0

Guest

Courses New

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

A particle moves along a parabolic path y=ax2 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) 2ac2
(C) -c​2/4a2
(D) -c/2a
 

Grade:12

1 Answers

Arun
25750 Points
6 years ago
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)

Think You Can Provide A Better Answer ?

Other Related Questions on Mechanics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More