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2 years ago

15

You are on a ParKour course. First you climb a angled wall up 9.5 meters. They you shimmy along the edge of a 3.5 meter long wal

k. Lastly you traverse 15 meters of obstacles to reach the end of the course. The trip takes 43 seconds . What is your average speed?

1 answer:

balu736 [363]2 years ago

5 0

1.53 meters per second

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Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.

skelet666 [1.2K]

Answer:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:

a) The vectorial product, a×b is:

$a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k$

b) The escalar product a·b is:

$a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43$

c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:

$(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17$

d) The component of a along the direction of b is:

$a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66$

I hope it helps you!

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