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Calculus IChapter 2: LimitsSection 2-1

Tangent Lines and Rates of Change — Question 6

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Question 6

The height hh (in meters) of a rocket launched vertically is given by the function h(t)=100+40t−5t2,h(t) = 100 + 40t - 5t^2, where tt is time in seconds after launch.

(a) Compute the average velocity of the rocket over the time interval [2,4][2, 4].

(b) Find the instantaneous velocity at t=2t = 2.

(c) Sketch the height function, the secant line over [2,4][2, 4], and the tangent line at t=2t = 2.

Original worksheet page 1: question and worked solution for 2-1-006
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Question 6 - Solution

Given: h(t)=100+40t−5t2h(t) = 100 + 40t - 5t^2

(a) Average velocity on [2,4][2, 4]: h(2)=100+80−20=160h(2) = 100 + 80 - 20 = 160 h(4)=100+160−80=180h(4) = 100 + 160 - 80 = 180 Average velocity=180−1604−2=202=10 m/s\text{Average velocity} = \frac{180 - 160}{4 - 2} = \frac{20}{2} = 10 \text{ m/s}

(b) Instantaneous velocity at t=2t = 2: h′(t)=40−10t⇒h′(2)=40−20=20 m/sh'(t) = 40 - 10t \quad \Rightarrow \quad h'(2) = 40 - 20 = 20 \text{ m/s}

(c) Interpretation: At t=2t = 2, the rocket is climbing at a rate of 20 m/s, which is faster than its average velocity between t=2t = 2 and t=4t = 4.

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Original worksheet page 2: question and worked solution for 2-1-006

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