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Calculus 1 : How to find position

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Example Question #905 : Spatial Calculus

At time t=0 a particle is at the origin at rest with no velocity. It then experiences an acceleration of

$a\left ( t\right ) = \sin{3t}$ .

After $\pi$ seconds, what is the particle's position?

Possible Answers:

$\frac{1}{3}\pi$

$\frac{-1}{9}-\frac{1}{3}\pi$

$\frac{1}{9}-\frac{1}{3}\pi$

$\frac{1}{3}\pi-\frac{1}{9}$

$-\frac{1}{3}\pi$

Correct answer:

$\frac{1}{3}\pi$

Explanation:

First, we integrate $a\left ( t\right )$ with respect to to get velocity:

$v\left ( t\right ) = -\frac{\cos{3t}}{3} + C$

We know that the particle is not moving at t=0 , so

v(0)=0 .

Solving this gives us

$v(t)=\frac{1}{3}-\frac{cos{3t}}{3}$ .

Now, we have to solve for position. To do this, we integrate v(t) with respect to . Thus, we get

$p(t)=\frac{1}{3}t-\frac{\sin{3t}}{9}+C$ .

We know that the particle starts at the origin, so we have to solve:

p(0)=0

This implies that C=0 , so we have:

$p(t)=\frac{1}{3}t-\frac{\sin{3t}}{9}$ .

As a special case:

$p(\pi)=\frac{1}{3}\pi-\frac{\sin{3\cdot \pi}}{9}=\frac{1}{3}\pi$ .

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Example Question #906 : Spatial Calculus

Given the initial velocity, initial position and acceleration of an object, find its position function.

v(0)=-12,p(0)=32,a(t)=5.8

Possible Answers:

p(t)=-6.2t+32

p(t)=2.9t^2-12t+32

p(t)=5.8t^2+20t

p(t)=5.8t^2-12t+32

Correct answer:

p(t)=2.9t^2-12t+32

Explanation:

We begin by integrating the acceleration function and using the initial condition to find the value of the constant of integration:

$\\ \int a(t)dt=\int5.8dt=5.8t+C=v(t)\\ v(0)=C=-12\Rightarrow v(t)=5.8t-12$

Now that we have the velocity function we repeat the process to find the position function:

$\\ \int v(t)dt=\int 5.8t-12dt=2.9t^2-12t+D=p(t)\\ p(0)=32=D\Rightarrow p(t)=2.9t^2-12t+32$

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Example Question #907 : Spatial Calculus

Which of the following is perpendicular to the vector (9,6) ?

Possible Answers:

(6,-9)

(-6,8)

(-9,-6)

(9,-6)

(-9,6)

Correct answer:

(6,-9)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (9,6) , we therefore know its perpendicular vector is (6,-9) .

We can further verify this by noting that the product of a vector and its perpendicular vector is .

Since $(9,6)\times (6,-9)=(9\times 6)+(6\times -9)=54+ (-54)=0$ , the two vectors are perpendicular to each other.

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Example Question #908 : Spatial Calculus

Which of the following is perpendicular to the vector (-5,-1) ?

Possible Answers:

(5,-1)

(5,1)

(-1,5)

(1,-4)

(-5,1)

Correct answer:

(-1,5)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (-5,-1) , we therefore know its perpendicular vector is (-1,5) .

We can further verify this by noting that the product of a vector and its perpendicular vector is .

Since $(-5,-1)\times (-1,5)=(-5\times -1)+(-1\times 5)=5+(-5)=0$ , the two vectors are perpendicular to each other.

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Example Question #909 : Spatial Calculus

Which of the following is perpendicular to the vector (9,7) ?

Possible Answers:

(7,-9)

(-7,8)

(-9,7)

(9,-7)

(-9,-7)

Correct answer:

(7,-9)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (9,7) , we therefore know its perpendicular vector is (-7,9) .

We can further verify this by noting that the product of a vector and its perpendicular vector is .

Since $(9,7)\times (7,-9)=(9\times 7)+(7\times -9)=63+(-63)=0$ , the two vectors are perpendicular to each other.

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Example Question #910 : Spatial Calculus

Which of the following is perpendicular to the vector (7,-6) ?

Possible Answers:

(7,6)

(-6,-7)

(-7,-6)

(-7,6)

(-6,7)

Correct answer:

(-6,-7)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (7,-6) , its perpendicular vector will be (-6,-7) . We can further verify this result by noting that the product of two perpendicular vectors is ; since $(7,-6)\times (-6,-7)=(7\times -6)+(-6\times -7)=-42+42=0$ , we know the two vectors are perpendicular to each other.

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Example Question #51 : How To Find Position

At what time is at a minimum if and ? and denote the position as a function of time and the velocity as a function of time, respectively. If needed, round to the nearest hundredth.

Possible Answers:

1.41

3.11

1.69

3.88

Correct answer:

1.41

Explanation:

In this question, we need to find the minimum of , so we need to take a derivative. But wait! We were given , so we're already one step ahead of the game because is the derivative of .

