Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 5, 5.1, Section 5.1, Problem 16

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Estimate the distance travelled by the car over a period of 30 seconds suppose that the car over a period of 30 seconds suppose that the car accelerates from rest to a speed of $120 km/h$ during this period.

*Refer to the graph at the book.

We must first convert the velocity with respect to time in seconds to be consistent with our units.

By estimating the distance using six approximating rectangles at right endpoints we obtain

$ \begin{equation} \begin{aligned} DR =& 50 \frac{km}{h} \left( \frac{1h}{3600s} \right) (5s) + 77 \frac{km}{h} \left( \frac{1h}{3600s} \right) (5s) + 95 \frac{km}{h} \left( \frac{1h}{3600s} \right) (5s) + 109 \frac{km}{h} \left( \frac{1h}{3600s} \right) (5s) + 117 \frac{km}{h} \left( \frac{1h}{3600s} \right) (5s) + 120 \frac{km}{h} \left( \frac{1h}{3600s} \right) (5s) \\ \\ DR =& 0.7889 km \end{aligned} \end{equation} $

Similarly, at left end points

$ \begin{equation} \begin{aligned} DL =& \frac{5}{3600} \left[ 0 + 50 + 77 + 95 + 209 + 117 \right] km \\ \\ DL =& 0.6222 km \end{aligned} \end{equation} $

To get an accurate answer, we take the average of both distance to obtain..

$\displaystyle \frac{0.7889 + 0.6222}{2} = 0.7056 km$