A particle moves along the parabolic path \[y = a{x^2}\] in such a way that the y-component of the velocity remains constant, say c. The x and y coordinates are in meters. Then acceleration of the particle at \[x = 1m\] is:
(A) \[ac\hat k\]
(B) \[2a{c^2}\hat j\]
(C) \[ - \dfrac{{{c^2}}}{{4{a^2}}}\hat i\]
(D) \[ - \dfrac{c}{{2a}}\hat i\]
Answer
249k+ views
Hint It is given that the particle is moving along a parabolic trajectory. It is also given that displacement is constant c. Now, we know that change in linear displacement per unit time is velocity and Change in velocity per unit time is acceleration. Using this logic, solve for acceleration.
Complete Step by Step Solution
It is given that a particle follows a parabolic trajectory given by the function\[y = a{x^2}\]. It is also mentioned that the y-component of the particle is said to be constant c. Now, velocity of the y-component is said to be constant c. Now, since the velocity of the particle is constant at the y-axis, acceleration will also remain constant and hence the acceleration experienced by the body will be along the x-component.
Thus , we can represent the equation as
\[y = a{x^2}\]
Now differentiating on both sides , we get
\[ \Rightarrow \dfrac{{dy}}{{dt}} = 2ax\dfrac{{dx}}{{dt}}\]
Here, differentiation of position with respect to time is equal to the velocity of the particle. Now it is given that the y-component of velocity remains constant throughout. Thus, \[\dfrac{{dy}}{{dt}} = c\]. Substituting this on the equation we get,
\[ \Rightarrow c = 2ax\dfrac{{dx}}{{dt}}\]
Now to get acceleration , we differentiate the terms with respect to time to get acceleration along x-direction value.
\[ \Rightarrow c = 2a\dfrac{{{d^2}x}}{{d{t^2}}}\]
\[ \Rightarrow \dfrac{c}{{2a}} = {a_x}\]
Now acceleration in the x-axis, is said to be opposite in direction of its velocity. Hence the final acceleration at x=1m will be ,
\[ \Rightarrow {a_x} = - \dfrac{c}{{2a}}\hat i\]
Hence, Option(d) is the right answer for the given question.
Note Acceleration of a body is defined as the change in velocity of a moving body with respect to its position , per unit time. Like velocity, acceleration is also a vector quantity which changes with respect to direction and magnitude.
Complete Step by Step Solution
It is given that a particle follows a parabolic trajectory given by the function\[y = a{x^2}\]. It is also mentioned that the y-component of the particle is said to be constant c. Now, velocity of the y-component is said to be constant c. Now, since the velocity of the particle is constant at the y-axis, acceleration will also remain constant and hence the acceleration experienced by the body will be along the x-component.
Thus , we can represent the equation as
\[y = a{x^2}\]
Now differentiating on both sides , we get
\[ \Rightarrow \dfrac{{dy}}{{dt}} = 2ax\dfrac{{dx}}{{dt}}\]
Here, differentiation of position with respect to time is equal to the velocity of the particle. Now it is given that the y-component of velocity remains constant throughout. Thus, \[\dfrac{{dy}}{{dt}} = c\]. Substituting this on the equation we get,
\[ \Rightarrow c = 2ax\dfrac{{dx}}{{dt}}\]
Now to get acceleration , we differentiate the terms with respect to time to get acceleration along x-direction value.
\[ \Rightarrow c = 2a\dfrac{{{d^2}x}}{{d{t^2}}}\]
\[ \Rightarrow \dfrac{c}{{2a}} = {a_x}\]
Now acceleration in the x-axis, is said to be opposite in direction of its velocity. Hence the final acceleration at x=1m will be ,
\[ \Rightarrow {a_x} = - \dfrac{c}{{2a}}\hat i\]
Hence, Option(d) is the right answer for the given question.
Note Acceleration of a body is defined as the change in velocity of a moving body with respect to its position , per unit time. Like velocity, acceleration is also a vector quantity which changes with respect to direction and magnitude.
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