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Q.

A particle starts from the origin at t=0 with a velocity of 8.0jm/s and moves in the x-y plane with a constant acceleration of (4.0i^+2.0j^)m/s2. At the instant the particle's x-coordinate is29 m, what is its y-coordinate?

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a

88.9 m

b

44.9 m

c

84.9 m

d

54.9 m

answer is B.

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Detailed Solution

x=uxx+12axt2 ∴  29=(0)(t)+12×(4.0)t2 t2=14.5 s2 t=3.8 s y=uyt+12ayt2 =(8)(3.8)+12×2×14.5 =44.9 m
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