Response Time Activity

In the response time activity my delta x was 42 centimeters/0.42 meters. The acceleration was -9.8m/s^2 and the initial velocity was 0m/s. To calculate the time I used the following procedure.
delta x=Vi delta t + 1/2 a(delta t)^2
-0.42=0(delta t) + 1/2 (-9.8)(delta t)^2
-0.42=0(delta t) + (-4.9)(delta t)^2
-0.42=-4.9(delta t)^2
-0.42/-4.9=detla t^2
0.292s=delta t

Vector Addition Activity

1. The simulation can be used to represent the path in this way; the beginning of the vector represents the starting point (0,0), the tip of the arrow represents where you arrived after 15 steps. The angle at with the vector is positioned also give a rough estimate of how the path looked.

The same graph can be used for demonstrating the scenario of a car driving at 15 mi/hr and 25 degrees because on the graph, the y-axis can represent miles while the x-axis represents the time in hours. Using that representation, the graph is able to show us that as time moved on, the distance increased. The angle at which it increased is the same as well which is why we can use the same graph.

If we were to create another scenario we could say that a cat moves 14 cm/min in the direction of north east. The graph for this scenario would look similar.

2. In the simulator, Rx and Ry are the components of the vector, they're sort of like coordinates. The absolute value of R is the magnitude of the vector based on the coordinates. It is an absolute value in case we go in a negative direction. The theta indicates an angle. I think component means part of something. In the simulation a component seems to be part of the bigger whole which may be a triangle if applicable. The components pair up with the vectors according to how they've been drawn or designed.

3. In this situation we are driving 14 mi/hr heading 35 degrees north of east. We are asked to find how fast the car is going north and east. Based on the simulation we could say that the car is traveling at 8mi/hr to the north and 12 mi/hr to the east. However, if we were to do this with no simulation we could use sin, cos, or tan equations.

To calculate north we use a sin equation which is basically sin theta= O/H
sin (35)= O/14
sin(35) * 14 = O
O= 5.99

For east we use Cos
cos (35) =A/14
cos(35)*14=A
A=12.65

The answers are slightly different than what is in the simulation, but that is because the simulation isn't exactly accurate either. That is how we could calculate without using a simulation.

4. I would estimate that the bird would have to fly at least 12 miles. The simulation says that the bird should travel about 11.7 miles. The bird would have to fly southwest to get to the shop. We could use the simulation to answer the question by first placing the vectors for south and west. Then we can place one more vector to complete the triangle. Once completed, we can look at the numbers on the top to get our answers. Using equations to figure it out we could figure out the bird's path by using either a sin or cos equation. Because we have both the adjacent and opposite leg, it is very easy to find the hypotenuse which is the bird's path.

5. I imagined that the bird would probably not have to go to far to get back to the dealership. This is because instead of turning or going in different directions, the bird would be able to directly to the dealership. If I didn't use the simulation, I would assume the distance that the bird travels would be close to 2 miles. This is because first the car goes 10 miles to the east them 8 miles backwards to the west. 10- 8 = 2 therefore the remaining distance to the car dealership is 2 miles. The simulation proves my theory right because if we look at the SUM vector, the magnitude is 2.8.

In the simulation in order to find the SUM of the vector we click the "show sum" button. The initial idea I had about vector addition was that when you add vectors opposite directions will result in subtraction. Therefore the direction determines positive or negative numbers. From testing the vectors, my theory about this seems to be correct. The numbers, might be slightly off because of the simulation not being to have exact numbers.

6. For this situation, the paper flew 7 m/s in the direction North of east and some wind pushed it along 8 m/s North of East. Using the simulation, we would take the two vectors and put them together. We can do this because they are going in the same direction. Based on the simulation, the Sum of the vectors would give us the speed. The speed is about 15.2 m/s in the direction of North of East. This conclusion was reached due to the simulation adding the vectors together.

Adding the vectors without the simulation would be quite easy. We have both of the vectors, 7 m/s and 8m/s going north of east. Due to the fact that they are not going in opposite directions, we can just add them right away and get the speed of 15 m/s North of east. we don't really need the simulation.

If we test this out, we can just add and subtract more vectors in different directions and we will see that the speed will change based on the directions of the vectors.

Conceptual Challenge Pg.50

2. Runaway Train   If a passenger train is traveling on a straight track with a Negative velocity and a positive acceleration, is it speeding up or slowing down?

The train is speeding up because it is accelerating while moving left. This is because the velocity is negative which means it is moving left. Since the acceleration is positive, it is speeding up. It is speeding up to the left.

3. Hike-and-Bike Trail   When Jennifer is out for a ride, she slows down on her bike as she approaches a group of hikers on a trail. Explain how her acceleration can be positive even though her speed is decreasing.  

I don't think it is possible for the acceleration to still be positive if the speed goes down. However, if she starts speeding up again when she's gone slow for awhile, the acceleration can go back to being positive.

Physics presentation