Question

1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is the time elapsed, measured in seconds. Compute the average velocity of the runner over the given time intervals. a. b. c. d. e. Use the preceding answers to guess the instantaneous velocity of the runner at sec. ( ) 3 4 6 t dt t = + [1.95, 2.05] [1.995, 2.005] [1.9995, 2.0005] [2, 2.00001] t = 2

Answer 1

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by $d(t)=\frac{t^{2}}{6}+4t$ where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

$V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}$

so let's take the first interval:

(a) [1.95, 2.05]

$V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}$

we get that:

$d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125$

$d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167$

so:

$V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s$

(b) [1.995, 2.005]

$V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}$

we get that:

$d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831$

$d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835$

so:

$V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s$

(c) [1.9995, 2.0005]

$V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}$

we get that:

$d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358$

$d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358$

so:

$V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s$

(d) [2, 2.00001]

$V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}$

we get that:

$d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333$

$d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333$

so:

$V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s$

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s