badge image

Enroll For Free Now & Improve Your Performance.

User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
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

3 years ago

Answers : (1)

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
Arun(askIITians forum expert)
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

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

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details