{"id":267,"date":"2021-02-03T23:36:49","date_gmt":"2021-02-03T23:36:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&p=267"},"modified":"2021-06-23T15:59:30","modified_gmt":"2021-06-23T15:59:30","slug":"putting-it-together-derivatives","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/putting-it-together-derivatives\/","title":{"raw":"Putting It Together: Derivatives","rendered":"Putting It Together: Derivatives"},"content":{"raw":"

Estimating Rate of Change of Velocity<\/h3>\r\nWe 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.\r\n

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<\/span>) as it approaches 229.9 mph.<\/p>\r\n

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.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]v(t)[\/latex] at different values of [latex]t[\/latex]<\/caption>\r\n
[latex]t[\/latex]<\/th>\r\n[latex]v(t)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
0<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
3.05<\/td>\r\n88<\/td>\r\n<\/tr>\r\n
5.88<\/td>\r\n147.67<\/td>\r\n<\/tr>\r\n
14.51<\/td>\r\n293.33<\/td>\r\n<\/tr>\r\n
19.96<\/td>\r\n337.19<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

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.\u00a0Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Average acceleration<\/caption>\r\n
[latex]t[\/latex]<\/th>\r\n[latex]\\frac{v(t)-v(19.96)}{t-19.96}=\\frac{v(t)-337.19}{t-19.96}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
0.0<\/td>\r\n16.89<\/td>\r\n<\/tr>\r\n
3.05<\/td>\r\n14.74<\/td>\r\n<\/tr>\r\n
5.88<\/td>\r\n13.46<\/td>\r\n<\/tr>\r\n
14.51<\/td>\r\n8.05<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft\/s).<\/p>\r\n ","rendered":"

Estimating Rate of Change of Velocity<\/h3>\n

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.<\/p>\n

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<\/span>) as it approaches 229.9 mph.<\/p>\n

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.<\/p>\n\n\n\n\n\n\n\n\n\n
[latex]v(t)[\/latex] at different values of [latex]t[\/latex]<\/caption>\n
[latex]t[\/latex]<\/th>\n[latex]v(t)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n
0<\/td>\n0<\/td>\n<\/tr>\n
3.05<\/td>\n88<\/td>\n<\/tr>\n
5.88<\/td>\n147.67<\/td>\n<\/tr>\n
14.51<\/td>\n293.33<\/td>\n<\/tr>\n
19.96<\/td>\n337.19<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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.\u00a0Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?<\/p>\n\n\n\n\n\n\n\n\n
Average acceleration<\/caption>\n
[latex]t[\/latex]<\/th>\n[latex]\\frac{v(t)-v(19.96)}{t-19.96}=\\frac{v(t)-337.19}{t-19.96}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n
0.0<\/td>\n16.89<\/td>\n<\/tr>\n
3.05<\/td>\n14.74<\/td>\n<\/tr>\n
5.88<\/td>\n13.46<\/td>\n<\/tr>\n
14.51<\/td>\n8.05<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft\/s).<\/p>\n

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Candela Citations<\/h3>\n\t\t\t\t\t
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