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.
MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade:

                        

A particle moves along the curve x^2/9+x^2/4=1 with constant speed v.express its velocity vectorially as a function of(x,y)

2 years ago

Answers : (1)

Khimraj
3007 Points
							

So, effectively, the displacement of the particle with respect to the reference co-ordinate axes is:-

s= (x^2)/9 + (y^2)/4 =1.

Now, velocity is ds/dt.

So, we partially differentiate the equation, first with respect to X, treating you as a constant. This gives the velocity of the particle. (Vx) in X direction.

Vx= (2/9)x (i)^

Similarly, to get Vy, partially differentiate with respect to y.

Vy= y/2 (j)^

So, the velocity vector is

V= (2/9)X (i)^ + y/2 (j)^.

2 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
Vouchers
To Win!!! Click Here for details