Velocity, Speed, Acceleration - AP Calculus BC

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The position of a car is given by the following function:

What is the velocity function of the car?

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The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

The derivative was found using the following rules:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} $e^u$$=e^u$ $\frac{\mathrm{du}$ }{\mathrm{d} x}$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

Let

Find the first and second derivative of the function.

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In order to solve for the first and second derivative, we must use the chain rule.

The chain rule states that if

and

then the derivative is

$\frac{dy}{dx}$=\frac{dy}{du}$$\frac{du}{dx}$

In order to find the first derviative of the function

we set

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

And differentiating we use the power rule which states

$\frac{\mathrm{d}$ }{\mathrm{d} $x}x^n$$=nx^{n-1}$

As such

$$\frac{du}{dx}$=\frac{\mathrm{d}$ }{\mathrm{d} x}[\pi x]$

$$\frac{du}{dx}$=\pi $x^{1-1}$=\pi$

And so

$f'(x)=\pi $e^{\pi x}$$

To solve for the second derivative we set

$y=\pi $e^{u}$$

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$$\frac{dy}{du}$=\frac{d}{du}$[\pi $e^u$]$

$\frac{dy}{du}$=\pi $e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d}$ }{\mathrm{d} $x}x^n$$=nx^{n-1}$

As such

$$\frac{du}{dx}$=\frac{\mathrm{d}$ }{\mathrm{d} x}[\pi x]$

$$\frac{du}{dx}$=\pi $x^{1-1}$=\pi$

And so the second derivative becomes

$\frac{dy}{dx}$=\pi $e^{\pi x}$$\pi=\pi^2$ $e^{\pi x}$

Find the velocity function of the particle if its position is given by the following function:

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The velocity function is given by the first derivative of the position function:

and was found using the following rules:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d}$ }{\mathrm{d} x} $x^n$$=nx^{n-1}$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} $\tan(t)=\sec^2$(t)$ , $\frac{\mathrm{d}$ }{\mathrm{d} x} $e^x$$=e^x$

Find the first and second derivatives of the function

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We must find the first and second derivatives.

We use the properties that

The derivative of is

The derivative of is

As such

$f'(x)=\frac{\mathrm{d}$ }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)$

To find the second derivative we differentiate again and use the product rule which states

$$\frac{\mathrm{d}$ }{\mathrm{d} x}[uv]=uv'+vu'$

Setting

and

we find that

$f''(x)=\frac{\mathrm{d}$ }{\mathrm{d} $x}[sec(x)tan(x)]=sec(x)sec^2$(x)+tan(x)sec(x)tan(x)$

As such

Given the velocity function

$v(x)=\int_${0}^{x^2+2x+1}$ $z^d$$\sin(z^5$) dz$

where is real number such that $d\in [0,1]$ , find the acceleration function

.

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We can find the acceleration function from the velocity function by taking the derivative:

We can view the function

$v(x)=\int_${0}^{x^2+2x+1}$ $z^d$$\sin(z^5$) dz$

as the composition of the following functions

$g(x)=\int_${0}^{x}$ $z^d$$\sin(z^5$) dz$

so that . This means we use the chain rule

to find the derivative. We have and , so we have

The position of an object is given by the equation $y = $\frac{1}{3}$ $t^3$ - $\frac{1}{2}$ $t^2$ -2t + 4$ . What is its acceleration at t = 2?

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If this function gives the position, the first derivative will give its speed and the second derivative will give its acceleration.

Now plug in 2 for t:

The equation $y=\frac{1}{2}$ $t^4$ +10t$ models the position of an object after t seconds. What is the acceleration at 3 seconds?

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If this function gives the position, the first derivative will give its speed and the second derivative will give its acceleration.

Plug in 3 for t:

The equation $y= \sin (2t)$ models the position of an object after t seconds. What is its speed after $\pi$ seconds?

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If this function gives the position, the first derivative will give its speed.

$y' = \cos (2t ) \cdot 2$

Plug in $\pi$ for t:

$2 \cos (2 \pi ) = 2\cdot 1 = 2$

The position of an object is modeled by the equation $y= $\frac{1}{2}$ \cos ^2 t$ What is the speed after $\frac{\pi}{4}$ seconds?

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If this function gives the position, the first derivative will give its speed. To differentiate, use the chain rule: $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, $f(x) = $\frac{1}{2}$ $x^2$$ and $g(x) = \cos x$ . Since and $g'(x ) = - \sin (x)$ , the first derivative is $\cos t \cdot - \sin t$ .

Plug in $\frac{\pi}{4}$ for t:

$\cos ($\frac{\pi}{4}$ ) \cdot - \sin ($\frac{\pi}{4}$ ) = $\frac{1}{\sqrt{2}$} \cdot - $\frac{1}{\sqrt2}$ = -$\frac{1}{2}$$

A particle's position on the -axis is given by the function $f(t)=\sin(t)-t+1$ from $0 < t <\pi$ .

When does the particle change direction?

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To find when the particle changes direction, we need to find the critical values of . This is done by finding the velocity function, setting it equal to , and solving for

$0=v(t) = f'(t) = \cos(t)-1$ .

Hence $\cos(t)=1$ .

The solutions to this on the unit circle are $t =0, \pi$ , so these are the values of where the particle would normally change direction. However, our given interval is $0 < t < \pi$ , which does not contain $0, \pi$ . Hence the particle does not change direction on the given interval.

