The position of an object at time $t$ is given by the parametric equations $\left\{\begin{array}{l}x=3 t^{2}+4 t \\ y=2 t^{2}+3\end{array}\right.$.
Find the horizontal velocity, the vertical velocity, and the speed at the moment where $t=1$. Do not worry about units in this problem.
Horizontal Velocity =
Vertical Velocity $=$
Speed $=$

Step 1 ：The position of an object at time $t$ is given by the parametric equations $\left\{\begin{array}{l}x=3 t^{2}+4 t \\ y=2 t^{2}+3\end{array}\right.$.

Step 2 ：We need to find the horizontal velocity, the vertical velocity, and the speed at the moment where $t=1$.

Step 3 ：The horizontal velocity is the derivative of the x-position with respect to time, which is $dx/dt = 6t + 4$.

Step 4 ：The vertical velocity is the derivative of the y-position with respect to time, which is $dy/dt = 4t$.

Step 5 ：At $t=1$, the horizontal velocity is $6*1 + 4 = \boxed{10}$.

Step 6 ：At $t=1$, the vertical velocity is $4*1 = \boxed{4}$.

Step 7 ：The speed is the magnitude of the velocity vector, which can be found by taking the square root of the sum of the squares of the horizontal and vertical velocities.

Step 8 ：At $t=1$, the speed is $\sqrt{(10)^2 + (4)^2} = \sqrt{100 + 16} = \sqrt{116} = 2\sqrt{29}$.

Step 9 ：So, the speed at $t=1$ is $\boxed{2\sqrt{29}}$.