
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
600.9k+ views
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.
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.
Recently Updated Pages
Questions & Answers - Ask your doubts

Master Class 9 Social Science: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Master Class 8 Science: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

State the laws of reflection of light

Difference Between Prokaryotic Cells and Eukaryotic Cells

Show that total energy of a freely falling body remains class 11 physics CBSE

