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Calculus 2 : Applications in Physics

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Example Questions

Example Question #51 : Integral Applications

If v(t)=4t+1 and x(2)=3 , what is the original position function?

Possible Answers:

x(t)=2t^2+t+7

x(t)=t^2+t-7

x(t)=2t^2+3t-7

x(t)=2t^2+t-17

x(t)=2t^2+t-7

Correct answer:

x(t)=2t^2+t-7

Explanation:

Recall that the integral of the velocity function is the position function:

$\int 4t+1dt$

Now, integrate:
$\frac{4t^2}{2}+t=2t^2+t+C$

Now plug in your initial conditions:

2(2^2)+2+C=3; C=-7

Plug your C back into your position function:

x(t)=2t^2+t-7

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Example Question #51 : Applications In Physics

If v(t)=4t-1 and x(2)=3 , what is the original position function?

Possible Answers:

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

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

x(t)=2t^2-4t-3

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

x(t)=2t^2-t

Correct answer:

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

Explanation:

Recall that the integral of velocity is the position function:

$\int 4t-1dt$

Now, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{4t^2}{2}-t=2t^2-t+C$

Now plug in your initial conditions to solve for C:

2(2^2)-2+C=3

C=-3

Now plug your C back in to your position function:

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

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Example Question #52 : Applications In Physics

If v(t)=6t+2 and x(1)=4 , what is the original position function?

Possible Answers:

x(t)=3t^2+2t-8

x(t)=3t^2+6t-1

x(t)=4t^2+2t-1

x(t)=3t^2+2t-1

x(t)=3t^2+2t+1

Correct answer:

x(t)=3t^2+2t-1

Explanation:

Recall that the integral of the velocity function is the position function:

$\int 6t+2dt$

Now, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{6t^2}{2}+2t=3t^2+2t+C$

Now, plug in your initial conditions:

3(1^2)+2(1)+C=4

C=-1

Now plug your C back into your position function:

x(t)=3t^2+2t-1

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Example Question #51 : Applications In Physics

In the vacuum of space, a particle is initially moving at a constant speed of only 6 meters/second away from you as you watch in your space capsule stationary relative to the particle. You track the object and then observe an acceleration that varies with time as,

$\small a(t)=2t^2$

After travelling for 20 minutes at this speed the particle is now $4\times 10^{9}$ km away from you. How far was it from you at the instant it began accelerating?

Possible Answers:

$x_o \approx 3.12\times 10^{9} m \: \: \:or\: \: \:3.12\times 10^{6} km$

$x_o \approx 2.12\times 10^{9} m \: \: \:or\: \: \:2.12\times 10^{6} km$

$x_o \approx 3.12\times 10^{12} m \: \: \:or\: \: \:3.12\times 10^{9} km$

$x_o \approx 3.65\times 10^{9} m \: \: \:or\: \: \:3.65\times 10^{6} km$

$x_o \approx 3.65\times 10^{12} m \: \: \:or\: \: \:3.65\times 10^{9} km$

Correct answer:

$x_o \approx 3.65\times 10^{12} m \: \: \:or\: \: \:3.65\times 10^{9} km$

Explanation:

$\small a(t)=2t^2$

To solve the problem we first need to find the position function by successively integrating. First find the velocity function by integrating the acceleration:

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

$\small v(t) = \frac{2}{3}t^3 +v_o$

The constant of integration $\small v_o$ is the initial velocity, which we are given as 6 meters-per-second. For convenience, we will omit the units in the equation.

$\small v(t)=\frac{2}{3}t^3 +6$

Now we integrate the velocity function to find the position function:

$\small \small x(t)=\int v(t) dt = \int \left(\frac{2}{3}t^3 +6 \right )dt = \frac{2}{12}t^4+6t+x_o$

$\small \small \small x(t)= \frac{1}{6}t^4+6t+x_o$

The constant of integration x_o represents the initial position when t=0 . After $\small t= 20 minutes = 1200 seconds$ the particle is $4\times 10^{9}$ km, or $4\times 10^{12}$ meters, away. Using standard S.I. units (meters and seconds), solve for x_o at t=1200 seconds where $x(t=1200)=4\times 10^{12}$ .

$\small x (t=1200)\: \: =\: \: \frac{1}{6}(1200)^4+6(1200)+x_o\: \: \: =\: \: \: 4\times 10^{12}$

Now solve for x_o

$x_o = 4\times 10^{12} -\frac{1}{6}(1200)^4-6(1200)$

$x_o \approx 3.65\times 10^{12} m \: \: \:or\: \: \:3.65\times 10^{9} km$

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Calculus 2 : Applications in Physics

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #51 : Integral Applications

Example Question #51 : Applications In Physics

Example Question #52 : Applications In Physics

Example Question #51 : Applications In Physics

All Calculus 2 Resources

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