the acceleration of an object is zero at t=3 , which of the following answers best represents the position of the object?

Possible Answers:

All of the numerical answers are correct.

None of the numerical answers are correct.

Correct answer:

All of the numerical answers are correct.

Explanation:

Given the acceleration is zero, integrate the acceleration equation twice to obtain the position equation.

a(t)=0

$s(t)=\int[\int a(t)dt]dt$

$\int a(t)dt=\int 0\: dt =C$

Integrate again to get the position function.

$s(t)=\int C\: dt=Ct$

Substitute t=3 .

s(t)=3C

Since is a constant, the position of the object can be anywhere when acceleration is zero.

Therefore, all of the numerical answers are correct.

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Example Question #11 : How To Find Position

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

Possible Answers:

Correct answer:

Explanation:

Given the velocity equation v(t) = 3t^2 - 2 , we can find the position by differentiating the velocity equation. This can be done using the power rule, which in general form is:

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

Therefore, the integral of the velocity equation is

x(t) = t^3-2t+C .

Using the initial position of 0, we can solve for the integration constant.

$x(0) = 0 - 0 +C = 0 \Rightarrow C= 0$

The complete position equation is then,

x(t) = t^3-2t .

Therefore, at t = 3 , the position will be,

$x(3) = 3^3-2\cdot 3 = 21$ .

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Example Question #871 : Calculus

The acceleration of an object is given by the equation $a(t) = -\cos(t) + 4$ . What is the position equation of the object, if the initial velocity of the object is with an initial position of ?

Possible Answers:

$x(t) = \cos(t) + 2t^2 + 3t +5$

$x(t) = \cos(t) + 2t^2$

$x(t) = \cos(t) + 2t^2 + 3t$

$x(t) = \cos(t) + 2t^2 + 3t +4$

Correct answer:

$x(t) = \cos(t) + 2t^2 + 3t +5$

Explanation:

The position equation of the onject can be found by integrating the acceleration equation $a(t) = -\cos(t) + 4$ twice.

To integrate the acceleration equation we must use the power rule for the second term of the acceleration where if

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

The acceleration equation can also be written as $a(t) = -\sin(t) +4t^0$ .

Applying the power rule to the acceleration equation with the knowledege that the integral of $\cos(t)$ is $\sin(t)$ gives us the velocity equation.

$v(t) = \int (-\cos(t)+4t^0)dt = -\sin(t) + \frac{4t^{0+1}}{0+1} = -\sin(t) +4t +C$

We now use the initial velocity of the object to solve for the velocity equation.

$v(0) = -\sin(0) + 4(0) +C = 0 + 0 +C = 3$

Therefore C = 3

The velocity equation is then $v(t) = -\sin(t) +4t +3$

Differientiating this equation will give the position equation of the object. We must again use the power rule and the knowledge that the integral of $\sin(t)$ is $-\cos(t)$ . Rewriting the velocity equation as

$v(t) = -\sin(t) +4t^1 + 3t^0$ .

Differentiating this equation gives us,

$x(t) = \int (-\sin(t)+4t^1+3t^0)dt$

$= -(-\cos(t)) +\frac{4t^{(1+1)}}{(1+1)} + \frac{3t^{0+1}}{(0+1)} +C \Rightarrow$

$\Rightarrow \cos(t) + 2t^2 + 3t +C$

Using the initial position of the object we can solve for .

$x(0) = \cos(0) + 2(0)^2 +3(0) +C =1+0+0+C= 5$

Therefore, C = 5 and the position equation is

$x(t) = \cos(t) + 2t^2 + 3t +5$

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Example Question #871 : Spatial Calculus

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

Possible Answers:

192

768

Correct answer:

768

Explanation:

The position of the object can be found by integrating the velocity equation and solving for x(4) . To integrate the velocity equation we first rewrite the equation.

v(t) = (4t^2)(3t) = 12t^3

To integrate this equation we must use the power rule where,

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

Applying this to the velocity equation gives us,

$x(t) = \int (12t^3)dt = \frac{12t^{3+1}}{3+1} = \frac{12t^4}{4} = 3t^4 +C$ .

