Question

A particle moves in a plane with a constant acceleration in a direction different from the initial velocity. The path of the particle is-
(A). A straight line
(B). An arc of circle
(C). A parabola
(D). An ellipse

Answer 1

Hint- A particle moves with constant acceleration in a direction different from initial velocity as given in the question. Since the particle has two different directions, it will also have two components. Let the velocity be $v$, acceleration be $a$, angle between them is $\theta [\theta < {90^ \circ }]$

Formula used: $s = ut + \frac{1}{2}a{t^2}$.

Complete step-by-step answer:

Component of Velocity in the direction of $a$(acceleration) is-
$\Rightarrow v\cos \theta$
Component of velocity in a direction perpendicular to $a$ is-
$\Rightarrow v\sin \theta$
As we know the second equation of motion is $s = ut + \frac{1}{2}a{t^2}$.
In the direction of acceleration, we get-
$\Rightarrow y = v\cos \theta + \frac{1}{2}a{t^2}$
Let this be equation 1-
$\Rightarrow y = v\cos \theta + \frac{1}{2}a{t^2}$ (equation 1)
In the direction perpendicular to acceleration, we get-
$\Rightarrow x = v\sin \theta t \\\ \\\ \Rightarrow t = \frac{x}{{v\sin \theta }} \\\$
Let this be equation 2-
$\Rightarrow t = \frac{x}{{v\sin \theta }}$ (equation 2)
Substituting the value of equation 2 in equation 1 we have-
$\Rightarrow y = v\cos \theta \left( {\frac{x}{{v\sin \theta }}} \right) + \frac{1}{2}a\left( {\frac{{{x^2}}}{{{{\sin }^2}\theta }}} \right) \\\ \\\ \Rightarrow y = \frac{x}{{\tan \theta }} + \frac{1}{2}a\frac{{{x^2}}}{{{{\sin }^2}\theta }} \\\$
The above equation is in the form of parabola: $y = A{x^2} + Bx + C$
Thus, option C is the correct option.

Note: We characterize a parabola as the locus of a point that moves such that its distance from a fixed straight line called the directrix which is equivalent to its distance from a fixed point called the focus. For a parabola whose axis is the x-axis and with vertex at the origin, the equation is, in which p is the distance between the directrix and the focus.