It doesn't fit so I will put it below but I am so confused on this question

Question

The displacement (in feet) of a particle moving in a straight line is given by s=1/2t^2-6t+23,

a)Find the average velocity over each time interval a) [4,8], b) [6,8], c) [8,10] d) [8,12]

b)Find the instantaneous velocity when t=8

c) Draw the graph of s as a function of t and draw the secant lines whose slopes are the average velocites in part a). Then draw the tangent line whose slope is the instantaneous velocity in part b).

Answer 1

Jessica,

 

Velocity is the change of displacement over change in time, so in part a) they are asking you to plug in those numbers into the equation and calculate the velocity. For example, for the interval [4,8] you get a displacement of 7 at t=4 and 7 at t=8. (7-7)/(8-4)=0. This is a pretty boring calculation but you get the point.

 

For part b, you can also interpret the differential of the same equation as instantaneous velocity at one point in time instead of over a period of time. This turns out to be something as simple as t-6. Plug in 8 and you get 2 (units are arbitrary here).

 

At this point graphing the behavior is a matter of plugging the equation into a calculator and drawing secant lines between where the points end up at t=4,6,8,10 and 12 for the time intervals required. For example, for [4,8] you'll draw a straight line at y=7 from t=4 to t=8. Once you draw all the lines it will get messy!  The tangent line at x=8 will be a line with a positive slope.

 

Hope that helps!