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AP Calculus AB : AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

A particle is moving in a straight path with a constant initial velocity. The particle is then subjected to a force causing a time-dependent acceleration given as a function of time:

a(t)=(a+b)t

After 10 seconds, the particle has a velocity equal to meters-per-second. Find the initial velocity in terms of the constants , and

Units are all in S.I. (meters, seconds, meters-per-second, etc.)

Possible Answers:

v_o =k-50(a+b)

v_o =b-5(a+k)

$v_o =\frac{b}{a}+k$

$v_o =\frac{k}{a+b}-50a$

$v_o =\frac{k}{a}+b$

Correct answer:

v_o =k-50(a+b)

Explanation:

a(t)=(a+b)t

Begin by finding the velocity function by integrating the acceleration function.

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

We use v_o as the constant of integration since the function v(t) is a velocity and at initially, at t = 0 , the velocity is equals the constant of integration.

$v(t)=\frac{1}{2}(a+b)t^2+v_o$

At t = 10 seconds we are told the velocity is equal to .
$v(t=10)=k\: \: \: \: \: \:\rightarrow\: \: \: \: \: \:\frac{1}{2}(a+b)100+v_o =k$

50(a+b)+v_o =k

v_o =k-50(a+b)

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Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Find the average value of the function f(x)=cosx on the interval $[0, \pi].$

Possible Answers:

$\frac{2}{\pi}$

$\frac{1}{\pi}$

$\frac{-1}{\pi}$

Correct answer:

Explanation:

The average value of a function on a given interval [a,b] is given by the following function:

$\frac{1}{b-a}\int_{a}^{b}f(x)dx.$

Now, let's simply input our values and function in:

$\frac{1}{\pi-0}\int_{0}^{\pi}-cosxdx=\frac{1}{\pi}[sin(\pi)-sin(0)]=\frac{0}{\pi}*0=0.$

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Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Determine the position function for a particle whose velocity is given by the equation

v(t)=t^2+t+16

and whose initial position is 10.

Possible Answers:

$\frac{t^3}{3}+\frac{t^2}{2}+16t+10$

${t^3}+{t^2}+16t+10$

$\frac{t^3}{3}+\frac{t^2}{2}+10$

$\frac{t^3}{3}+\frac{t^2}{2}+16t+C$

Correct answer:

$\frac{t^3}{3}+\frac{t^2}{2}+16t+10$

Explanation:

The position function describing any object is the antiderivative of the velocity function (in other words, velocity is the derivative of position).

So, we first integrate the velocity function:

$s(t)=\int v(t)dt= \int (t^2+t+16)dt=\frac{t^3}{3}+\frac{t^2}{2}+16t+C$

The following rule was used for integration:

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

Now, to determine the constant of integration, we use our initial condition given,

s(0)=10

Plugging this into our function, we get

0+0+0+C=10

C=10

Our final answer is

$\frac{t^3}{3}+\frac{t^2}{2}+16t+10$

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Example Question #2 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

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.

Possible Answers:

2t^2+3

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

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

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

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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Example Question #4 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Given a particle with an acceleration at time to be a(t)=4 . With initial conditions v(0)=4 and p(0)=-4 where v(t) is the velocity at time , and p(t) is position of the particle at time .

Find the position at time .

Possible Answers:

p(t)=t^2+t+4

p(t)=2t^2-4t

p(t)=2t^2+4t-4

p(t)=2t^2

p(t)=2t^2+t-4

Correct answer:

p(t)=2t^2+4t-4

Explanation:

We first must establish the following relationship

p''(t)=a(t) and p'(t)=v(t)

We now may note that

$\int a(t)dt=v(t)+C$ or $\int4dt=4t+C=v(t)$

Since v(t)=3t+C

We must plug in our initial condition v(0)=4

Therefore our new velocity equation is v(t)=4t+4

We now may similarly integrate the velocity equation to find position.

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

Plugging in our second initial condition p(0)=-4

We find our final equation to be:

p(t)=2t^2+4t-4

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Example Question #881 : Ap Calculus Ab

Find the velocity function given the following information:

The acceleration function is $a(t)=3t+\cos(t)$ ;

v(0)=3

Possible Answers:

$\frac{t^2}{2}+\sin(t)+3$

$\frac{3t^2}{2}+\sin(t)+3$

$\frac{3t^2}{2}-\sin(t)+3$

$\frac{3t^2}{2}+\sin(t)+C$

Correct answer:

$\frac{3t^2}{2}+\sin(t)+3$

Explanation:

To find the velocity function, we integrate the acceleration function (the acceleration is the antiderivative of the velocity):

$v(t)=\int a(t)dt=\int [3t +\cos(t)]dt =\frac{ 3t^2}{2}+\sin(t)+C$

The rules of integration used were

$\int x^n dx = \frac{x^{n+1}}{n+1}+C$ , $\int \cos(t)dt= \sin(t)+C$

To solve for the integration constant, we plug in the given initial condition:

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

3=C

Our final answer is

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

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Example Question #6 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

What is the position function if the initial position is 0 and the velocity function is given by v(t)=10t+1 ?

