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Question

(c) A particle moves with constant acceleration in a direction different from initial

velocity. Let the velocity be $v,$ acceleration be $a,$ angle

Vedclass · Accepted Answer

A parabola

Vedclass · Answer

(c) A particle moves with constant acceleration in a direction different from initial

velocity. Let the velocity be $v,$ acceleration be $a,$ angle between them is $	heta\left[	heta<90^{0}ight]$

Component of Velocity in the Direction of $a$ is $v \cos 	heta$ and in a direction perpendicular to $a$ is vsin $	heta$

From equations of Kinematics, In the direction of acceleration, $y=v \cos 	heta t+\frac{1}{2} a t^{2}-----(1)$

In the direction perpendicular to acceleration,

$x=v \sin 	heta t$

$t=\frac{x}{v \sin 	heta}--------(2)$

Substituting 2 in $1,$ we get:

$y=v \cos 	heta\left(\frac{x}{v \sin 	heta}ight)+\frac{1}{2} a\left(\frac{x^{2}}{\sin ^{2} 	heta}ight)$

$y=\frac{x}{	an 	heta}+\frac{1}{2} a \frac{x^{2}}{\sin ^{2} 	heta}$

This is in the form of a parabola: $y=A x^{2}+B x+C$

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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