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At t = 0, a particle leaves the origin with a velocity of 8 m/s in the positive x direction and moves in the x?
At t = 0, a particle leaves the origin with a velocity of 8 m/s in the positive x direction and moves in the xy plane with a constant acceleration of (-2 i - 4 j) m/s2. What is the maximum horizontal displacement that the particle reaches during its motion?
1 m
2 m
4 m
16 m
2 Answers
- ?Lv 710 years agoFavorite Answer
it reaches its max disp. when Vx = 0 , hence :
Vx^2 = 8^2 + 2(-2) X
=> - 4X = - 64
=> X = 16 m <<<<<<<<<<<<< D
- mikesellLv 45 years ago
You do have a 'c' fee. 'c' isn't unavoidably the third term in the equation, it rather is the single that doesn't have a variable linked to it. subsequently, this is 8. The term which you're lacking is 'b'. that might assist you visualize this, you are able to rewrite the equation as "X^2+0X+8". through fact that any variety situations 0 is likewise 0, and including 0 does not replace the variety, the two equations are the comparable. So your a-fee is a million, your b-fee is 0, and your c-fee is 8. purely as an aside, through fact which you do not have a b-fee, you are able to resolve this plenty extra actual with algebra than with the quadratic formulation, so until the subject tells you to apply the formulation, do not.
