Question

A particle crossing the origin at time \(t = 0\), moves in the xy-plane with a constant acceleration ‘a’ in y-direction. If the equation of motion of the particle is \(y = bx^2\) (where \(b\) is a constant), then its velocity component in the x-direction is

• A. \(\sqrt{\frac{2b}{a}}\)
• B. \(\sqrt{\frac{a}{2b}}\)
• C. \(\sqrt{\frac{a}{b}}\)
• D. \(\sqrt{\frac{b}{a}}\)

Hint:

To find components of velocity along different directions, relate parametric equations using derivatives and eliminate intermediate variables using known motion formulas.

Answer 1

The Correct Option is B

Step 1: Given the equation of trajectory: \(y = bx^2\)
Differentiate both sides with respect to time \(t\):
\[ \frac{dy}{dt} = \frac{d}{dt}(bx^2) = 2bx\frac{dx}{dt} \] Step 2: Let \(v_y = \frac{dy}{dt}\), \(v_x = \frac{dx}{dt}\)
So, \(v_y = 2bxv_x\) 
Step 3: In y-direction, the particle has constant acceleration \(a\)
Since \(v_y = at\), substitute in previous equation:
\[ at = 2bxv_x \Rightarrow v_x = \frac{at}{2bx} \] Step 4: Eliminate \(x\) using the original trajectory equation: \(y = bx^2\)
In y-direction: \(y = \frac{1}{2}at^2\) (since \(a\) is constant and initial velocity in y is 0)
\[ \frac{1}{2}at^2 = bx^2 \Rightarrow x = \sqrt{\frac{at^2}{2b}} = t\sqrt{\frac{a}{2b}} \] Step 5: Substitute \(x\) in the expression for \(v_x\): \[ v_x = \frac{at}{2b \cdot t\sqrt{\frac{a}{2b}}} = \frac{a}{2b \sqrt{\frac{a}{2b}}} = \sqrt{\frac{a}{2b}} \] Step 6: Select the correct option.
So, the x-component of velocity is \(\sqrt{\frac{a}{2b}}\), which matches option (2).