Question

derive the equation of motion for a particle moving in a plane and show that the motion can be resolved in two independent motions in mutally perpendicular direction

Answer 1

Answer:

hey mate

Explanation:

here is your answer

solution :

let u be the initial velocity of a particle at an initial time ti = 0 a it's uniform acceleration v its final velocity at a final time tf = t r0 the position vector at time ti r the position vector at time and s the displacement of the particle in the elapsed time ∆t = tf - ti = t

then

$a = \frac{v - u}{t}$

v = u + at. (1)

s = (Vav)t = 1/2 (u + v)t = 1/2 (u + u + at)t

= s = ut + 1/2 at²

The dot product of EQ (1) with itself gives

v . v = (u + at) . (u + at)

= u . u + 2u . at + a . at²

= v² = u² + 2a ( ut + 1/2 at²)

substituting from EQ (2)

v² = u² + 2a . s

eqn 1 2 and 3 are the kinematical equation in vector form for a motion in an X - y plane the component equation are

VX = UX + axt. (eqn 1a)

By = uy + ayt. (eqn 1b)

SX = X - x0 = uxt + 1/2. axt². (eqn 2a)

sy = y - y0 = uyt +1/2 ayt². (eqn 2b)

v²x = u²x + 2axsx. (eqn 3a)

v²y = u²y + 2aysy. (eqn 3b)

we can see that eqs 1a 2a and 3a involve only the x-components of displacement velocity and acceleration while eqs 1b 2b and 3b involve the y-components of these qualities thus a two dimensional motion can be described by two exclusive components along perpendicular direction

$thanks$