The initial condition is actually superfluous information because we do not need to integrate.

Setting we get

0=t^3-2t

0=t(t^2-2)

Next we need to set each part equal to zero and solving for x.

t=0

0=t^2-2

2=t^2

$\pm\sqrt2=t$

Therefore, $-\sqt2$ and $\small 0$ can be disregarded because the problem asks for $\small t>0$ .

To check if $\sqrt{2}$ is a minimum, we can use the second derivative test. The derivative of is 3t^2-2 , which is positive at $^{\sqrt{2}}$ , so $^{\sqrt{2}}$ is a minimum.

$\sqrt2\approx1.41$

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Example Question #52 : How To Find Position

Mark runs to his school and back in $\small 8$ seconds. His position can be described by x(t)=-t^2+8t , where $\small t$ is in seconds. During his $\small 8$ second journey, Mark will be furthest from his home when he is at the school. His home is located at $\small x=0$ , and he begins running at $\small t=0$ . Where is his school?

Possible Answers:

$\small x=8$

$\small x=4$

$\small x=0$

$\small x=64$

$\small x=16$

Correct answer:

$\small x=16$

Explanation:

Mark will be furthest from his home when he is at the school, so we need to maximize this function. We must therefore take the derivative and set it equal to zero.

Use the rule,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ to find the derivative.

This result in,

$\small -2t+8=0 \therefore t=4$ .

Using the second derivative test, we can show it's a max because the second derivative is $\small -2$ (which is negative at $\small t=4$ , thus $\small t=4$ is a maximum).

We can substitute $\small 4$ back into the $\small x(t)$ to find where his school is.

$-(4)^2+8\cdot 4=16$ .

So his school is at $\small x=16$ . No distance units were mentioned in the problem, so leaving the answer unitless is acceptable.

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Example Question #53 : How To Find Position

Given the acceleration of a fruit falling from a tree is $-10\frac{m}{s^2}$ , the initial velocity of the fruit is zero, and the initial poisiton of the fruit is meters, find the position of the fruit at two seconds.

Possible Answers:

2 meters

4 meters

3 meters

0 meters

1 meters

Correct answer:

0 meters

Explanation:

To solve this problem, we first have to understand that the derivative of position with respect to time is velocity and the derivative of velocity with respect to time is acceleration. By understanding that concept, we then are able to perform a double intergration to find the position function for the fruit. The general formula for integration is $\int x^{n}=\frac{x^{n+1}}{n+1}+C$ and if integrating a constant $\int n = nx+C$ , where C can be any real number.

The first integration of the acceleration yields

v(t)=-10t+C , however the question states the intial velocity of the fruit is 0, therefore the final velocity equation is .

Performing the second integration yields

S(t)=-5t^2+C ; we are given the initial position of the fruit to be 20 meters, therefore the final position equation becomes S(t)=-5t^2+20 .

We are asked to find the position of the fruit at t=2 seconds , so by plugging in into our position equation, we find that the position of the fruit at is 0 meters.

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Example Question #54 : How To Find Position

A baseball is thrown with an initial velocity of $100\frac{m}{s}$ . Given that the position of the baseball at seconds is meters and the acceleration of baseball is $-1\frac{m}{s^2}$ all the time, find the position equation for the ball.

Possible Answers:

$S(t)=\frac{-t^2}{2}+10t+52$

Not enough information is given.

$S(t)=\frac{-t^2}{2}+100t-268$

$S(t)=\frac{-t^2}{2}+100t-128$

$S(t)=\frac{-t^3}{3}+100t+128$

Correct answer:

$S(t)=\frac{-t^2}{2}+100t-128$

Explanation:

For this problem, it is important to understand that velocity is the derivative of position with respect to time and acceleration is the derivative of velocity with respect to time. Given the initial acceleration equation, we can take the double integral to obtain the position equation. The general formula for integration is

$\int x^{n}=\frac{x^{n+1}}{n+1}+C$ and if integrating a constant $\int n = nx+C$ , where C can be any real number.

Therefore the integral of the acceleration gives

v(t)=-t+C , however we can solve for C given that the inital velocity is $100\frac{m}{s}$ .

This makes the velocity euqation v(t)=-t+100 . Taking the integral of the velocity equation, we obtain the position equation $S(t)=\frac{-t^2}{2}+100t+C$ . We are given the position of the ball at 2 seconds to be 70 meters. Plugging those two values in will allow us to solve for C=-128 .

Therefore the final position equation is

$S(t)=\frac{-t^2}{2}+100t-128$ .

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Calculus 1 : How to find position

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #905 : Spatial Calculus

Example Question #906 : Spatial Calculus

Example Question #907 : Spatial Calculus

Example Question #908 : Spatial Calculus

Example Question #909 : Spatial Calculus

Example Question #910 : Spatial Calculus

Example Question #51 : How To Find Position

Example Question #52 : How To Find Position

Example Question #53 : How To Find Position

Example Question #54 : How To Find Position

All Calculus 1 Resources

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