We must solve for the value of by using the initial position of the object.

x(0) = 3(0)^4 + C =0 + C= 0

Therefore, C = 0 and x(t) = 3t^4 .

x(4) = 3(4)^4 = 3(256) = 768

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Example Question #22 : How To Find Position

The velocity of an object is given by the equation $v(t)= 2\sin(t)$ . What is the equation for the position of the object if the object has an initial position of ?

Possible Answers:

$x(t) = -2\cos(t)$

$x(t) = 2\cos(t)$

$x(t) = -2\cos(t) +2$

$x(t) = -2\sin(t)$

Correct answer:

$x(t) = -2\cos(t) +2$

Explanation:

The position of the object can be found by integrating the equation for the object's velocity. Knowing that the derivative of $-\cos(t)$ is $\sin(t)$ , the integral of $\sin(t)$ must be $-\cos(t)$ .

Integrating the velocity equation gives us,

$x(t) = \int (2\sin(t))dt = 2(-\cos(t)) +C = -2\cos(t) +C$

To find the complete solution of the position equation we must use the initial position.

$x(0) = -2\cos(0) +C = -2 +C = 0$

Therefore C = 2 and $x(t) = -2\cos(t) +2$

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Example Question #871 : Spatial Calculus

Find a vector perpendicular to (5,7) .

Possible Answers:

(7,-5)

(-5,7)

(5,-7)

(-7,5)

(-5,-7)

Correct answer:

(-7,5)

Explanation:

By definition, a vector (a,b) has a perpendicular vector (-b,a) .

Therefore, the vector (5,7) has a perpendicular vector (-7,5) .

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Example Question #872 : Spatial Calculus

Find a vector perpendicular to (2,6) .

Possible Answers:

(2,-6)

(-2,6)

(-6,2)

(-2,-6)

(6,2)

Correct answer:

(-6,2)

Explanation:

By definition, a vector (a,b) has a perpendicular vector (-b,a) .

Therefore, the vector (2,6) has a perpendicular vector (-6,2) .

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Example Question #25 : How To Find Position

Find a vector perpendicular to (3,12) .

Possible Answers:

(12,3)

(3,-12)

(-12,3)

(-3,-12)

(-3,12)

Correct answer:

(-12,3)

Explanation:

By definition, a vector (a,b) has a perpendicular vector (-b,a) .

Therefore, the vector (3,12) has a perpendicular vector (-12,3) .

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Example Question #1 : Integration

The acceleration of an object is given by the equation a(t) = 4t+1 . What is the equation for the position of the object, if the object has an initial velocity of and an initial position of ?

Possible Answers:

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 5t$

x(t) = 2t^3 + t^2 + 10t

x(t) = 2t^3 - t^2 + 10t

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

Correct answer:

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

Explanation:

To find the position of the object we must use the power rule to integrate the acceleration equation twice. The power rule is such that

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

Therefore integrating the acceleration equation gives us

$v(t) = \int (4t+1)dt = \frac{4t^{1+1}}{1+1} +\frac{1t^{0+1}}{0+1} +C = 2t^2 + t +C$

We can solve for the value of by using the initial velocity of the object.

v(0) = 2(0)^2 +(0)+C = 10

Therefore C = 10 and v(t) = 2t^2 + t + 10

To find the position of the object we integrate the velocity equation.

$x(t) = \int (2t^2+t+10)dt = \frac{2t^{2+1}}{2+1} + \frac{t^{1+1}}{1+1} + \frac{10t^{0+1}}{0+1} +C = \frac{2t^3}{3}+\frac{t^2}{2}+10t +C$

We can solve for this new value of by using the object's initial position

$x(0) = \frac{2(0)^3}{3} + \frac{(0)^2}{2} + 10(0) +C = 0 + C = 0$

Therefore C = 0 and $x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #15 : Position

Example Question #16 : Position

Example Question #11 : How To Find Position

Example Question #871 : Calculus

Example Question #871 : Spatial Calculus

Example Question #22 : How To Find Position

Example Question #871 : Spatial Calculus

Example Question #872 : Spatial Calculus

Example Question #25 : How To Find Position

Example Question #1 : Integration

All Calculus 1 Resources

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