User:Jhs190/sandbox

Team 1 Example for 1-D Motion
You are driving along the road at $$30\frac{m}{s}$$, when you notice a duck start to cross the road $$100m$$ ahead of you, so you start to deccelerate. How long will it take you to come to a complete stop? After $$3s$$ of deccelerating the duck flys away so you accelerate back to $$30\frac{m}{s}$$ at $$12\frac{m}{s^2}$$. Once you reach $$30\frac{m}{s}$$, how far have you travled in total since first seeing the duck?

For the first part of this problem, to find the the time it would take for the car to come to a complete stop. We will need two kinematic equations to solve this. Equation 1 will be used to find the acceleration necessary to come to a complete stop. Equation 2 will be used to find the time.

Equation 1: $$v_f^2=v_i^2+2a(x_f-x_i)$$

The initial velocity, final velocity and total distance traveled are given. We will set the final position equal to the total distance traveled and the initial poistion to zero. We can manipulate Equation 1 and plug in the given values to find the acceleration. An example on how to set this up in IPython (JupyterLab) is shown below. The acceleration is calcualted to be. We will now use this value in Eqaution 2 in order to find the time taken to come to a complete stop.

Equation 2: $$v_f = v_i +at$$

We can manipulate Equation 2 and plug in the given values to find the time. An example on how to set this up in IPython (JupyterLab) is shown below. It takes for the car to come to a complete stop.

Now we will solve the second part of this problem. First the car deccelerates at $$-9\frac{m}{s}$$, for. We can use equation 3 to find the distance traveled for this portion.

Equation 3: $$x_f=x_i+v_i t+\frac{1}{2}a t^2$$ The car travels $$49.5m$$ during this time. We use Equation 1 to calculate the velocity after the car has traveled this distance and just before it begins to accelerate again. This velocity is found to be $$3\frac{m}{s}$$. Next we use Equation 2 to find the time it takes to accelerate from $$3\frac{m}{s}$$ to $$30\frac{m}{s}$$ at $$12\frac{m}{s^2}$$. This time is caluclated to be 2.25 seconds. Finally we can calculate the total distance traveled using Equation 3. Notice, we set the initial position to be the $$49.5m$$ that the car has alread traveled. The total distance traveled by the car is found to be $$86.62m$$.

Example 2
$$x(t) = (t^2+1)\mathbf{\hat{i}} +6t^3\mathbf{\hat{j}}+(8t+4)\mathbf{\hat{k}}$$

$$v(t)=\frac{dx(t)}{dt}$$

$$v(t)=2t\mathbf{\hat{i}}+18t^2\mathbf{\hat{j}}+8\mathbf{\hat{k}}$$

$$a(t)=\frac{dv(t)}{dt}=\frac{d^2x(t)}{dt^2}$$

$$a(t)=2\mathbf{\hat{i}}+36t\mathbf{\hat{j}}$$

$$\lVert r \rVert=\sqrt{r_x\mathbf{\hat{i}}+r_y\mathbf{\hat{j}}+r_z\mathbf{\hat{k}}}$$

Example 3
An astronaut is floating in space looking down at the earth. The coordinates of the astronaut's space craft are $$\langle14,6.5,7.8\rangle$$ with respect to the astronaut's reference frame. the The astronaut's partner informs him of an asteroid passing by. In order to see it, the astonaut must rotate $$67^\circ$$ to the left (around the z-axis), then $$54^\circ$$down (around the new y-axis), and then $$38^\circ$$ on his side (the new x-axis). What are the coordinates of the space craft in the astronauts new refernce frame.