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Flashcards: Calculus AB

The function describing the acceleration of a spacecraft with respect to time is

a(t)=4t

Determine the function describing the position of a spacecraft given that the initial acceleration is 0, the initial velocity is 3, and the initial position is 9.

$\frac{2t^3}{3}+3t$

$\frac{2t^3}{3}+9$

$\frac{2t^3}{3}+3t+9$

2t^2+3

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Flashcards: Calculus AB

The function describing the acceleration of a spacecraft with respect to time is

a(t)=4t

Determine the function describing the position of a spacecraft given that the initial acceleration is 0, the initial velocity is 3, and the initial position is 9.

Your answer:

$\frac{2t^3}{3}+3t$

Your answer:

$\frac{2t^3}{3}+9$

Your answer:

2t^2+3

Correct answer:

$\frac{2t^3}{3}+3t+9$

Explanation:

To find the position function from the acceleration function, we integrate the acceleration function to find the velocity function, and integrate again to get the position function:

$v(t)=\int a(t)dt=\int 4t dt= 2t^2+C$

The integral was found using the following rule:

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

To find the constant of integration, we use the initial velocity condition given:

v(0)=3=0+C=C

Now, after replacing C with the known value, we integrate the velocity function to get the position function:

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

The same rule of integration was used as above.

We use the same procedure to solve for C, too, only this time using the initial position condition:

s(0)=9=0+0+C=C

Our final answer is

$s(t)=\frac{2t^3}{3}+3t+9$

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