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Calculus 1 : Position

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Example Questions

Example Question #872 : 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:

768

192

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 #21 : How To Find Position

Find a vector perpendicular to (2,6) .

Possible Answers:

(-2,-6)

(-6,2)

(2,-6)

(6,2)

(-2,6)

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 #21 : 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 #873 : Spatial Calculus

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

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

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

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

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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Example Question #2 : 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 : 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:

285

193

345

510

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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Calculus 1 : Position

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #872 : Spatial Calculus

Example Question #22 : How To Find Position

Example Question #871 : Spatial Calculus

Example Question #21 : How To Find Position

Example Question #21 : How To Find Position

Example Question #873 : Spatial Calculus

Example Question #2 : Integration

Example Question #21 : Position

Example Question #882 : Spatial Calculus

Example Question #883 : Spatial Calculus

All Calculus 1 Resources

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