Question
The following data shows the position (x), the speed (v), and the deceleration (v 2 ) of a car from the moment that driver hit
The following data shows the position (x), the speed (v), and the deceleration (v 2 ) of a car from the moment that driver hit the break until the moment that the car came to a stop.
X | v | V 2 |
2.00 | 6.90 | 47.61 |
4.00 | 6.00 | 36.00 |
6.00 | 4.90 | 24.01 |
8.00 | 3.40 | 11.56 |
10.00 | 0.00 | 0.00 |
Required:
1. Please utilize an Excel spreadsheet to find the trend of deceleration versus distance, and the parameters involved. Show your Excel formulation (using Excel Statistics functions) to find the parameters and trend of this deceleration.
Hint: Please plot Deceleration versus Distance, i.e., v 2 –values [ on the y-axis] and x-values [on the x-axis] to find the best fit linear equation:
v 2=2 ax + v 2 i
Notice this is the deceleration equation between the driver hitting the break until the moment that the car comes to a stop. Note that 2a is the slope of deceleration line, which is called the deceleration rate. The v i 2 in this equation is the y-axis intercept at x=0 that is given in your trend equation, when you fit a linear trend to your data in Excel.
2. How long did it take for the car to stop from the moment the driver hit the break?
Hint: Please notice the driver hit the break at an initial position assumed to be x=0.0 meter and with an initial speed of v i = unknown at that position (i.e., at x=0.0, v i =?). The car then went through this deceleration process until it came to stop. The data is only given to you from the position of x=2 meters, and not from the initial position of x=0.0 meter. So, you can find these values through the trend and parameters that you will find in Excel. For example, you should be able to show the initial speed when the driver hit the break, the decelerating rate (slope), and etc.
To find the time it took from x=0 to x=10 meters for the car to stop: you need to think that at x=10 the deceleration is still the same as the slope of your trend line, which is also the same as distance travelled divided by the square of time it took to travel this distance. Therefore, by equating these two, the time it took from x=0 to x=10 meters can be calculated, i.e., 10/S 2 is = to the deceleration slope of the trend to find time, S.
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