HardPhysicsfluid
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-- 6. A particle moves in space along the path z = ax³ + by² in such a way that dx/dt = dy/dt = c. The acceleration of the particle is:
A
(6acx + 2bc)k̂
B
(2ax² + 6by²)k̂
C
(6ac²x + 2bc²)k̂
D
(4bc²x + 3ac²)k̂
Explanation
vx = vy = c. ax = ay = 0. vz = dz/dt = 3ax²(dx/dt) + 2by(dy/dt) = 3acx² + 2bcy. az = d(vz)/dt = 6acx(dx/dt) + 2bc(dy/dt) = 6ac²x + 2bc². Acceleration vector is (6ac²x + 2bc²)k̂.
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