MCQ
A particle moves in a plane with constant acceleration in a direction different from the initial velocity. The path of the particle will be
- AA straight line
- BAn arc of a circle
- ✓A parabola
- DAn ellipse
✓
Answer
Correct option: C.
A parabola
c
(c) A particle moves with constant acceleration in a direction different from initial
(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 $\theta\left[\theta<90^{0}\right]$
Component of Velocity in the Direction of $a$ is $v \cos \theta$ and in a direction perpendicular to $a$ is vsin $\theta$
From equations of Kinematics, In the direction of acceleration, $y=v \cos \theta t+\frac{1}{2} a t^{2}-----(1)$
In the direction perpendicular to acceleration,
$x=v \sin \theta t$
$t=\frac{x}{v \sin \theta}--------(2)$
Substituting 2 in $1,$ we get:
$y=v \cos \theta\left(\frac{x}{v \sin \theta}\right)+\frac{1}{2} a\left(\frac{x^{2}}{\sin ^{2} \theta}\right)$
$y=\frac{x}{\tan \theta}+\frac{1}{2} a \frac{x^{2}}{\sin ^{2} \theta}$
This is in the form of a parabola: $y=A x^{2}+B x+C$
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