A biologist studying the motion of bacteria notes a bacterium at a position r1 vct = (2.2i vct+3.7j vct-1.2k vct ) mu metre after 6.2s the bacterium is at r2vct =( 4.6i vct+ 1.9k vct) mu metre Wat is the average velocity express the answer in vector notation and cal the magnitude *vct=vector .bcz the symBol is not available pls do ans within today it's urgent
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r1 = (2.2i + 3.7j - 1.2k) μm
r2 = (4.6i + 1.9k) μm
r2 - r1 = 4.6i + 1.9k - 2.2i - 3.7j + 1.2k
r2 - r1 = (2.4i - 3.7j + 3.1k) μm
Average velocity = Total displacement / Total time
= (r2 - r1) / T
= (2.4i - 3.7j + 3.1k) μm / 6.2 s
= (0.387i - 0.596j + 0.5k) μm/s
Magnitude of average velocity = sqrt(0.387^2 + 0.596^2 + 0.5^2)
= 0.8688 μm/s
r2 = (4.6i + 1.9k) μm
r2 - r1 = 4.6i + 1.9k - 2.2i - 3.7j + 1.2k
r2 - r1 = (2.4i - 3.7j + 3.1k) μm
Average velocity = Total displacement / Total time
= (r2 - r1) / T
= (2.4i - 3.7j + 3.1k) μm / 6.2 s
= (0.387i - 0.596j + 0.5k) μm/s
Magnitude of average velocity = sqrt(0.387^2 + 0.596^2 + 0.5^2)
= 0.8688 μm/s
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