The displacement (in meters) of a particle moving in a straight line is given by

Question

The displacement (in meters) of a particle moving in a straight line is given by s= t^2 - 8t + 18, where t is measured in seconds. A. Find the average velocity over each time interval: a. [3, 4] b. [3.5, 4] c. [4, 5] d. [4, 4.5] For a I got v= 7. b I got v = 7.5 c I got v =9 and d is v= 8.5 B. find the instantaneous velocity when t= 4. I thing the answer is v= 8. C. Draw the graph of s as a function of t and draw the secant line whose slopes are the average velocities in part A and the tangent line whose slope is the instantaneous velocity in part B. Help don't know what to draw here. Thanks!

Explanation / Answer

your velocity equation becomes V= 2*t -8 and acceleration becomes a =2... therefore a) 2 m/s b) 1 m/s c) 2m/s d) 1 m/s .... and instantaneous velocity at t=4 becomes v =0 m/s. you can draw graph of s vs t .. it's a parabola symmetric wrt x =0 and have equation (s-2) = (t-4)^2 ... therefore it's virtex is (4,2). And accordingly you can find tangent and secant of this parabola.. but check velocity results first!!.. i fear that your results have some issue there!

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