Question

explain path length and displacement for the two dimensional motion of a particle hence sate the expression for average velocity and instantaneous velocity ​

Answer 1

Answer:

hey mate

Explanation:

here is your answer

solution:

consider a two dimensional motion in which A (X1 y1) and B (X2 y2) are the position of a particle at instants t1 and t2 respectively

the motion of the particle is in general along a curved path shown by the dashed line in the figure

path length : the length of this curved path between A and B is the distance travelled by the particle

displacement:

suppose R1 = x1i + y1i is the position vector of a particle vector of a particle at an initial the T1 and r2 = x2i + y2j is the position vector at the final time t2

then the displacement s of the particle during that time interval ∆t = t2 - t1 is

s = ∆r = r2 - R1 = (X2 - X1)I + (y2 - y1)I

= (∆x)i + (∆ Y)J

Average velocity during that time interval is

|Vav| = √(Vav)²x + (Vav)²y

and makes an angle theta with the +x-axis given by

tan theta = (Vav)y/(Vav)x

=∆y/∆t/∆x/∆t

= ∆y/∆x

which is the slope of the secant joining the initial and final position

Instantaneous velocity

v = |v| = √(dx/dt)² + (Dy/Dt)²

and makes an angle theta with the + x-axis given by

tan teta = Dy/dt/DX/dt

= Dy/DX

which is the slope of the tangent to the curve at the point at which the instantaneous velocity is determined.

$thanku$