Question

the coordinates of a bird flying in the xy plane are given by ( x(t) = alpha t ) and ( y(t) = 3.0 , \text{m} - \beta t^2 ), where ( alpha = 2.4 , \text{m/s} ) and ( \beta = 1.2 , \text{m/s}^2 ). for help with math skills, you may want to review: differentiation of polynomial functions, vector magnitudes, determining the angle of a vector. for general problem - solving tips and strategies for this topic, you may want to view a video tutor solution of calculating average and instantaneous accelerations. part c: calculate the magnitude of the bird’s velocity at ( t = 2.0 , \text{s} ). express your answer in meters per second. view available hint(s) input box with units m/s and buttons

Explanation:

Step1: Find \( v_x \) and \( v_y \)

The velocity components are the derivatives of the position functions. For \( x(t)=\alpha t \), the derivative \( v_x=\frac{dx}{dt}=\alpha \). Given \( \alpha = 2.4\space m/s \), so \( v_x = 2.4\space m/s \) (constant, since derivative of linear term is constant). For \( y(t)=3.0\space m-\beta t^2 \), the derivative \( v_y=\frac{dy}{dt}=- 2\beta t \). Given \( \beta = 1.2\space m/s^2 \) and \( t = 2.0\space s \), so \( v_y=-2\times1.2\times2.0=- 4.8\space m/s \).

Step2: Calculate the magnitude of velocity

The magnitude of velocity \( \vec{v} \) is given by \( v=\sqrt{v_x^2 + v_y^2} \). Substitute \( v_x = 2.4\space m/s \) and \( v_y=-4.8\space m/s \) into the formula:
\[
v=\sqrt{(2.4)^2+(- 4.8)^2}=\sqrt{5.76 + 23.04}=\sqrt{28.8}\approx5.37\space m/s
\]

Answer:

\( \approx5.4\space m/s \) (or more precisely \( \sqrt{28.8}\approx5.37\space m/s \))