Physics Challenge- Wind cars

In this lab, we used motion sensors to get two equations of different wind powered cars. These cars both accelerated from rest. In the lab, we tried to predict where the cars would have to start in order to intersect at the origin. Our work is as follows:

Constant Acceleration Particle Model

Position vs Time Graphs: Previously we have known that when a position vs time graph has a straight line on it, the object is moving at a constant velocity. With the CAPM model, we learned that these graphs can also have a curved line, which indicates acceleration. Also, we know that steeper slopes indicate that an object is moving fast, and that when an object moves in the positive direction it is going forwards and when it is going towards the negative side of the graph that it is going in the negative direction. We can also apply this information to our new curved lines.

As the line moves down as the vector
indicates, the velocity is increasing.
As the vector going up indicates, the
velocity is also increasing. 

If a velocity line is going up in the direction
 of the top vector, the object is moving in
the positive direction, and if the line is going
downward like the bottom vector, then the
object is moving in the negative direction. 

If the line moves toward the origin over
 time line the top vector does, the object
is slowing down. Also, if the line moves
towards the origin like the bottom vector,
the line is also slowing down. 

If the line on the graph moves towards
the origin like the top vector, the
object is still moving in the positive
direction. Similarly, on the bottom vector,
the object is moving in the negative
direction. 

In this example, the object starts with
a velocity of 10m/s. From seconds 0-2, the
object is slowing down. Then, as it crosses
the origin, it changes direction and
begins speeding up in the negative direction.

In this graph, the object is increasing in speed in the positive direction. To calculate displacement,
we can calculate the area between the line and the x-axis. This area is in the shape of a triangle, so
we can use Displacement = 1/2 (Base)(Height) to find displacement. By substituting values in from
the triangle that the line makes with the x-axis, we create an equation that should look like this:
1/2(10)(10) = displacement. From there, we can conclude that the object finished 50m from where it started.
To take a different approach, we can also use the equation ΔX = 1/2aT^{2 + initial velocity} to find the displacement. Initial velocity is the velocity the velocity at zero seconds. In this case, initial velocity is zero. Also, we can find the acceleration by using the equation (ΔV)/(Δt) = a. We can only use this equation for acceleration when our velocity vs time graphs are a straight time, indicating constant acceleration. In this case, the acceleration is 1m/s^2. After we find acceleration, we can plug in our numbers and solve for the displacement. ΔX = 1/2(1)(100) + 0 which comes out to equal 50m, the same as our first calculation.

Average Acceleration:

Formula: 

Average acceleration is the average rate at which an object is speeding up. Average acceleration may be very useful for many other equations and formulas in physics. 

In this graph, to find the instantaneous velocity at five seconds we can
find the slope between two pints that will make 5s the midpoint.
 For example, we can find the slope between the points at 4s and 6s.
This line is also tangent at five seconds. The fourth second falls at 1m and
the sixth at 2m. Therefore, we can use the equation (change in position)/(change in time) 
to find the line tangent. This tangent line will be the instantaneous velocity at that point. 

Example: 

In the position vs time graph above, the object has four distinct motions. From point A to B, the object is moving in the positive direction. Also, since the line is curved, we can determine that it is either slowing down or speeding up. The way to differentiate between the two is look at the slope of the line. At the beginning of the line, the slope is very steep. It gradually becomes less steep until the slope becomes zero at point B. With this information, we can tell that the object is initially going very fast and then slows down. From point B to C, the object is moving towards the negative side of the graph, telling us that the object is moving backwards. The slope closest to point B is not very steep, but gets steeper as it gets closer to point C, indicating that the object is speeding up. At point C something interesting happens. The object crosses the origin with a very steep slope (high velocity) and then begins to slow down again. This can be seen with the gradual decline of slope. At point D, the object switches direction again and begins to travel in the positive direction again. Starting with a small slope, the object begins traveling slowly and then speeds up until it reaches the origin, finishing at the same position  that it started at. 

Velocity vs Time Graphs: Constant velocities are shown on a velocity vs time graph as a horizontal line. When a velocity vs time graph has a diagonal line, the velocity is either increasing or decreasing. An object with an increasing or decreasing speed has an acceleration. As seen on the velocity vs time graphs below, as the line moves away from the origin, the object is increasing in speed. Also, if the line is going in the positive side of the graph, it is moving forwards (above the origin) and if it is in the negative side of the graph, it is moving backwards (under the origin) 

Example: 

 Displacement and Velocity vs  Time graphs: We can use two methods to find the displacement from a velocity vs time graph. The first is to find the area of the space between the line and the x-axis. This space represents how far the object has traveled. Also, we can use the equation ΔX = 1/2aT^2 where ΔX = displacement, a = acceleration, and t = time. 

Example problem: 

Acceleration vs Time Graphs: Acceleration vs Time graphs do not tell us much other than the slope of the velocity vs time graph. a vs t graphs are directly related to the slope of v vs t graphs. For example, if the slope of a velocity vs graph is negative, the acceleration line will also be on the negative side of the graph. Keep in mind that there is no acceleration when the velocity vs time graph has a straight horizontal line with no slope. 

Examples: 

In this velocity vs time graph, the line has a positive slope from 0-5 seconds, no slope from 5-15 seconds, and then a negative slope from 15-25 seconds. Because of this, we can conclude that the acceleration for the first five seconds is positive, there is no acceleration for the next ten seconds, and then there is a negative acceleration for the last ten seconds. The acceleration vs time graph would look as follows: 

Motion Maps: We have now learned how to make both acceleration and velocity motion maps. In velocity motion maps, the distance between the points represents how far the object has traveled and the vector represents the velocity. The larger the vector, the larger the velocity. In acceleration motion maps, the distance between dots represents distance traveled, but the vectors represent acceleration. Constant acceleration is represented by vectors of the came length as seen below. In the image below, the object's velocity is increasing with constant velocity. 

Average Velocity: Average velocity is calculated by the equation (change in position)/(change in time) and represents the average velocity that an object travels at over a given amount of time. 

Instantaneous Velocity: Instantaneous velocity is the velocity at a given time. To find instantaneous velocity on a position vs time graph, you can find a line tangent to the point at which that time is located (also the same as making that time the midpoint of two other points on the line). 

Real World Examples: Physics is ever present in the real world. One example of a real world situation where acceleration is present is as follows: A dog runs down his driveway with an initial speed of 5m/s for 8s and then uniformly increases his speed to 10m/s in the next five seconds. How long is his driveway? 

How to solve: So, we are given both time and velocity. Since we are trying to solve for displacement, we can use the equations ΔX = 1/2aT^{2 + initial velocity} and x = Vt. In the first eight seconds of the run, the dog has a constant velocity. Because of this, we use the exertion ΔX = Vt. After substitution, this becomes ΔX = (5m/s)(8s), simplified to 40m. Keep in mind that this is only the first half of the dog's journey. The second half can be calculated with the equation ΔX = 1/2aT^{2 + (initial velocity)(time)}

. We can find acceleration with the equation ΔV/Δt which comes out to be 1m/s^2. After substitution, this equation becomes ΔX = 1/2(5)^2 + (5)(5) This can be simplified to ΔX = 37.5. So, now that we have both the first part of the journey and the second, we can add them together in order to find the total displacement (length of the driveway). 40m + 37.5m = 77.5m, the length of the driveway.