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```
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

```
3 years ago

## Answers : (1)

Arun
25768 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 RegardsArun(askIITians forum expert)
```
3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions