Q37P Here are the positions at three ... [FREE SOLUTION]

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Chapter 1: Q37P (page 40)

Here are the positions at three different times for a bee in flight (a bee鈥檚 top speed is about 7 m/s ). $Time 6.3 s 6.8 s 7.3 s Position < - 3.5, 9.4, 0 > m < - 1.3, 6.2, 0 > m < 0.5, 1.7, 0 > m$ (a)Between 6.3 s and 6.8 s , what was the bee鈥檚 average velocity? Be careful with signs.
(b)Between6.3 s and 7.3 s, what was the bee鈥檚 average velocity? Be careful with signs.
(c)Of the two average velocities you calculated, which is the best estimate of the bee鈥檚 instantaneous velocity at time 6.3 s ?
(d)Using the best information available, what was the displacement of the bee during the time interval from 6.3 s to 6.33 s ?

Short Answer

Expert verified

(a) The bee's average velocity between 6.3 s and 6.8 s was $(4.4 \hat{i} - 6.4 \hat{j}) m / s$ .

(b) The bee's average velocity between 6.3 s and 7.3 s was $(4 \hat{i} - 7.7 \hat{j}) m / s$ .

(d) The displacement of the bee during the time interval from 6.3 s to 6.33 s was $(0.132 \hat{i} - 0.192 \hat{j}) m$ .

Step by step solution

Given data

Position of the bee at $t_{1} = 6.3 s : < - 3.5, 9.4, 0 > m$

Position of the bee at $t_{2} = 6.8 s : < - 1.3, 6.2, 0 > m$

Position of the bee at $t_{3} = 7.3 s : < 0.5, 1.7, 0 > m$

Position vector and average velocity

The position vector of a location $< x_{1}, y_{1}, z_{1} >$ with respect to another location $< x_{2}, y_{2}, z_{2} >$ is

$\overset{鈫�}{r} = (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k} . . . . . (l)$

The average velocity for a displacement $\overset{鈫�}{r}$ in time $t$ is

$\overset{鈫�}{v} = \frac{\overset{鈫�}{r}}{t} . . . . . (ll)$

Determining the average velocity between 6.3 s and 6.8 s

The displacement of the bee from time $t_{1}$ to time $t_{2}$ is the position vector of the location of the bee at time $t_{2}$ with respect to the location of the bee at time $t_{1}$ . From equation (I), this displacement is

${\overset{鈫�}{r}}_{21} = \{(- 1.3 - (- 3.5)) \hat{i} + (6.2 - 9.4) \hat{j} + (0 - 0) \hat{k}\} m = (2.2 \hat{i} - 3.2 \hat{j}) m$

This displacement took place in time $t_{2} - t_{1}$ . From equation (II), the average velocity is

${\overset{鈫�}{v}}_{21} = \frac{{\overset{鈫�}{r}}_{21}}{t_{2} - t_{1}} = \frac{(2.2 \hat{i} - 3.2 \hat{j}) m}{(6.8 s - 6.3 s)} = \frac{(2.2 \hat{i} - 3.2 \hat{j}) m}{0.5 s} = (4.4 \hat{i} - 6.4 \hat{j}) m / s$

Thus, the required average velocity is $(4.4 \hat{i} - 6.4 \hat{j}) m / s$ .

Determining the average velocity between 6.3 s and 7.3 s

The displacement of the bee from time $t_{1}$ to time $t_{3}$ is the position vector of the location of the bee at time $t_{3}$ with respect to the location of the bee at time $t_{1}$ . From equation (I), this displacement is

${\overset{鈫�}{r}}_{31} = \{(0.5 - (- 3.5)) \hat{i} + (1.7 - 9.4) \hat{j} + (0 - 0) \hat{k}\} m = (4 \hat{i} - 7.7 \hat{j}) m$

This displacement took place in time $t_{3} - t_{1}$ . From equation (II), the average velocity is

${\overset{鈫�}{v}}_{31} = \frac{{\overset{鈫�}{r}}_{31}}{t_{3} - t_{1}} = \frac{(4 \hat{i} - 7.7 \hat{j}) m}{(7.3 s - 6.3 s)} = \frac{(4 \hat{i} - 7.7 \hat{j}) m}{1 s} = (4 \hat{i} - 7.7 \hat{j}) m / s$

Thus, the required average velocity is $(4 \hat{i} - 7.7 \hat{j}) m / s$ .

Determining the best estimate of the bee's instantaneous velocity at 6.3 s

More the bee travels, more will its velocity change. So the average estimate calculated at a small time interval will be a better estimate of instantaneous velocity. Hence ${\overset{鈫�}{v}}_{21}$ will be a better estimate of the bee's instantaneous velocity at 6.3 s .

Determining the displacement of the bee between 6.3 s and 6.33 s

Consider the velocity of the bee to remain constant at ${\overset{鈫�}{v}}_{21}$ in between 6.3 s and $t_{4} = 6.33 s$ . From equation (II), the displacement in this time period is

Thus, the required displacement is $(0.132 \hat{i} - 0.192 \hat{j}) m$ .