AP Calculus BC - Velocity, Speed, Acceleration

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?

Explanation

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 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?

Explanation

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:

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?

It doesn't change direction within the given bounds

$x= \pi$

$x=\frac{\pi}{2}$

$x= \frac{\pi}{4}$

$x=2\pi$

Explanation

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.

$48\alpha + \frac{\beta}{4}$

$48\alpha +2\beta$

$3\alpha t^2+\frac{\beta}{2\sqrt{t}}$

Explanation

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:

$a(4)=3\alpha(4^2)+\frac{\beta}{2\sqrt{4}}=48\alpha+\frac{\beta}{4}$

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

Explanation

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.

$v'(t)=a(t)=(3)3t^{3-1}-(2)4t^{2-1}-5=9t^2-8t-5$

So, our acceleration function is:

Now, plug in 3 for t and solve.

So, our answer is 52.

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

Explanation

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

Explanation

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.

$v'(t)=a(t)=(3)3t^{3-1}-(2)4t^{2-1}-5=9t^2-8t-5$

So, our answer is:

Find the first and second derivatives of the function

Explanation

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

Find the velocity function from an acceleration function given by

and the condition

Explanation

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

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

$s(t)=e^t\tan(t)+t^5+3$

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

$e^t\tan(t)-e^t\sec^2(t)+5t^4$

$te^t\tan(t)+e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+t^5$

Explanation

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

$v(t)=s'(t)=e^t\tan(t)+e^t\sec^2(t)+5t^4$

and was found using the following rules:

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

Velocity, Speed, Acceleration

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