A particle moves in the xy plane with a velocity (which depends on the position) v ⃗=ai ̂+bx j ̂ , where a and b are constants. At the initial instant the particle is located at the origin (x(0) = y(0) = 0). Find the equation of the trajectory y(x).

1. **Integrate \( v_x \) to find \( x(t) \):**

The velocity component in the \( x \)-direction is given by

v_x = a

which leads to the differential equation

\frac{dx}{dt} = a

Integrating with respect to \( t \):

\int dx = \int a \, dt

x(t) = at + C

Given the initial condition \( x(0) = 0 \):

x(t) = at

2. **Substitute \( x(t) \) into \( v_y \) and find \( y(t) \):**

Substitute \( x = at \) into the expression for \( v_y \), where

v_y = bx

so

v_y = b(at) = bat

Now we integrate with respect to \( t \):

\frac{dy}{dt} = b(at)

\int dy = \int bat \, dt

y(t) = \frac{ba}{2} t^2 + C

With the initial condition \( y(0) = 0 \):

y(t) = \frac{ba}{2} t^2

3. **Eliminate \( t \) to get \( y(x) \):**

Since \( x = at \), solve for \( t \):

t = \frac{x}{a}

Substitute back into \( y(t) \):

y = \frac{ba}{2} \left( \frac{x}{a} \right)^2

y = \frac{ba}{2} \cdot \frac{x^2}{a^2}

y = \frac{b}{2a} x^2

Thus, the trajectory of the particle is

y(x) = \frac{b}{2a} x^2