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Specimen Lab Report

Decent Essays

Car Braking Times and Variations
IB Mathematics SL
Research Report- Specimen Paper A

Given the stimulus word, Life, I started thinking of ways that I can use calculus in looking at vehicular dilemmas. I decided to see how long and how far it takes cars to stop. Approximately 1.3 million people die in road crashes each year, on average 3,287 deaths per day in the entire world. Within the first week of getting my first car I was rear-ended twice. Both situations could have easily been avoided by proper braking. This made me wonder how long it takes peoples cars to stop before a collision occurs. We all know that there are many factors that contribute to this such as rain, wear and tear on tires or brake pads, and human error. These variables …show more content…

The measure of velocity is distance divided by time. We also need to establish that the term “vehicle” in our case means a sedan or light truck. We also need to know the relationship between mph and fps. Mph =miles per hour, fps = feet per second. We use mph mutiplied by (1.467)=fps. Now we need to see what an average stopping time is. Over all cars, an average stopping ratio is about 15 fps. This is used when the street surface is dry and the vehicle has good tires (little wear). For example, if the initial velocity is 60 mph (88 fps), and the car is decelerating at an average of 15 fps, after 1 second the velocity would have been reduced to 73 fps. This would keep decreasing until the vehicle comes to a complete stop after about 6.87 seconds (this includes an extra second for delayed reaction time of the driver). This is calculated by (initial velocity)/ 15fps +1, so (88)/15+1=6.87seconds. Now that we have our stopping time we can find the distance it takes for a vehicle to stop. This is determined by the formula of ½ the initial velocity (in fps) multiplied by the time required to stop. In this case it works out to be ½(88)(6.87)= 302.28 feet. Remember, fps = …show more content…

86.53
40 58.7 144.21
50 73.4 216.29
60 88 302.28
70 102.7 402.93

To find the rate that a car decelerates we will use a deceleration function. This is just the opposite of the acceleration function. In this situation we are working with feet and seconds. The corresponding units for velocity (speed) will be feet per second (f/s) and for acceleration f/s2. The acceleration is in the opposite direction to the motion, and so the acceleration is negative. The following formula only relates the magnitudes of the distance travelled, initial velocity and acceleration.

The simple formula for the stopping distance is: d=u^2/2a • u is the initial velocity
• v is the final velocity
• a is the stopping distance of a car
• t is the time of motion
• d is the deceleration To find the rate of deceleration of the car at a given speed we can use the equation above. Let’s use our example from before of 60 mph. Remember that 60 mph is 88 fps. For this situation the measured stopping distance is 302.28 feet and the velocity at which the brakes are applied is 88

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