\[y = 2x^2 = 2(2t^2)^2 = 8t^4\]
\[a_x = rac{dv_x}{dt} = 4\]
The acceleration of the particle is given by: \[y = 2x^2 = 2(2t^2)^2 = 8t^4\] \[a_x
\[a(2) = 4i + 36j\] A particle moves along a curve defined by \(y = 2x^2\) . The \(x\) -coordinate of the particle varies with time according to \(x = 2t^2\) . Determine the velocity and acceleration of the particle at \(t = 1\) s. Solution The \(y\) -coordinate of the particle is given by: \[y = 2x^2 = 2(2t^2)^2 = 8t^4\] \[a_x
\[v_y = rac{dy}{dt} = 32t^3\]