AP Calculus AB - Acceleration

The velocity of a particle is given by the function $v(t)=4e^{cos(3t)}$ . What is its acceleration function?

$-12sin(3t)e^{cos(3t)}$

$-12e^{cos(3t)}$

$-12cos(3t)e^{cos(3t)}$

$12sin(3t)e^{cos(3t)}$

$12cos(3t)e^{cos(3t)}$

Explanation

Acceleration is given by the time derivative of velocity.

In this case we will use the rules that the derivative of

is $e^u \cdot \frac{du}{dx}$

and the derivative of

is $-sin(u)\cdot \frac{du}{dx}$ .

Applying this knowledge we can find the acceleration function to be the following.

$a(t)=v'(t)=\frac{d}{dt}(4e^{cos(3t)})$

$a(t)=-12sin(3t)e^{cos(3t)}$

Given that the position function of a paper airplane is and its initial veloctiy is $6\frac{ft}{sec}$ find its acceleration at seconds.

$-4 \frac{ft}{s^2}$

$4 \frac{ft}{s^2}$

$-5 \frac{ft}{s^2}$

$5 \frac{ft}{s^2}$

None of the above.

Explanation

In order to solve this proble, it must first be realized that the derivative of a position function is the velocity function and the derivative of the velocity function becomes the acceleration function. In this problem we are given the position function and asked to find the acceleration after a certain amount of time. By taking the double derivative of the function and plugging in the time, we will be able to find the acceleration of the paper plane at that time. In order to take the derivative of the position function, we must first know the power rule, $\frac{d}{dx}x^n=nx^{n-1}$ .

Applying the power rule to the position function, we find the first derivative to be .

Applying the power rule a second time to the veloctiy function, we find the second derivative to be .

Now, by simply plugging seconds, we find that the acceleration of the paper airplane at that time is $-4\frac{ft}{s^2}$ .

Suppose a particle travels in a circular motion in the xy-plane with

$x(t) = R \cos(\omega t)$

$y(t) = R \sin(\omega t)$

for some constants $\omega, R$ . Notice that this is circular because .

What is the total magnitude of the acceleration, $a = \sqrt{(x'')^2 + (y'')^2}$ ?

$R \omega^2$

$2R\omega^2$

$R^2 \omega^4$

$2 \pi R \omega$

Explanation

We know that and so the acceleration components are:

$x''(t) = - R \omega^2 \cos(\omega t)$

$y''(t) = - R \omega^2 \sin(\omega t)$

Plugging these into the formula for the acceleration and again recognizing that $\sin^2(\theta) + \cos^2(\theta) = 1$ , we get $a^2 = R^2 \omega^4$ , or, after the square root, $a = R \omega^2$ .

The velocity of an object is given by the following equation:

$v(t)=\frac{3}{2}t^2-\frac{1}{3}t+16$

Find the equation for the acceleration of the object.

$a(t)=3t-\frac{1}{3}$

$a(t)=\frac{1}{3}t-3$

$a(t)=\frac{1}{2}t-16$

$a(t)=16t+\frac{1}{3}$

Explanation

Acceleration is the derivative of velocity, so in order to find the equation for the object's acceleration, we must take the derivative of the equation for its velocity:

$v(t)=\frac{3}{2}t^2-\frac{1}{3}t+16$

We will use the power rule to find the derivative which states:

$f(x)=ax^n \rightarrow f'(x)=a\cdot nx^{n-1}$

$a(t)=v'(t)=\frac{3}{2}\cdot 2t^1-\frac{1}{3}t^0=3t-\frac{1}{3}$

The velocity of a particle is given by the function . What is the acceleration of the particle at time ?

Explanation

Acceleration of a particle can be found by taking the derivative of the velocity function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of velocity with respect to time, we are evaluating how velocity changes over time; i.e acceleration! This is just like finding velocity by taking the derivative of the position function.

$a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$

Taking the derivative of the function

The acceleration function is

At time

The velocity of an object is given by the following equation:

$v(t)=\frac{3}{7}t^2-\frac{1}{14}t+5$

Find the acceleration of the object at seconds.

$\frac{23}{14}$

$\frac{6}{7}$

$\frac{1}{14}$

$\frac{3}{7}$

Explanation

Acceleration is the derivative of velocity, so we must take the derivative of the given equation to find an equation for acceleration:

$v(t)=\frac{3}{7}t^2-\frac{1}{14}t+5$

$a(t)=v'(t)=\frac{6}{7}t-\frac{1}{14}$

Now we can plug in t=2 to find the acceleration of the object at 2 seconds:

$a(2)=\frac{6}{7}(2)-\frac{1}{14}=\frac{12}{7}-\frac{1}{14}=\frac{24}{14}-\frac{1}{14}=\frac{23}{14}$

Find the acceleration at given the following position function.

Explanation

To find acceleration at a point, simply differentiate the position function twie and then plug in . Thus,

Function gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Find the acceleration (in meters per second per second) of the particle at seconds.

$\small 4601\frac{2}{3}$

$\small 3125\frac{2}{3}$

$\small \small 4625$

$\small 23\frac{1}{3}$

$\small 3125$

Explanation

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

To derive a polynomial, simply decrease each exponent by one and bring the original number down in front to multiply.

So this

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$