Question

the coordinates of a bird flying in the xy plane are given by x(t) = αt and y(t) = 3.0 m − βt², where α = 2.4 m/s and β = 1.2 m/s². 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 e calculate the magnitude of the bird’s acceleration at t = 2.0 s. express your answer in meters per second squared. view available hint(s) input box m/s²

Explanation:

Step1: Find x - acceleration

The x - coordinate is given by \(x(t)=\alpha t\). The acceleration \(a_x\) is the second derivative of \(x(t)\) with respect to time. The first derivative of \(x(t)\) (velocity in x - direction) \(v_x=\frac{dx}{dt}=\alpha\). The second derivative (acceleration in x - direction) \(a_x = \frac{d^2x}{dt^2}=0\ m/s^2\) (since \(\alpha\) is a constant).

Step2: Find y - acceleration

The y - coordinate is given by \(y(t)=3.0\ m-\beta t^2\). The first derivative (velocity in y - direction) \(v_y=\frac{dy}{dt}=- 2\beta t\). The second derivative (acceleration in y - direction) \(a_y=\frac{d^2y}{dt^2}=-2\beta\). We know that \(\beta = 1.2\ m/s^2\), so \(a_y=-2\times1.2=- 2.4\ m/s^2\). The magnitude of the y - acceleration is \(|a_y| = 2.4\ m/s^2\) (since acceleration is a vector, but we are interested in the magnitude of the net acceleration. Since \(a_x = 0\), the net acceleration is just the magnitude of \(a_y\)).

Step3: Calculate the magnitude of net acceleration

The magnitude of the acceleration vector \(\vec{a}\) is given by \(a=\sqrt{a_x^2 + a_y^2}\). Since \(a_x = 0\) and \(a_y=- 2.4\ m/s^2\) (the negative sign indicates direction, but for magnitude we take the absolute value), we have \(a=\sqrt{0^2+( - 2.4)^2}=\sqrt{5.76}=2.4\ m/s^2\).

Answer:

\(2.4\)