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12.6: Vector Magnitude & Distance
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Warmup If velocity of a particle is given by π£ π‘ =arcsinβ‘(2π₯), calculate the distance the particle traveled over the interval (0,.5).
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A vector is a quantity have both ______________ and ______________.
So if we wanted to find the magnitude of both the x(t) and y(t) vectors together. What would we do? (Campbell: for visual help draw a graph with two vectors)
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π₯ π‘ ,π¦ π‘ = (π₯ π‘ ) 2 +(π¦(π‘) ) 2 Determining vector magnitude
π₯ π‘ ,π¦ π‘ = (π₯ π‘ ) 2 +(π¦(π‘) ) 2 Does this vector magnitude give you direction?
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π₯β²β² π‘ ,π¦β²β² π‘ = (π₯β²β² π‘ ) 2 +(π¦β²β² π‘ ) 2 =
So what would each of these vector magnitudes give you? π₯ π‘ ,π¦ π‘ = (π₯ π‘ ) 2 +(π¦ π‘ ) 2 = π₯β² π‘ ,π¦β² π‘ = (π₯β² π‘ ) 2 +(π¦β² π‘ ) 2 = π₯β²β² π‘ ,π¦β²β² π‘ = (π₯β²β² π‘ ) 2 +(π¦β²β² π‘ ) 2 =
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π·ππ π‘ππππ= π π (π₯β² π‘ ) 2 +(π¦β² π‘ ) 2 ππ₯
How would you calculate distance traveled? π·ππ π‘ππππ= π π (π₯β² π‘ ) 2 +(π¦β² π‘ ) 2 ππ₯
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A particle moves along a curve defined by the function π¦=(π₯β3 ) β π₯+2 . The x-coordinate of the particle is given by the function π₯(π‘), which is satisfied by the equation ππ₯ ππ‘ = 2 π‘β3 for π‘β₯0 with the initial condition π₯ 4 =1. Find the particular solution for π₯ π‘ . (by hand) Find ππ¦ ππ‘ in terms of t. (by hand)
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Find the speed of the particle at t=6 (calculator)
Find the total distance traveled by the particle from 4,6 . (calculator) Find the acceleration vector at t=6. (calculator)
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