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Question: A particle moves along a parabolic path \( y = 9{x^2} \) in such a way that the \( x \) component of...

A particle moves along a parabolic path y=9x2y = 9{x^2} in such a way that the xx component of velocity remains constant and has a value 0.333m/s0.333m/s . The magnitude of the acceleration of the particle is:
(A) 1\left( A \right){\text{ 1}}
(B) 2\left( B \right){\text{ 2}}
(C) 3\left( C \right){\text{ 3}}
(D) 4\left( D \right){\text{ 4}}

Explanation

Solution

As we know that the xx component of the velocity is the differentiation of xx with respect to tt . So differentiating the parabolic path with respect to time tt and by doing so we will get the yy component of velocity. And in this way, we can get the magnitude of the acceleration.

Complete Step By Step Answer:
We have the parabolic path equation given by y=9x2y = 9{x^2}
So by the hint, we have the xx component of the velocity vx=dxdt{v_x} = \dfrac{{dx}}{{dt}} and is given by 0.333m/s0.333m/s .
Now on differentiating the above question equation with respect to time tt , we get the equation as
⇒dydt=18xdxdt\Rightarrow \dfrac{{dy}}{{dt}} = 18x\dfrac{{dx}}{{dt}}
And since, dydt=vy\dfrac{{dy}}{{dt}} = {v_y} , therefore the above equation can be written as
⇒vy=18xvx\Rightarrow {v_y} = 18x{v_x}
Now on substituting the values, we will get the equation as
⇒vy=18x(0.333)\Rightarrow {v_y} = 18x\left( {0.333} \right)
And on solving it, the equation will be
⇒vy=5.994x\Rightarrow {v_y} = 5.994x
From the above we can see that the xx component velocity is constant, So the acceleration will have the only yy component.
From the above statement, we have acceleration given by
⇒a=ay=dvydt\Rightarrow a = {a_y} = \dfrac{{d{v_y}}}{{dt}}
And on substituting the values, we will get the equation as
⇒5.994dxdt\Rightarrow 5.994\dfrac{{dx}}{{dt}}
And it can also be written as
⇒5.994vx\Rightarrow 5.994{v_x}
And on substituting the values, we get
⇒a=5.994(0.333)\Rightarrow a = 5.994\left( {0.333} \right)
On solving it we will get
⇒a=1.996∼2ms−2\Rightarrow a = 1.996 \sim 2m{s^{ - 2}}
Hence, the option (B)\left( B \right) is correct.

Note:
Acceleration can be defined as the rate of change of velocity. It is a vector quantity which is just the rate of change of velocity. But the velocity may be positive or negative or zero. Also, the S.I unit of the velocity will be m/sm/s . And the SI unit of acceleration is m/s2m/{s^2} .