91Ó°ÊÓ

A velocity vector is a critical component in understanding motion in physics, especially when dealing with objects moving in a plane, such as a bird in flight. Velocity vectors are composed of components along the coordinate axes, often denoted as

• ¾±Ì‚ (along the x-axis)
• ÂáÌ‚ (along the y-axis)

In our exercise, the velocity vector \[ \vec{v} = (\alpha - \beta t^2)\hat{\imath} + \gamma\hat{\jmath} \]represents both the speed and direction of the bird's flight. Each part of this vector tells a different part of the motion story. The component \((\alpha - \beta t^2)\hat{\imath}\) describes how the bird's velocity changes over time in the horizontal direction while \(\gamma\hat{\jmath}\) describes it in the vertical direction. By analyzing this vector, we can determine the bird's flight path and how it changes over time due to different forces like gravity or air resistance.

Integration plays a fundamental role in physics as it allows us to find variables from their rates of change. If you know the rate at which something changes, such as velocity, integration can help you find the original quantity, like position. In our problem, the velocity vector given to us is: \[ \vec{v} = (\alpha - \beta t^2)\hat{\imath} + \gamma\hat{\jmath} \]To find the position vector \(\vec{r}(t)\), we integrate each component of the velocity vector:

• For the x-component: \[ x(t) = \int (\alpha - \beta t^2) \, dt = \alpha t - \frac{\beta}{3} t^3 + C_1 \]
• For the y-component: \[ y(t) = \int \gamma \, dt = \gamma t + C_2 \]

Given the initial condition that the bird starts at the origin, the constants \(C_1\) and \(C_2\) are zero. This process of integration helps us reconstruct the bird's trajectory from its velocity.

With integration complete, we can express the bird's position as a function of time. Its position vector is given by:\[ \vec{r}(t) = \left( \alpha t - \frac{\beta}{3} t^3 \right) \hat{\imath} + \gamma t \hat{\jmath} \]This describes the bird's location at any time \(t\).
The next step is to determine the acceleration vector \(\vec{a}(t)\), which tells us how the bird's velocity changes over time. Acceleration is obtained by differentiating the velocity vector:

• \( a_x(t) = \frac{d}{dt} \left( \alpha - \beta t^2 \right) = -2\beta t \)
• \( a_y(t) = \frac{d}{dt} \gamma = 0 \)

Thus, the acceleration vector is:\[ \vec{a}(t) = -2\beta t \hat{\imath} \]This tells us that the bird experiences a varying horizontal acceleration and no vertical acceleration, simplifying the analysis of its motion.

A crucial part of understanding the bird's flight is determining its altitude, particularly when it returns to a given x-position. In our case, we need the bird’s altitude when it returns to \(x=0\) for the first time after \(t=0\).
To solve this, set the x-component of the position vector to zero:\[ \alpha t - \frac{\beta}{3} t^3 = 0 \]Factoring gives:\[ t (\alpha - \frac{\beta}{3} t^2) = 0 \]Apart from the trivial solution \(t=0\), solve for \(t^2 = \frac{3\alpha}{\beta}\). Calculating this gives the first positive time \(t\):\[ t = \sqrt{\frac{3 \times 2.4}{1.6}} = \sqrt{4.5} = 2.12 \, \text{s} \]Finally, substitute \(t=2.12\) back into the equation for the y-component to find the bird's altitude:\[ y(t) = \gamma t = 4.0 \times 2.12 = 8.48 \, m \]This calculated altitude allows us to understand the bird’s flight path and behavior successfully as it intersects the specific point on its trajectory.