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Answer the following · Q10

Q.Derive equations of motion for a particle moving in a plane and show that the motion can be resolved into two independent motions in mutually perpendicular directions.

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Let a particle in a plane have velocity u⃗\vec{u} at t=0t=0 and constant acceleration a⃗\vec{a}. Since the acceleration is constant, average acceleration equals instantaneous acceleration, so a⃗=v⃗−u⃗t\vec{a}=\dfrac{\vec{v}-\vec{u}}{t}, giving v⃗=u⃗+a⃗t\vec{v}=\vec{u}+\vec{a}t — the vector first equation of motion. For displacement, use the average velocity v⃗av=u⃗+v⃗2\vec{v}_{av}=\dfrac{\vec{u}+\vec{v}}{2} (valid for constant acceleration), so s⃗=v⃗av t=u⃗+(u⃗+a⃗t)2t=u⃗t+12a⃗t2\vec{s}=\vec{v}_{av}\,t=\dfrac{\vec{u}+(\vec{u}+\vec{a}t)}{2}t=\vec{u}t+\tfrac12\vec{a}t^2 — the vector second equation. Resolving both vector equations into x and y components gives four scalar equations: vx=ux+axtv_x=u_x+a_xt, vy=uy+aytv_y=u_y+a_yt, sx=uxt+12axt2s_x=u_xt+\tfrac12a_xt^2, sy=uyt+12ayt2s_y=u_yt+\tfrac12a_yt^2. The x-equations involve only x-components of velocity, displacement and acceleration, and the y-equations involve only y-components — the two pairs share no common variable, so they can be solved completely separately. This proves that any motion in a plane can be treated as two simultaneous, indepe …

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