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AP Calculus AB : How to find position

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Example Question #1 : How To Find Position

Given the velocity function $v(t)=t^3-t+7$ , find the position at time $x=2$ with initial position $x(0)=2$ .

Possible Answers:

Correct answer:

Explanation:

First, we find the position function $x(t)=\int (t^3-t+7) dt=\frac{t^4}{4}-\frac{t^2}{2}+7t+constant$

$x(0)=2\Rightarrow constant=2\Rightarrow x(t)=\frac{t^4}{4}-\frac{t^2}{2}+7t+2$

Therefore, $x(2)=\frac{2^4}{4}-\frac{2^2}{2}+7*2+2=18$ .

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Example Question #2 : How To Find Position

Given the acceleration function is $a(t)=2t^2-t+3$ , intial velocity is 2 m/s, and the displacement at t=1 is x(1)=4 .

Find its initial position.

Possible Answers:

0.5

-0.5

Correct answer:

0.5

Explanation:

First, we find the velocity function by integrating the acceleration function $v(t)=\int 2t^2-t+3 = \frac{2}{3}t^3-\frac{1}{2}t^2+3t+c$

Since the initial velocity is 2 m/s, we have c=2 .

Then we integrate again to get the displacement function.

$x(t)=\int \frac{2}{3}t^3-\frac{1}{2}t^2+3t+2=\frac{1}{6}t^4-\frac{1}{6}t^3+\frac{3}{2}t^2+2t+k$

$x(1)=1.5+2+k=3.5+k=4$

This implies k=0.5

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AP Calculus AB : How to find position

Study concepts, example questions & explanations for AP Calculus AB

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Example Question #1 : How To Find Position

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