Question

The coordinates of a bird flying in the xy plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2. Calculate the velocity vector of the bird as a function of time.

Answer 1

     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 2

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 $2.4\sqrt{1+t^2}$