Question

A particle moves in the xy-plane with constant acceleration a directed along the negative y-axis. The equation of path of the particle has the form \[y=bx-c{{x}^{2}}\], where b and c are positive constants. The velocity v of the particle at the origin of coordinates will be
A. \[\sqrt{\dfrac{a}{2c}(1+{{b}^{2}})}\]
B. \[\sqrt{\dfrac{a}{4c}(1+{{b}^{2}})}\]
C. \[\sqrt{\dfrac{3a}{2c}(1+{{b}^{2}})}\]
D. None of these

Answer 1

Hint: We can do this problem by finding the x and y components of velocity of the particle. Differentiation of the displacement can give the velocity and the acceleration of the particle. By combining the velocities of the particle, we can find out the total velocity of the particle with the projectile motion.

Formula used:
\[\begin{align}
  & \dfrac{dx}{dt}=v \\
 & \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=\dfrac{dv}{dt}=a \\
\end{align}\]
a is the acceleration, v is the velocity, x is the displacement and t is the time.
\[v=\sqrt{v_{x}^{2}+v_{y}^{2}}\]

Complete step-by-step answer:
According to the question a particle moves with constant acceleration (-a). Since it is accelerated along the negative y-axis. The equation of the path of the particle is already given. This is similar to the equation of projectile motion.
\[y=bx-c{{x}^{2}}\]……………………(1)
To find the velocity, first we have to differentiate this equation with respect to time (t).
\[\dfrac{dy}{dt}=b\dfrac{dx}{dt}-2cx\dfrac{dx}{dt}\]
We can call \[\dfrac{dy}{dt}\] as velocity along the y-axis \[({{v}_{y}})\] and \[\dfrac{dx}{dt}\] as velocity along the x-axis \[({{v}_{x}})\].
\[{{v}_{y}}=b{{v}_{x}}-2cx{{v}_{x}}\]……………(2)
As we know, at the origin x and y will be zero. Therefore, equation (2) can be written as,
\[{{v}_{y}}=b{{v}_{x}}\]………………………..(3)
The particle possesses negative acceleration along the y-axis. As we know, the derivative velocity will be the acceleration. This acceleration is happening only in the direction along the y-axis.
\[\begin{align}
  & \dfrac{d{{v}_{y}}}{dt}=-a \\
 & \dfrac{d{{v}_{x}}}{dt}=0 \\
\end{align}\]…………………..(4)
We can differentiate equation (2) with respect to time t.
\[\dfrac{d{{v}_{y}}}{dt}=b\dfrac{d{{v}_{x}}}{dt}-2c{{v}_{x}}^{2}-2cx\dfrac{d{{v}_{x}}}{dt}\]
We can plug equation (4) in this.
\[-a=b\times 0-2c{{v}_{x}}^{2}-2cx\times 0\]
From this equation, we can find out the \[{{v}_{x}}\] component.
\[{{v}_{x}}=\sqrt{\dfrac{a}{2c}}\]…………………………(5)
From these calculations, we found the x and y-components of the velocity of the particle. Now we can find out the total velocity of the particle.
\[v=\sqrt{v_{x}^{2}+v_{y}^{2}}\]
First, we are going to put equation (3) into this.
\[v=\sqrt{v_{x}^{2}+{{(b{{v}_{x}})}^{2}}}\]
\[v=\sqrt{v_{x}^{2}(1+{{b}^{2}})}\]
Now we can put the x component of velocity that we have already found.
\[v=\sqrt{\dfrac{a}{2c}(1+{{b}^{2}})}\]
Therefore the correct answer is option A.

Note: We can also do this problem by comparing the equation of projectile motion with the equation. In this problem, the negative value of acceleration has very importance. If you are taking that as positive, the entire calculations will be wrong. Do not forget to combine the x and y components of velocity to find the total velocity of the particle.