A particle moves in the x-y plane with a velocity v_ x =8 t-2 and v _ y = Z . If it…
Physics · JEE Main · NTA Exams — Kinematics
A particle moves in the x-y plane with a velocity \(v_{x}=8 t-2\) and \(\mathbf{v}_{\mathbf{y}}=\mathbf{Z}\). If it passes through the point x = 14 and y = 4 at t = 2 s, the equation of the path is
- \(x=y^{2}-y+z\)
- \(x=y+z\)
- \(\mathbf{x}=\mathbf{y}^{2}+\mathbf{z}\)
- \(x=y^{2}+y+z\)
Answer
(D) x=y^ 2 +y+z
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