Possible Answers:

5t^2+t

$\frac{5t^2}{2}+t$

5t^2+1

Correct answer:

5t^2+t

Explanation:

To find the position function, we must integrate the velocity function, as velocity is the antiderivative of position:

$s(t)=\int v(t) dt = \int (10t+1)dt = 5t^2+t+C$

The following rule of integration was used:

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

Finally, we use the initial condition to solve for the integration constant:

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

C=0

Our final answer is

s(t)=5t^2+t

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Example Question #181 : Integrals

A particle at the origin has an initial velocity of . If its acceleration is given by a = -24t^2 + 6t - 8 , find the position of the particle after 1 second.

Possible Answers:

29m

27m

24m

32m

18m

Correct answer:

27m

Explanation:

In this problem, letting s(t) denote the position of the particle and v(t) denote the velocity, we know that s''(t) = v'(t) = a(t) . Integrating and working backwards we have,

$v(t) = \int a(t)dt = \int (-24t^2 + 6t - 8)dt = -8t^3 + 3t^2 -8t + C$

Plugging in our initial condition, v(0) = 32 , we see immediately that C = 32 .

Repeating the process again for s(t) , we find that

$s(t) = \int v(t)dt = \int (-8t^3 + 3t^2 -8t +32)dt = -2t^4 + t^3 - 4t^2 + 32t + C$

Plugging in our initial condition, s(0) = 0 (we started at the origin) we see that C = 0 . This gives us a final equation

s(t) = -2t^4 + t^3 - 4t^2 + 32t . The problem asks for s(1) which is simply s(t) = -2(1)^4 + (1)^3 - 4(1)^2 + 32(1) = -4 + 1 - 4 + 32 = 27

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Example Question #8 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Find the integral which satisfies the specific conditions of $(\frac{\pi}{2}, 4)$

$\int4sin(t)-12cos(t)+3t^2dt$

Possible Answers:

y=-16tan(t)+t^3+12.12

y=4cos(t)+12sin(t)+t^3

y=-4cos(t)-12sin(t)+t^3+12.12

y=-4tcos(t)-12tsin(t)+t^3

Correct answer:

y=-4cos(t)-12sin(t)+t^3+12.12

Explanation:

Find the integral which satisfies the specific conditions of $(\frac{\pi}{2}, 4)$

$\int4sin(t)-12cos(t)+3t^2dt$

To do this problem, we need to recall that integrals are also called anti-derivatives. This means that we can calculate integrals by reversing our integration rules.

Furthermore, to find the specific answer using initial conditions, we need to find our "c" at the end.

Thus, we can have the following rules.

$\int sin(t)dt = -cos(t)+c$

$\int cos(t)dt = sin(t)+c$

$\int t^ndt=\frac{t^{n+1}}{n+1}+c$

Using these rules, we can find our answer:

$\int4sin(t)-12cos(t)+3t^2dt$

Will become:

$-4cos(t)-12sin(t)+\frac{3t^3}{3}+c=-4cos(t)-12sin(t)+t^3+c$

And so our anti-derivative is:

-4cos(t)-12sin(t)+t^3+c

Now, let's find c. First set our above expression equal to y

y= -4cos(t)-12sin(t)+t^3+c

Next, plug in for y and t. Then solve for c

$4= -4cos(\frac{\pi}{2})-12sin(\frac{\pi}{2})+(\frac{\pi}{2})^3+c$

Looks a bit messy, but we can clean it up to get:

$4= 0-12+3.88+c\rightarrow 4=-8.12+c\rightarrow c=12.12$

Now, to solve, simply replace c with 12.12

y=-4cos(t)-12sin(t)+t^3+12.12

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Example Question #881 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,13]:\\&f(-2)=5\\&f(1)=-19\\&f(4)=5\\&f(7)=-6\\&f(10)=-18\\&f(13)=0\\&\\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align*}$

Possible Answers:

495

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=5\\&f(1)=-19\\&f(4)=5\\&f(7)=-6\\&f(10)=-18\\&f(13)=0\\&\\&\text{Although the total interval has a length of }15\\&\text{Notice the points are equidistant. This distance is our subinterval }3\\&\text{and since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{13}f(x)\approx3[(-19)+(5)+(-6)+(-18)+(0)]\\&\int_{-2}^{13}f(x)\approx-114\end{align*}$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #1 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #2 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #4 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #881 : Ap Calculus Ab

Example Question #6 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #181 : Integrals

Example Question #8 : Finding Specific Antiderivatives Using Initial Conditions, Including Applications To Motion Along A Line

Example Question #881 : Ap Calculus Ab

All AP Calculus AB Resources

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