Position - AP Calculus AB

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Question

Find a vector perpendicular to (4,3).

Answer

In general, if we have a vector (a,b), a perpendicular vector is (b,-a).

So here, the perpendicular vector is (3,-4).

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Question

if a=i + 2j - 3k and b=4i + 7k, express the vector 3a + 2b.

Answer

To express the vector in terms of i, j, and k, we need to combine like terms and distribute.

3a + 2b

= 3(i + 2j - 3k) + 2(4i + 7k)

= 3i + 6j - 9k + 8i + 14k

= 11i +6j + 5k

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Question

The veloctiy of a particle at time is given by . What is its change in position between time and time ?

Answer

The position function is the intergral of the velocity function. So here, position is given by where is the constant of integration. Because only a difference in position is asked, and not an absolute position, the constant of integration cancels out.

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Question

The velocity of a particle is given by the function $v(t) = 3\sqrt{t}+2t$ . What is it's position at time if it's starting position was 4

Answer

To find the position from velocity, the function must be integrated. This gives $s(t)=2t^{3/2}+t^2+C$ . substituting 4 for and using the given initial condition gives the answer

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Question

Find the position at if the acceleration function is: .

Answer

To find the position from the acceleration function, integrate the acceleration function twice.

$v(t)= \int a(t)dt = t$

$s(t) = \int v(t) dt = \frac{t^2}{2}$

Substitute to find the postion.

$s(t=2)= \frac{2^2}{2} = 2$

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Question

Find the position at if the acceleration is: .

Answer

To find the position function, integrate the acceleration function twice.

$v(t)= \int a(t) dt= \int (t^2+2t+1 )dt = \frac{1}{3}t^3 + t^2 +t$

$s(t)=\int v(t)dt=\int (\frac{1}{3}t^3+t^2+t) :dt=\frac{1}{12}t^4+\frac{1}{3}t^3+\frac{1}{2}t^2$

Evalute the position at .

$s(t=2)=\frac{1}{12}(2)^4+\frac{1}{3}(2)^3+\frac{1}{2}(2)^2=\frac{4}{3}+\frac{8}{3}+2=6$

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Question

The velocity of an object is given by the following equation:

If , find the equation for the position of the object at any time .

Answer

Velocity is the derivative of position, so in order to obtain an equation for position, we must integrate the given equation for velocity:

$s(t)=\int v(t)dt=\int (7t+4)dt=\frac{7}{2}t^2+4t+C$

The next step is to solve for C by applying the given initial condition, s(0)=5:

$s(t)=\frac{7}{2}t^2+4t+C$

$s(0)=0+0+C=5\rightarrow C=5$

So our final equation for position is:

$s(t)=\frac{7}{2}t^2+4t+5$

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Question

The position function of a ball from the ground when it is thrown by a pitcher is .

Where is the ball located at ?

Answer

To find the position of the ball, we plug in

So turns into:

$p(1)=4\cdot 1+2$

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Question

Harold is sitting in a parked car 10 feet from a wall. He begins to accelerate the car at a rate of 14 feet per second squared away from the wall. How far away from the wall will he be in 6 seconds?

Answer

To begin, let's recognize that since acceleration is the second derivative of position with respect to time, we can in turn, integrate an acceleration function twice with respect to time to find position.

Since accleration is a constant, 14 feet per second, our acceleration function is:

Integrating this once gives us a velocity function

$v(t) = (14 feet/s^2) \cdot t + v_0$

where

is our initial velocity. Since the car started at rest, v0 will be equal to zero, leaving us with:

$v(t) = (14 feet/s^2) \cdot t$

To get our position function, we can in turn integrate our velocity function:

$x(t) = (1/2)(14 feet/s^2) \cdot t^2 + x_0$

where x0 is our initial position. We are told that our initial position is 10 feet away from the wall, so we can rewrite this equation as:

$x(t) = (1/2)(14 feet/s^2) \cdot t^2+ 10 feet$

To find where the car is at 6 seconds, we need only plug in the value of 6 seconds where ever our time variable t occurs, giving our answer of 262 feet.

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Question

Function gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Find the equation that models the particle's postion as a function of time.

Answer

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

We are starting with the following

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

We need to perform the following:

$\small \small \small \small\int v(t)dt=\int t^5+3t^4-\frac{7t^2}{3}dt$

Recall that to integrate, we add one to each exponent and divide by the that number, so we get the following. Don't forget your +c as well.

