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Home > CALC > Chapter 4 > Lesson 4.4.1 > Problem 4-127

  1. A log going over a waterfall. As a log falls in a waterfall, its velocity is v(t) = −32t − 18 in feet per second. The position of the log at time t = 0 was at the top of the waterfall, 500 feet above sea level. Homework Help ✎

    1. Where is the log after 1 second? 2 seconds? 3 seconds?

    2. Where is the log after t seconds? This is the position function s(t). What is its relationship to v(t) ?

You need to find the position function, s(t).

Antidifferentiate v(t).

Solve for C, by evaluating the given point s(0)=500.

Check your work:
s(1) = 466 ft
s(2) = 400 ft
s(3) = 302 ft

Refer to the hint in part (a).

Use the eTool below to visualize the problem.
Click the link at right for the full version of the eTool: Calc 4-127 HW eTool