# What are speed velocity, acceleration of particle which move on ellipse at x=pi/4 ?position x=(2cost,sint)

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that if x(t) denotes the position of particle moving along a curve then the velocity of particle at the time t is `(dx(t))/(dt).`

Hence, evaluating the velocity of the particle moving along the ellipse at `x = pi/4`  yields:

`v(t)|_(pi/4) = ((d(2cos t, sin t))/(dt))|_(pi/4)`

`v(t)|_(pi/4) = ((-2sin t,cos t))/(dt))|_(pi/4)`

`v(pi/4) = (-2sin (pi/4), cos(pi/4))=gtv(pi/4) =(-2 sqrt2/2 , sqrt2/2)`

`v(pi/4) = (-sqrt2,sqrt2/2)`

Hence, the speed of particle moving along a curve is the absolute value of velocity  such that s(t)=|v(t)|.

`s(pi/4) = sqrt(4sin^2 (pi/4) + cos^2(pi/4))`

`s(pi/4) = sqrt(2 + 1/2) =gt s(pi/4) = sqrt(5/2)`

The acceleration of particle moving along a curve is the derivative of velocity with respect to t such that: `a(t) = (dv(t))/(dt)` .

`a(t)|_(pi/4) = ((d(-2sin t,cos t))/(dt))|_(pi/4)`

`a(t)|_(pi/4) = ((-2cos t,-sin t)/(dt))|_(pi/4)` `a(pi/4) = (-2cos (pi/4), -sin(pi/4))=gta(pi/4) =(-2 sqrt2/2 ,-sqrt2/2)`

`` `a(pi/4) =(-sqrt2 ,-sqrt2/2)`

Hence, evaluating the velocity, speed and acceleration of a particle moving on an ellipse, at

` x = pi/4=> v(pi/4) = (-sqrt2,sqrt2/2), s(pi/4) = sqrt(5/2) and a(pi/4) =(-sqrt2 ,-sqrt2/2)`