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Calculus 1 : Position

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Example Questions

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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Example Question #891 : Spatial Calculus

Find a vector perpendicular to (2,4) .

Possible Answers:

(-2,4)

(4,-2)

(-2,-4)

(2,-4)

(-4,2)

Correct answer:

(4,-2)

Explanation:

By definition, any vector (a,b) has a perpendicular vector (b,-a) . Given a vector (2,4) , the perpendicular vector is (4,-2) .

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 (2,4) and (4,-2) , we get:

$(2,4)\times (4,-2)=(2\times 4)+(4\times -2)=8+(-8)=0$

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Example Question #892 : Spatial Calculus

Find a vector perpendicular to (7,-12) .

Possible Answers:

(-12,-7)

(-7,12)

(-12,7)

(7,12)

(-7,-12)

Correct answer:

(-12,-7)

Explanation:

By definition, any vector (a,b) has a perpendicular vector (b,-a) . Given a vector (7,-12) , the perpendicular vector is (-12,-7) .

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 (7,-12) and (-12,-7) , we get:

$(7,-12)\times (-12,-7)=(7\times -12)+(-12\times -7)=-84+84=0$

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Example Question #7 : Integration

If the velocity function of a car is v(t)= 3cos(t) , what is the position when t=1 ?

Possible Answers:

3sin(1)

3cos(1)

$-\frac{9\pi}{2}$

$\frac{3\pi}{2}$

Correct answer:

3sin(1)

Explanation:

To find the position from the velocity function, take the integral of the velocity function.

$\int v(t)\:dt=\int 3cos(t)\:dt$

s(t)=3sin(t)

Substitute t=1 .

s(t=1)=3sin(1)

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Example Question #893 : Spatial Calculus

For this question, remember that velocity is $\frac{position}{time}$ .

A particle's velocity is given by the function y=x . What is the particle's position at x = 4 ?

Possible Answers:

$\frac{x^2}{2}$

Correct answer:

Explanation:

Position is the anti-derivative of velocity, so we find the anti-derivative of the given velocity function to find the position function.

The velocity function is given as y =x , and the anti-derivative of this using the power rule is

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

$v(t)=x \rightarrow f(x)=\frac{x^{1+1}}{2}=\frac{x^2}{2}$ .

We then evaluate that function at the given point x = 4 to get,

$\frac{4^2}{2}=\frac{16}{2}=8$ .

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Calculus 1 : Position

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

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

Example Question #891 : Spatial Calculus

Example Question #892 : Spatial Calculus

Example Question #7 : Integration

Example Question #893 : Spatial Calculus

All Calculus 1 Resources

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