$\small \small \int t^5+3t^4-\frac{7t^2}{3}dt=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Which makes our position function, h(t), the following:

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

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Question

Consider the velocity function modeled in meters per second by v(t).

Find the position of a particle whose velocity is modeled by after seconds.

Answer

Recall that velocity is the first derivative of position, so to find the position function we need to integrate .

$V(t)=\int 4t^3-5t^2+6dt$

Becomes,

$V(t)=t^4-\frac{5t^3}{3}+6t+c$

Then, we need to find

$V(10)=10^4-\frac{5\cdot10^3}{3}+6\cdot10+c=8393.3\bar{3}+c$

So our final answer is:

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

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Question

If the velocity of an object is represented by , what is the position of the object at ?

Answer

To find the position given the velocity curve, take the antiderivative of .

$s(t)=\int v(t)dt=30\int t^2=30\left[\frac{t^{2+1}}{2+1} \right ]=30\left[\frac{t^3}{3} \right ]=\frac{30t^3}{3}=10t^3$

Solve for,

.

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Question

Find the position function given the velocity function:

Answer

To find the position from the velocity function, integrate

by increasing the exponent of each t term and then dividing that term by the new exponent value.

$s(t)=\int v(t)dt= \frac{3}{6}t^6+2t^2=\frac{1}{2}t^6+2t^2$

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Question

If the acceleration function of an object is , what is the position of the object at ? Assume the initial velocity and position is zero.

Answer

To find the position function from the acceleration function, integrate twice.

$v(t)=\int a(t)dt=t^2+11t$

When integrating, remember to increase the exponent of the variable by one and then divide the term by the new exponent. Do this for each term.

$s(t)=\int v(t)dt=\int\int a(t)dt = \frac{1}{3}t^3+\frac{11}{2}t^2$

Solve for .

$\frac{1}{3}(1)^3+\frac{11}{2}(1)^2=\frac{1}{3}+\frac{11}{2} = \frac{2}{6}+\frac{33}{6}= \frac{35}{6}$

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Question

Find the position function if and .

Answer

In order to find the position function from the velocity function we need to take the integral of the velocity function.

$s(t)=\int v(t)dt$ .

When taking the integral, we will use the inverse power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Applying this rule to each term we get,

$\s(t)= \int v(t)=\int (2t+3)dt \ \s(t)=\frac{2t^2}{2}+3t+c\ \s(t)= t^2+3t+c$

To find the value of the constant c we will use the initial condition given in the problem.

Setting the initial condition ,

yields .

Therefore the position function becomes,

.

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Question

Suppose the velocity function of an object is . What is the position of the object at time ?

Answer

To find the position function given the velocity, the velocity function needs to be integrated.

$s(t)=\int v(t)dt= \int (6t+2)dt$

Solve for .

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Question

The velocity of an object is given by the equation . What is the position of the object at , if the object has an initial position of ?

Answer

The position of the object can be found by integrating the velocity equation 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.

We can solve for the constant using the initial position,

Therefore

Now we can solve for the position at t=2

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Question

The velocity of an object is given by the equation . What is the position of the object at if it has an initial position of zero?

Answer

Given the velocity equation , 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

.

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,

.

Therefore, at , the position will be,

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

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Question

Suppose an object's acceleration function is . The object's velocity is . What should be the object's position at that velocity?

Answer

Integrate to obtain the velocity function, .

$v(t)=\int a(t)dt=\int (2t-1) dt=t^2-t$

Since velocity is 6, . Substitute and solve for time.

Since negative time does not exist, the only valid solution for t at this velocity is .

Integrate the velocity function to find the position function, .

$s(t)=\int v(t)dt=\frac{t^3}{3}-\frac{t^2}{2}$

Substitute to to obtain the position.

$\frac{3^3}{3}-\frac{3^2}{2}= \frac{27}{3}-\frac{9}{2}=9-\frac{9}{2}=\frac{9}{2}$

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Question

Consider the velocity function given by :

Find the equation which models the position of a particle if its velocity can be modeled by .

Answer

Recall that velocity is the first derivative of position and acceleration is the second derivative of position. Therfore, we need to integrate v(t) to find p(t)

$p(t)=V(t)=\int 5t^4-343t^6+51t^2dt$

So our answer is:

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