Physics Challenge

In this challenge Hank and I paired up with another group to predict where two wind cars would intersect. First, we used a motion sensor to find the initial velocity and the acceleration of cart D. We tested this three times and then averaged our slopes to get an average acceleration of: .1203 m/s^2.
Our initial velocity was Vi: -.18.We recorded 1m away from the sensor. 
Prediction: In order to make a prediction, we used the formula ∆x=1/2a(∆t)^2+(-.18)(t)
∆x=1/2(.1203)^2+(.18)(t)
We solved that equation to find time and got, 4.3s.
Next, we plugged time back into the equation to find our predicted point of intersection. 
Prediction
The distance we found was 1.072 m. 
Cart D: 1.072 m 
Cart C: 97 cm
Results
Actual Point of intersection: 1.072 m
After we tested our prediction we discovered that it was correct. Therefor, our percent error is 0. 

Constant Acceleration Particle Model

Constant Acceleration Particle Model
This unit we learned how to calculate instantaneous velocity, how to find displacement, how to determine the acceleration of an object, create a velocity vs. time graph, an acceleration vs. time graph, and a motion map solely from a position vs. time graph, and write a mathematical model to describe the motion of an object. 

Very important equations
x=1/2a•t^2+xfinal 
average velocity=∆v/∆t 
acceleration=∆v/t
y=∆v/∆t (t) +starting point
velocity=∆x/∆t (xfinal-xinitial/tfinal-tinitial)
∆x=1/2b•h
*∆x is also = to the displacement: final-xinitial*
*Instantanenous velocity at the midtime is = to the average velocity*

CAPM.1
There are two ways to determine the instantaneous velocity of an object. 
1) You can determine the slope of the tangent to an x vs. t graph at a given point (my preferred way)
or
2) You can use the equation: instantaneous velocity=a•t+vinitial 
For the first way, you can pick a point on the graph and look at the points on either side of it. 

For this graph, if you are trying to find the velocity of the object at 3 seconds, you would look at the position of the object at 2 seconds and 4 seconds. At 2s the object had traveled 10m and at 4s it had traveled 20m. Next, you set up your equation like so: 10-20m/2-4s. The distances of the positions at 4 and 2 seconds go on the numerator and the time goes on the bottom. Subtract on the top and bottom and voila, you have found the instantaneous velocity! At 3 seconds on this position vs. time graph, the instantaneous velocity at 3 seconds=5m/s.

CAPM.2

There are also two ways to find the displacement of an object. The displacement is the object's overall change in position. 

1) You can find the area under a v vs. t curve

or

2) You can use the equation: x=1/2a•t^2+vinitial•t

CAPM.4

If I am given a position vs. time graph, I can describe the motion of the object, create a v vs. t graph, an a vs. t graph, a motion map, find the average and instantaneous velocity, and find the displacement of the object. 

The first column contains the position vs. time graphs, the second column contains the velocity vs. time graphs, and lastly, the acceleration vs. time graphs. For the first example, the object stays at a constant velocity and does not change position. The v vs. t graph and the a vs. t graph are identical because the velocity is 0 and so is the acceleration. For the second example, the object is moving at a constant velocity. In the v vs. t graph, the line shows that the object is moving constantly and in the positive direction. The object does not accelerate so the line on the a vs. t graph stays on the x-axis at 0. For the last example, the line slopes upwards, constantly accelerating. The v vs. t graph depicts an object with a velocity that changes every second and the a vs. t graph shows the object is accelerating upwards. 

v vs. t graph: line above the x-axis=moving forward & line below the x-axis=moving backwards 

a vs. t graph: line above the x-axis= positive slope on v vs.t graph& line below the x-axis= negative slope on v vs. t graph. 

If the object's velocity has a negative slope then the acceleration line will also be negative (under the origin). 

Practice Problems

In the graph above, the object starts at 0m and ends at 50m over the course of 10 seconds. The slope of the line gets less and less steep which means the object slowed down. It is moving in a positive direction. 

What is the instantaneous velocity at 3 seconds?

Well, first we should look at the two points on either side of 3. At 2 seconds the object had traveled 18m and at 4 seconds it had traveled 31m. 

18-31m/2-4s

 -13/-2

instantaneous velocity at 3 s=6.5 m/s

The initial velocity of the object was 5 m/s. Find the acceleration of the object. 

a=∆v/∆t

6.5-0/3s

2.16m/s^2

Make a corresponding v vs. t graph and an a vs. t graph.

In this graph, the object’s velocity slowed down constantly, from -15m/s to 0m/s after 8 seconds. The object also moved backwards since the line is under the x-axis.

Calculate the acceleration

A=∆v/∆t

0-15/0-8

1.88ms^2

What is the velocity of the object at 1 second and 5 seconds?

V=a•t+vinitial

=(1.88)(1) + (-15)

=-13.12 m/s

V=a•t+vinitial

=(1.88)(5) + (-15)

= -5.6m/s

What is the displacement from 1 to 5 seconds?

∆x=1/2a•t^2+vinitial•t

=1/2(1.88)(5)^2+(-13.12)(5)

=23.5+ (-65.6)

=-42.6m

Physics in every day life

We may not take the time to find the average velocity or acceleration of an object on a daily basis but we do have devices that do it for us instantaneously. After learning about this constant acceleration particle model, I can calculate my average velocity on my run if I were to measure the number of steps I take or my displacement if I kept track of my acceleration & velocity. Although unlikely to use while running, these skills are useful in other areas of life. These concepts were seemingly abstract until we applied them to our everyday lives with the extra credit opportunity of finding the height of 3^rd Mitchell or the displacement of a driveway after a dog ran from one end to the other.

Helpful Reminders