Verified Answer :
The derivative of the function space as a function of time is equal to a function of speed as a function of time.
\(x(t)=2.4t \\ \frac{x(t)}{t}=t \\ v_{x}(t)=t \\ \\ y(t)=3-1.2\times t^2 \\ \frac{y(t)}{t}=1.2\times2t \\ v_{y}(t)=2.4t\)
The velocity vector is given by the vector sum of the velocities of both axes.
\(v_{R}^2(t)=v_{x}^2(t)+v_{y}^2(t) \\ v_{R}^2(t)=(t)^2+(2.4t)^2 \\ \boxed {v_R(t)=2.6t}\)
If you notice any mistake in my english, please let me know, because I am not native.
Answer : The velocity vector of the bird as a function of time \(2.4\sqrt{1+t^2}\)
Solution :
\(x(t)=\alpha t\\\\\frac{dx}{dt}=\alpha\)
The x-component is, \(\frac{dx}{dt}=\alpha\)
\(y(t)=3.0m-\beta t^2\\\\\frac{dy}{dt}=-2\beta t\)
The y-component is, \(\frac{dy}{dt}=-2\beta t\)
Now we have to calculate the velocity vector of the bird as a function of time.
\(\overset{\rightarrow}V=\sqrt{x^2+y^2}\)
Now put the values of x-component and y-component in this equation, we get
\(\overset{\rightarrow}V=\sqrt{x^2+y^2}=\sqrt{(\alpha)^2+(-2\beta t)^2}=\sqrt{(2.4)^2+(-2\times 1.2\times t)^2}=2.4\sqrt{1+t^2}\)
Therefore, the velocity vector of the bird as a function of time \(2.4\sqrt{1+t^2}\)