Q.
A particle leaves the origin with an initial velocity u=(3i^) and a constant acceleration a=(-1.0i^-0.5j^) m/s2. Its velocity v and position vector r when it reaches its maximum x coordinate are
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a
v=-2j^
b
v=(-1.5j^)m/s
c
r=(4.5i^-2.25j^)m
d
r=(3i^-2j^)m
answer is B.
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Detailed Solution
ux=3m/s, ax=−1.0m/s2∴ vx2=ux2+2ax⋅x or 0=(3)2+2(−1)(x) or x=4.5m Also vx=ux+axt 0=3−1.0t or t=3sy=uyt+12ayt2=0+12(−0.5)(3)2 =−2.25m and vy=ayt=(−0.5)(3)=−1.5m/s∴ v=vxi^+vyj^=0−1.5j^=(−1.5j^)m/s and r=xi^+yj^=(4.5i^−2.25j^)m
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