Fan Cart Post

Question: In this challenge, the question we were trying to answer was, “Where will two fan carts with different accelerations collide .”

What we did: Each individual group used a motion sensor and a fan cart running at a constant desired acceleration, and performed a test to find out the initial velocity and acceleration, (which was the slope of a velocity time graph), by letting the cart run down a table towards the sensor. We ran the test twice with the fan at the highest speed just to make sure the values we collected were similar in range, but we only ended up using one of the trials for the final prediction since we knew the data from the motion sensor would be accurate. 

Data: For this particular experiment, all that was needed from each group to make the prediction was the acceleration of the fan cart. My group used Cart A, and the acceleration of the first trial was .184, and the acceleration of the second trial was .0175. We found initial velocities as well (.0765 and .182), but they weren't needed to make our prediction and they were extremely small as well. The group we partnered with, Serenity and Connor, used Cart C, and their accelerations were .0219 and .0205.

How we made the prediction: Our groups decided that we would use the formula 1/2at^2 to predict the point at which the carts would collide since that is the distance/displacement formula ({vi}{t}) was not needed because the initial velocity was practically 0). 

The first step we took was to calculate the time at which the carts would collide using the first velocities stated above from both groups as the acceleration. 2 meters or 200 centimeters represents the distance the carts would be placed from each other on opposite sides of the origin:

2=1/2at^2+1/2at^2

2=1/2(.184)(t)^2+1/2(.0219)

2=(.10295)t^2

2/.10295=(.10295)t^2/.10295

t^2=19.46 (take the root)

t=4.41 seconds= the time it will take the carts to collide

To make the prediction for where the carts would collide we used the same formula only with 4.40 seconds as our time:

1/2at^2=distance cart traveled

CART C: 1/2(.0219)(4.41)^2=.21295m or 21.295 cm

CART A: 1/2(.184)(4.41)^2=-1.79m or 179 cm

Prediction: We predicted that the Cart C would travel 21.295 cm from the starting point of 100cm towards the origin, and Cart A would travel 179 cm from the starting point of 100 cm towards the origin from the other side to collide after traveling those distances.

Results: Our predictions were pretty accurate as the carts collided right around the distances predicted. Cart C traveled 28cm and Cart A traveled 175cm. Our percent error calculated for the predicted distance is:

 179-172/172x100%= 4.06% error