Home
Class 11
PHYSICS
A particle moves along the curve (x^(2))...

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

Text Solution

AI Generated Solution

To express the velocity of a particle moving along the curve \(\frac{x^2}{9} + \frac{y^2}{4} = 1\) as a function of \(x\) and \(y\), we can follow these steps: ### Step-by-Step Solution 1. **Understanding the Curve**: The given equation represents an ellipse. We need to find the velocity vector of a particle moving along this curve. 2. **Implicit Differentiation**: We start by differentiating the equation of the ellipse with respect to time \(t\): \[ ...
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Solved Examples|9 Videos

  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Exercise 4.1|17 Videos

  • GRAVITATION

    CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos

  • KINEMATICS-2

    CENGAGE PHYSICS|Exercise Exercise Integer|9 Videos

Similar Questions

Explore conceptually related problems

A particle moves along the curve 12y=x^(3) . . Which coordinate changes at faster rate at x=10 ?

A particle moves along the curve y^(2) = 2x where x = (t^(2))/(2) & y gt 0 . What is the acceleration of the particle at t = 2s ?

A particle moves along the positive branch of the curve y = (x^(2))/(2) where x = (t^(2))/(2),x and y are measured in metres and t in second. At t = 2s , the velocity of the particle is

A particle moves along the curve x^(2)=2y . At what point, ordinate increases at the same rate as abscissa increases ?

A particle moves along 1 ositive branch of the curve y=(x)/(2) where x=(t^(3))/(3) , x and y are measured in metres and t in seconds, then-

A particle moves along the curve y=(4)/(3)x^(3)+5 . Find the points on the curve at which the y - coordinate changes as fast as the x - coordinate.

A particle moves along the curve y=((2)/(3))x^(3)+1. Find the points on the curve at which the y-coordinate is changing twice as fast as the x-coordinate

A particle moves along the curve y=(2)/(3) x^(3)+1 . Find the point on the curve at which y-coordinate is changing twice as fast as x-coordinate.

A particle moves in a plane along an elliptic path given by (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 At point (0, b), the x-component of velocity is u. The y-component of acceleration at this point is-

A particle moves in the x-y plane according to the law x=t^(2) , y = 2t. Find: (a) velocity and acceleration of the particle as a function of time, (b) the speed and rate of change of speed of the particle as a function of time, (c) the distance travelled by the particle as a function of time. (d) the radius of curvature of the particle as a function of time.