Question:
A PARTICLE MOVES ALONG THE PARABOLIC PATH y=ax^2 IN SUCH A WAY THAT THE X COMPONENT OF THE VELOCITY REMAINS CONSTANT, SAY c . THE ACCELERATION OF THE PARTICLE IS
Answer:
Differentiate y=ax^2 on both sides with respect to time(t)...
dy/dt = 2axdx/dt
Vy = 2ax(Vx)....where Vy is y component of velocity and Vx is x component of velocity.
Given Vx is constant (say some c)
Vy = 2axc
then again differentiating on both sides we get
d(Vy)/dt = 2acdx/dt
Ay = Anet = 2ac(Vx) = 2ac(c) = 2ac^2
As Ay( Acceleration in y direction) is along y axis unit vector associated with it is j
ANSWER is 2ac^2 j
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