If p(t) models the position of an electron as a function of time. Find the velocity of the electron when \(\displaystyle t=\frac{3\pi}{4}\)

\(\displaystyle p(t)=4cos(t)+4\)

If p(t) models the position of an electron as a function of time. Find v(t), the velocity function of the electron:

\(\displaystyle p(t)=4cos(t)+4\)

Recall that velocity is the derivative of position. To find v(t), we need to find p'(t)

Remember that the derivative of cosine is negative sine, so we get...

\(\displaystyle p(t)=4cos(t)+4\)

\(\displaystyle p'(t)=-4sin(t)\)

We are almost there, but we need to find the velocity at the given time, not just the velocity function. 

\(\displaystyle p'\left(\frac{3\pi}{4}\right)=-4sin\left(\frac{3\pi}{4}\right)=-2\sqrt{2}\)

So our answer is:

\(\displaystyle -2\sqrt{2}\)

The velocity function is found by taking the antiderivative of acceleration. 

\(\displaystyle v(t)=\int-9.8dt=-9.8t+c\)

The initial velocity is the velocity when t = 0. We can use v(0) = 15 to find the constant c.

\(\displaystyle v(0)=15=-9.8(0)+c\)

\(\displaystyle 15=c\)

Therefore, the function describing velocity is

\(\displaystyle v(t)=-9.8t+15\)

Velocity can be found as the integral of acceleration with respect to time:

\(\displaystyle v(t)= \int a(t)dt =\int (3t^2+4t-1)dt\)

\(\displaystyle v(t)=t^3+2t^2-t+C\)

To find the constant of integration, utilize the initial condition, i.e. the initial velocity:

\(\displaystyle v(0)=15=3(0^3)+2(0^2)-0+C\)

\(\displaystyle C=15\)

Thus

\(\displaystyle v(t)=t^3+2t^2-t+15\)

And

\(\displaystyle v(2)=(2^3)+2(2^2)-2+15=8+8-2+15=29\)

Velocity is given by the derivative of the position function with respect to time:

\(\displaystyle v(t)=\frac{d}{dt}p(t)=\frac{d}{dt}\left(20-20e^{-\frac{1}{4}t}cos\left(\frac{\pi}{8}t\right)\right)\)

Utilizing the product rule of derivatves, which follows the form:

\(\displaystyle \frac{d}{dt}uv=\frac{du}{dt}v+\frac{dv}{dt}u\)

We find velocity to be:

\(\displaystyle v(t)=-20\left(-\frac{1}{4}e^{-\frac{1}{4}t}cos\left(\frac{\pi}{8}t\right)-\frac{\pi}{8}e^{-\frac{1}{4}t}sin\left(\frac{\pi}{8}t\right)\right)\)

\(\displaystyle v(t)=5e^{-\frac{1}{4}t}cos\left(\frac{\pi}{8}t\right)+\frac{5}{2}\pi e^{-\frac{1}{4}t}sin\left(\frac{\pi}{8}t\right)\)

\(\displaystyle v(4)=5e^{-1}cos\left(\frac{\pi}{2}\right)+\frac{5}{2}\pi e^{-1}sin\left(\frac{\pi}{2}\right)\)

\(\displaystyle v(4)=\frac{5\pi}{2e}\)

Velocity can be found by integrating acceleration with respect to time:

\(\displaystyle v(t)=\int a(t)dt=\int (6t^2+4t^3)dt\)

\(\displaystyle v(t)=2t^3+t^4+C\)

To find the integration of constant, utilize the initial velocity:

\(\displaystyle v(0)=2(0)^3+(0)^4+C=0\)

\(\displaystyle C=0\)

\(\displaystyle v(t)=2t^3+t^4\)

Now, to find the average velocity, integrate the velocity function once more over the interval and divide by the interval:

\(\displaystyle v_{avg}=\frac{1}{4-2}\int_{2}^{4}(2t^3+t^4)dt=\frac{1}{4-2}(\frac{t^4}{2}+\frac{t^5}{5}|_2^4)\)

\(\displaystyle v_{avg}=\frac{1}{2}((\frac{256}{2}+\frac{1024}{5})-(\frac{16}{2}+\frac{32}{5}))\)

\(\displaystyle v_{avg}=\frac{1}{2}(\frac{1592}{5})=\frac{796}{5}\)

Velocity can be found as the time derivative of position:

\(\displaystyle v(t)=x'(t)=\frac{d}{dt}(3t^4+4t^3+6t^2+25)\)

\(\displaystyle v(t)=12t^3+12t^2+12t\)

Therefore, at \(\displaystyle t=2\), the velocity of the particle is:

\(\displaystyle v(2)=12(2)^3+12(2)^2+12(2)=168\)

The instantaneous velocity is simply the derivative of the position function with respect to time:

\(\displaystyle v(t)=\frac{ds}{dt}\)

\(\displaystyle s=s(t)=5t^4-7t\)

Therefore:

\(\displaystyle v(t)=\frac{d}{dt}[5t^4-7t]=20t^3-7\)

To find the instantaneous velocity at t=3 seconds:

\(\displaystyle v(3)=20(3)^3-7=533ft/sec\)

Recall that the velocity of moving body can be computed from its position function by taking the first derivative of that function.  Thus, we know that:

\(\displaystyle v(t) = s'(t)\)

This is relatively easy for our data:

\(\displaystyle v(t) = 8t + 3\)

Thus, the velocity of the particle is 

\(\displaystyle v(6) = 8*6 + 3=51\).

To find the velocity of a moving body based on its position function, you need to take the first derivative of that position function.  Thus, we know, using the product rule that:

\(\displaystyle v(t) = s'(t)= 15sin(t) + 15tcos(t)\)

To calculate the velocity, you merely need to find the value of \(\displaystyle v(20)\):

\(\displaystyle v(20)=15sin(20) + 15*20cos(20)=136.12\)

Recall that the velocity of a body can be calculated by taking the integral of its acceleration function, thus, we know:

\(\displaystyle v(t) = $\int 30sin(2t)+\frac{4}{t}dt$\)

This would be:

\(\displaystyle -15cos(2t)+4*ln(t)\)

Thus, we can find the velocity of this body by solving for \(\displaystyle v(4)\):

\(\displaystyle v(4)=-15cos(8)+4*ln(4)=7.728...\)