Question

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

Answer 1

Hint: As we know that the $ x $ component of the velocity is the differentiation of $ x $ with respect to $ t $ . So differentiating the parabolic path with respect to time $ t $ and by doing so we will get the $ y $ 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 = 9{x^2} $
So by the hint, we have the $ x $ component of the velocity $ {v_x} = \dfrac{{dx}}{{dt}} $ and is given by $ 0.333m/s $ .
Now on differentiating the above question equation with respect to time $ t $ , we get the equation as
 $ \Rightarrow \dfrac{{dy}}{{dt}} = 18x\dfrac{{dx}}{{dt}} $
And since, $ \dfrac{{dy}}{{dt}} = {v_y} $ , therefore the above equation can be written as
 $ \Rightarrow {v_y} = 18x{v_x} $
Now on substituting the values, we will get the equation as
 $ \Rightarrow {v_y} = 18x\left( {0.333} \right) $
And on solving it, the equation will be
 $ \Rightarrow {v_y} = 5.994x $
From the above we can see that the $ x $ component velocity is constant, So the acceleration will have the only $ y $ component.
From the above statement, we have acceleration given by
 $ \Rightarrow a = {a_y} = \dfrac{{d{v_y}}}{{dt}} $
And on substituting the values, we will get the equation as
 $ \Rightarrow 5.994\dfrac{{dx}}{{dt}} $
And it can also be written as
 $ \Rightarrow 5.994{v_x} $
And on substituting the values, we get
 $ \Rightarrow a = 5.994\left( {0.333} \right) $
On solving it we will get
 $ \Rightarrow a = 1.996 \sim 2m{s^{ - 2}} $
Hence, the option $ \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/s $ . And the SI unit of acceleration is $ m/{s^2} $ .