A particle moves in the xy plane wi...
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A particle moves in the xy plane with only an x-component of acceleration of 2 ms-2. The particle starts from the origin at t = 0 with an initial velocity having an x-component of 8 ms-1 and y-component of -15 ms-1. Velocity of particle after time t is (A) [(8 + 2t) i - 15j]m s-1 (B) zero (C) 2tî + 151 (D) directed along Z-axis.

JEE/Engineering Exams
Physics
Solution
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( begin{aligned} a_{x} &=2 m s-2 u_{x} &=8 m s^{-1} y_{x} &=u_{x}+a_{x} t &=(8+2 t) &=left(y_{y}=-15 m s^{-1}right. &left.0^{0} cdot u^{-1}=(8+2 t) hat{imath}-15 hat{jmath}right) m s^{-1} text {ans } & mathbb{A} end{aligned} )
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