Find the parallel and perpendicular components of the acceleration. 10/4/201Y 14

Question

Find the parallel and perpendicular components of the acceleration.

10/4/201Y 14. (10 points) Find the parallel and perpendicular components of the acceleration (that is relative to the velocity vector) given that u=-5 + 5 + 10 k and (a) (2.5 points) Find the unit vector associated with the velocity. Ilok (b) (2.5 points) Determine the magnitude of the parallel component of the acceleration relative to the velocity vector. (e) (2.5 points) Determine the vector a, 0.4( -11 +1.6 (d) (2.5 points) Determine the vector a. ,6 2. .ws 1 +.6yHJ .3

Explanation / Answer

14. given velocity vector

v = -5i + 5j + 10 k

acceleration vector

a = -1 i - 1 j + 2k

|a| = sqroot(1 + 1 + 4) = sqroot(6)

a. velocity unit vector v' = v/|v| = (-5i + 5j + 10k)/sqroot(25 + 25 + 100) = (-i + j + 2k)/sqroot(6)

b. magnitude of parallel component of acceleration relative to velocity component = a.v/|v| = ap

ap = (-i - j + 2k).(-i + j + 2k)/sqr4oot(6) = 4/sqroot(6)

c. ap vectro = 4*v/sqroot(6)|v| = 4(-i + j + 2k)/6 = (-2i + 2j + 4k)/3

d. magnitude of perpendicular compoentn of acceleration, at = sqroot(|a|^2 - ap^2) = sqroot(6 - 16/6) = sqroot(20/6) = sqroot(10/3)

unit vector in this direction = at' = mi + nj + pk

(mi + nj + pk).(-5i + 5j + 10k) = 0

-m + n + 2p = 0

let m = 0.3

then n + 2p = 0.3

also, sqroot(m^2 + n^2 + p^2) = 1

0.09 + (0.3 - 2p)^2 + p^2 = 1

5p^2 - 1.2p -0.82 = 0

solving for p

p = -0.302

hence

n = 0.9048

so vector at' = sqroot(10/3)[0.3i + 0.9048j - 0.302k]

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