A particle moves in a plane with constant acc | vedclass

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(C) particle moves with constant acceleration in a direction different from the initial velocity. Let the initial velocity be $v$ and the constant acc

Vedclass · Accepted Answer

$A$ parabola

Vedclass · Answer

(C) particle moves with constant acceleration in a direction different from the initial velocity. Let the initial velocity be $v$ and the constant acceleration be $a$,with an angle $	heta$ between them $(0 Resolve the velocity into two components: one parallel to the acceleration $(v \cos 	heta)$ and one perpendicular to the acceleration $(v \sin 	heta)$.Let the direction perpendicular to the acceleration be the $x$-axis and the direction of acceleration be the $y$-axis.In the $x$-direction,there is no acceleration,so the displacement is $x = (v \sin 	heta) t$,which gives $t = \frac{x}{v \sin 	heta}$.In the $y$-direction,the initial velocity is $v \cos 	heta$ and the acceleration is $a$. The displacement is $y = (v \cos 	heta) t + \frac{1}{2} a t^2$.Substituting the expression for $t$ into the equation for $y$:$y = (v \cos 	heta) \left( \frac{x}{v \sin 	heta} ight) + \frac{1}{2} a \left( \frac{x}{v \sin 	heta} ight)^2$$y = x \cot 	heta + \frac{a}{2 v^2 \sin^2 	heta} x^2$.This equation is of the form $y = Ax^2 + Bx$,which represents a parabola.

Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

$A$ particle moves in a plane with constant acceleration in a direction different from the initial velocity. The path of the particle will be

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