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Calculus 1 : Position

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Example Questions

Example Question #861 : Spatial Calculus

If the velocity of an object is represented by v(t)=30t^2 , what is the position of the object at t=2 ?

Possible Answers:

240

120

Correct answer:

Explanation:

To find the position given the velocity curve, take the antiderivative of v(t)=30t^2 .

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

Solve for,

s(t=2) .

s(t=2)=10(2^3)=80

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Example Question #1 : Integration

Find the position function given the velocity function:

v(t)=3t^5+4t

Possible Answers:

$\frac{1}{2}t^6+\frac{1}{2}t^2$

$\frac{1}{2}t^6+2t^2$

15t^4+4

t^6+t^2

3t^6+4t^2

Correct answer:

$\frac{1}{2}t^6+2t^2$

Explanation:

To find the position from the velocity function, integrate

v(t)=3t^5+4t 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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Example Question #862 : Spatial Calculus

Suppose the velocity function of an object is v(t)=6t+2 . What is the position of the object at time t=2 ?

Possible Answers:

Correct answer:

Explanation:

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

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

s(t)= 3t^2+2t

Solve for s(t=2) .

s(t=2)=3(2)^2+2(2)= 12+4=16

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Example Question #13 : How To Find Position

Suppose an object's acceleration function is a(t)=2t-1 . The object's velocity is . What should be the object's position at that velocity?

Possible Answers:

$\frac{9}{2}$

There is not enough information.

$\frac{2}{3}$

Correct answer:

$\frac{9}{2}$

Explanation:

Integrate a(t)=2t-1 to obtain the velocity function, v(t) .

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

Since velocity is 6, v(t)=6 . Substitute and solve for time.

6=t^2-t

0=t^2-t-6

0=(t-3)(t+2)

t=3,-2

Since negative time does not exist, the only valid solution for t at this velocity is t=3 .

Integrate the velocity function v(t) to find the position function, s(t) .

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

Substitute t=3 to s(t) 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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Example Question #863 : Spatial Calculus

Consider the velocity function given by v(t) :

v(t)=5t^4-343t^6+51t^2

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

Possible Answers:

p(t)=t^5-49t^7+17t^3

p(t)=60t^2-10290t^4+102

p(t)=120t-30870t^3

p(t)=t^5-49t^7+17t^3+c

p(t)=20t^3-2058t^5+102t

Correct answer:

p(t)=t^5-49t^7+17t^3+c

Explanation:

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)

v(t)=5t^4-343t^6+51t^2

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

So our answer is:

p(t)=t^5-49t^7+17t^3+c

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Example Question #2 : Integration

Consider the velocity function given by v(t) :

v(t)=5t^4-343t^6+51t^2

Find the position of a particle after seconds if its velocity can be modeled by v(t) and the graph of its position function passes through the point (2,2) .

Possible Answers:

-38289.79

-3828.979

3828979

Correct answer:

Explanation:

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)

v(t)=5t^4-343t^6+51t^2

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

So we get:

p(t)=t^5-49t^7+17t^3+c

What we ultimately need is p(5), but first we need to find c: Use the point (2,2)

p(t)=t^5-49t^7+17t^3+c

$2=2^5-49\cdot 2^7+17\cdot 2^3+c$

c=6106

So our position function is:

p(t)=t^5-49t^7+17t^3+6106

$p(5)=5^5-49\cdot 5^7+17\cdot 5^3+6106=-3816769$

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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 #12 : How To Find Position

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:

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

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

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

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

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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Calculus 1 : Position

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #861 : Spatial Calculus

Example Question #1 : Integration

Example Question #862 : Spatial Calculus

Example Question #13 : How To Find Position

Example Question #863 : Spatial Calculus

Example Question #2 : Integration

Example Question #15 : Position

Example Question #16 : Position

Example Question #11 : How To Find Position

Example Question #12 : How To Find Position

All Calculus 1 Resources

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