Distance travelled by a car in feet is given by: #t^3-9t^2+24t+5#, how do you determine the average velocity over the interval (1,2)?

Question

Samantha Adkison · Accepted Answer

#4# feet/time unit (or less) in some unknown direction.

If #f(t)=t^3-9t^2+24t+5# is the distance (in feet) traveled after #t# (?seconds): #f(1)=1^3-9(1^1)+24(1)+5 = 21# #f(2)=2^3-9(2^2)+24(2)+5 = 8-36+48+5 = 25# Time interval length (#Deltat#) #= 2-1=1 # (time unit; maybe seconds) Average speed = ((f(2)-f(1))/(2-1) = 4 # feet/(?second?) It should be noted that velocity has to have a direction component. The vehicle's velocity in the direction of travel will be 4 feet/(time unit) if it is moving straight ahead. Its velocity, if it is not traveling in a straight line, will be less than 4 feet/time unit in a direction that is not discernible from the information provided.

Emilia Aguilar · Answer

undefined To determine the average velocity over the interval (1,2) for the function $ t^3 - 9t^2 + 24t + 5 $, use the formula for average velocity, which is the change in position divided by the change in time. In this case, the change in time is 2 - 1 = 1 (since the interval is from 1 to 2).

To find the change in position, evaluate the function at the endpoints of the interval and subtract the values:

At $ t = 1 $: $ 1^3 - 9(1)^2 + 24(1) + 5 $
At $ t = 2 $: $ 2^3 - 9(2)^2 + 24(2) + 5 $

Then, subtract the value at $ t = 1 $ from the value at $ t = 2 $.

After finding the change in position, divide it by the change in time (1) to get the average velocity over the interval (1,2).