By: Ross R, Adam C, Jacob K, Jerry H

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Presentation transcript:

By: Ross R, Adam C, Jacob K, Jerry H I Was Going How Fast? By: Ross R, Adam C, Jacob K, Jerry H

Question #1 A police officer investigating an accident finds a skid mark 115 feet long. Approximately how fast was the car going when the driver applied the brakes? 1.y=√21x 2.y=√21(115) 3.y=√2415 4.y=49.143 Answer:The driver was going about 49 mph at the time that the brakes were applied.

Question #2 If a car is traveling at 60 MPH and the driver applies the brakes in an emergency situation, how much distance does your model say is required for the car to come to a complete stop? 60=√(21)(d) 3600=(21)(d) ≈171.428=d 171.428 feet are required for the car to completely stop.

Question #3 3. What is a realistic domain and range for this situation? Domain: {x∈R l x≥0} [0,∞) Range: {y∈R l y≥0} [0,∞)

Question #4 Does doubling the length of the skid double the speed the driver was going? Justify your response using tables, symbols, and graphs. No, doubling the length of the skid does NOT double the speed the driver was going. The equation of the problem is s=√21d, with s representing the speed of the driver and d representing the distance of the skid mark. Due to the square root in the function, you do not see a constant increase that would lead to the distance and speed being proportional when you graph the function. Instead, you see the graph gradually get more and more gradual. To test my theory, I decided to graph the function on desmos and use the original distance of the skid as 2 feet. I then doubled the length to make it 4, and the speed did not double with it.