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Calculus IChapter 2: LimitsSection 2-1

Tangent Lines and Rates of Change — Question 2

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Question 2

The displacement of a car traveling along a straight road is given by s(t)=10t−0.5t2s(t) = 10t - 0.5t^2 where ss is in meters and tt is in seconds.

(a) Find the average velocity of the car over the interval [2,6][2, 6].

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

(c) Sketch the displacement function and the secant and tangent lines described above.

Original worksheet page 1: question and worked solution for 2-1-002
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Question 2 - Solution

We are given: s(t)=10t−0.5t2s(t) = 10t - 0.5t^2

(a) Average velocity on [2,6][2, 6]: s(2)=10(2)−0.5(4)=20−2=18s(2) = 10(2) - 0.5(4) = 20 - 2 = 18 s(6)=10(6)−0.5(36)=60−18=42s(6) = 10(6) - 0.5(36) = 60 - 18 = 42 Average velocity=42−186−2=244=6 m/s\text{Average velocity} = \frac{42 - 18}{6 - 2} = \frac{24}{4} = 6 \text{ m/s}

(b) Instantaneous velocity at t=2t = 2: s′(t)=10−t⇒s′(2)=10−2=8 m/ss'(t) = 10 - t \quad \Rightarrow \quad s'(2) = 10 - 2 = 8 \text{ m/s}

(c) Interpretation: The average velocity between t=2t = 2 and t=6t = 6 is 6 m/s. At t=2t = 2, the car is moving faster than average at 8 m/s. We now sketch the curve with the secant line between t=2t=2 and t=6t=6, and the tangent line at t=2t=2.

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 2-1-002

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