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you walk in the direction of the unit vector ⟨1, 0, 0⟩ a distance of 18 m at a constant speed of 2.22 m/s, then turn and walk in the direction of the unit vector ⟨1, 0, 1⟩/√2 a distance of 14 m at a constant speed of 2.9 m/s. what was your average velocity?
average velocity = ⟨,, ⟩ m/s
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Explanation:

Step1: Find time for first segment

Time \( t_1 = \frac{d_1}{v_1} \), where \( d_1 = 18 \, \text{m} \), \( v_1 = 2.22 \, \text{m/s} \).
\( t_1 = \frac{18}{2.22} \approx 8.108 \, \text{s} \)

Step2: Find time for second segment

Time \( t_2 = \frac{d_2}{v_2} \), where \( d_2 = 14 \, \text{m} \), \( v_2 = 2.9 \, \text{m/s} \).
\( t_2 = \frac{14}{2.9} \approx 4.828 \, \text{s} \)

Step3: Find total displacement vectors

First displacement: \( \vec{r_1} = d_1 \cdot \langle 1, 0, 0
angle = \langle 18, 0, 0
angle \, \text{m} \)
Second displacement: \( \vec{r_2} = d_2 \cdot \frac{\langle 1, 0, 1
angle}{\sqrt{2}} = 14 \cdot \frac{\langle 1, 0, 1
angle}{\sqrt{2}} \approx \langle 9.899, 0, 9.899
angle \, \text{m} \)

Total displacement: \( \vec{r}_{\text{total}} = \vec{r_1} + \vec{r_2} = \langle 18 + 9.899, 0 + 0, 0 + 9.899
angle = \langle 27.899, 0, 9.899
angle \, \text{m} \)

Step4: Find total time

\( t_{\text{total}} = t_1 + t_2 \approx 8.108 + 4.828 = 12.936 \, \text{s} \)

Step5: Calculate average velocity

Average velocity \( \vec{v}_{\text{avg}} = \frac{\vec{r}_{\text{total}}}{t_{\text{total}}} \)
\( \vec{v}_{\text{avg}} = \frac{\langle 27.899, 0, 9.899
angle}{12.936} \approx \langle 2.16, 0, 0.766
angle \, \text{m/s} \)

Answer:

The average velocity is approximately \( \langle 2.16, 0, 0.77
angle \, \text{m/s} \) (components rounded appropriately).