A particle moves in space with velocity given by

$v(t)=\alpha $t^3$+\beta $\sqrt{t}$$

where $\alpha, \beta$ are constant parameters.

Find the acceleration of the particle when t=4.

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To find the acceleration of the particle, we must take the first derivative of the velocity function:

$a(t)=v'(t)=3\alpha $t^2$+\frac{\beta}{2\sqrt{t}$}$

The derivative was found using the following rule:

$\frac{\mathrm{d}$ }{\mathrm{d} x} $x^n$$=nx^{n-1}$

Now, we evaluate the acceleration function at the given point:

Find the velocity function from an acceleration function given by

and the condition

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Acceleration is the rate of change of velocity, so we must integrate the acceleration function to find the velocity function:

$v(t)=\int a(t)dt=\int (4t+5 ) $dt=2t^2$+5t+C$

The integration was performed using the following rules:

$\int $x^ndx$$=\frac{x^{n+1}$$}{n+1}+C$ , $\int a dx=ax+C$

To find the integration constant C, we must use the initial condition given:

Our final answer is

The velocity of a particle is given by v(t). Find the function which models the particle's acceleration.

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The velocity of a particle is given by v(t). Find the function which models the particle's acceleration.

To find the acceleration from a velocity function, simply take the derivative.

In this case, we are given v(t), and we need to find v'(t) because v'(t)=a(t).

To find v'(t), we need to use the power rule.

For each term, simply multiply by the exponent, and then subtract one from the exponent. Constant terms will drop out, linear terms will become constants, and so on.

So, our answer is:

The velocity of a particle is given by v(t). Find the particle's acceleration when t=3.

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The velocity of a particle is given by v(t). Find the particle's acceleration when t=3.

To find the acceleration from a velocity function, simply take the derivative.

In this case, we are given v(t), and we need to find v'(t) because v'(t)=a(t).

To find v'(t), we need to use the power rule.

For each term, simply multiply by the exponent, and then subtract one from the exponent. Constant terms will drop out, linear terms will become constants, and so on.

So, our acceleration function is:

Now, plug in 3 for t and solve.

So, our answer is 52.

The position of a car is given by the following function:

What is the velocity function of the car?

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The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

The derivative was found using the following rules:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} $e^u$$=e^u$ $\frac{\mathrm{du}$ }{\mathrm{d} x}$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

Let

Find the first and second derivative of the function.

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In order to solve for the first and second derivative, we must use the chain rule.

The chain rule states that if

and

then the derivative is

$\frac{dy}{dx}$=\frac{dy}{du}$$\frac{du}{dx}$

In order to find the first derviative of the function

we set

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

And differentiating we use the power rule which states

$\frac{\mathrm{d}$ }{\mathrm{d} $x}x^n$$=nx^{n-1}$

As such

$$\frac{du}{dx}$=\frac{\mathrm{d}$ }{\mathrm{d} x}[\pi x]$

$$\frac{du}{dx}$=\pi $x^{1-1}$=\pi$

And so

$f'(x)=\pi $e^{\pi x}$$

To solve for the second derivative we set

$y=\pi $e^{u}$$

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$$\frac{dy}{du}$=\frac{d}{du}$[\pi $e^u$]$

$\frac{dy}{du}$=\pi $e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d}$ }{\mathrm{d} $x}x^n$$=nx^{n-1}$

As such

$$\frac{du}{dx}$=\frac{\mathrm{d}$ }{\mathrm{d} x}[\pi x]$

$$\frac{du}{dx}$=\pi $x^{1-1}$=\pi$

And so the second derivative becomes

$\frac{dy}{dx}$=\pi $e^{\pi x}$$\pi=\pi^2$ $e^{\pi x}$

Find the velocity function of the particle if its position is given by the following function:

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The velocity function is given by the first derivative of the position function:

and was found using the following rules:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d}$ }{\mathrm{d} x} $x^n$$=nx^{n-1}$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} $\tan(t)=\sec^2$(t)$ , $\frac{\mathrm{d}$ }{\mathrm{d} x} $e^x$$=e^x$

Find the first and second derivatives of the function

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We must find the first and second derivatives.

We use the properties that

The derivative of is

The derivative of is

As such

$f'(x)=\frac{\mathrm{d}$ }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)$

To find the second derivative we differentiate again and use the product rule which states

$$\frac{\mathrm{d}$ }{\mathrm{d} x}[uv]=uv'+vu'$

Setting

and

we find that

$f''(x)=\frac{\mathrm{d}$ }{\mathrm{d} $x}[sec(x)tan(x)]=sec(x)sec^2$(x)+tan(x)sec(x)tan(x)$

As such

Given the velocity function

$v(x)=\int_${0}^{x^2+2x+1}$ $z^d$$\sin(z^5$) dz$

where is real number such that $d\in [0,1]$ , find the acceleration function

.

Tap to see back →

We can find the acceleration function from the velocity function by taking the derivative:

We can view the function

$v(x)=\int_${0}^{x^2+2x+1}$ $z^d$$\sin(z^5$) dz$

as the composition of the following functions

$g(x)=\int_${0}^{x}$ $z^d$$\sin(z^5$) dz$

so that . This means we use the chain rule

to find the derivative. We have and , so we have

The position of an object is given by the equation $y = $\frac{1}{3}$ $t^3$ - $\frac{1}{2}$ $t^2$ -2t + 4$ . What is its acceleration at t = 2?

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If this function gives the position, the first derivative will give its speed and the second derivative will give its acceleration.

Now plug in 2 for t: