Estimating Rate of Change of Velocity
We were introduced to a super fast car in the beginning of this module and now have the tools to calculate its acceleration at various times as it speeds up in a race.
Reaching a top speed of 270.49 mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from 0 to 60 mph in 3.05 seconds, from 0 to 100 mph in 5.88 seconds, from 0 to 200 mph in 14.51 seconds, and from 0 to 229.9 mph in 19.96 seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration) as it approaches 229.9 mph.
First observe that 60 mph = 88 ft/s, 100 mph [latex]\approx 146.67[/latex] ft/s, 200 mph [latex]\approx 293.33[/latex] ft/s, and 229.9 mph [latex]\approx 337.19[/latex] ft/s. We can summarize the information in a table.
[latex]t[/latex] | [latex]v(t)[/latex] |
---|---|
0 | 0 |
3.05 | 88 |
5.88 | 147.67 |
14.51 | 293.33 |
19.96 | 337.19 |
Now compute the average acceleration of the car in feet per second on intervals of the form [latex][t,19.96][/latex] as [latex]t[/latex] approaches 19.96, by creating a table for average acceleration. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?
[latex]t[/latex] | [latex]\frac{v(t)-v(19.96)}{t-19.96}=\frac{v(t)-337.19}{t-19.96}[/latex] |
---|---|
0.0 | 16.89 |
3.05 | 14.74 |
5.88 | 13.46 |
14.51 | 8.05 |
The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft/s).