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Grade: 12

                        

A particle moves along a parabolic path y=ax 2 in such a way that the y-component of the velocity remains constant, say c. The x and y coordinates are in meters. The the accleration of the particle at x=1m is (A) ac (B) 2ac 2 (C) -c ​2 /4a 2 (D) -c/2a

2 years ago

Answers : (1)

Arun
24742 Points
							
Dear Chaitanya
 
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 
Option B
 
Regards
Arun(askIITians forum expert)
2 years ago
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