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Q.

A particle leaves the origin with an initial velocityv=(3.00i^)m/s and a constant accelerates a=(-1.00i^-0.500j^)m/s2. When the particle reaches its maximum x coordinate, what is its position vector?

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a

r=(4.5i^-2j^)m

b

r=(4.5i^-2.25j^)m

c

r=(5.8i^-225j^)m

d

r=(5i^-225j^)m

answer is B.

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Detailed Solution

x=u2t+12axt2 =3×3+12(-10)(3)2=4.5 m y=uyt+12ayt2 =0-12(0.5)(3)2=-2.25 m ∴ r=(4.5i^-2.25j^)m
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