Bird Flight: Calculating Time to Reach Specific Height

This problem involves calculating the time it takes for a bird to reach a certain altitude, given its velocity and starting position. We need to use vector concepts and unit conversions to find the solution. The bird is flying in a straight line, which simplifies the calculation as we can use basic kinematic equations for constant velocity.

Given Velocity Vector and Starting Point

The velocity vector of the bird is given as $ \mathbf{v} = 10\hat{i} + 6\hat{j} + \hat{k} $ km/hr.

The starting point (initial position) of the bird is $ P_0 = (1, 2, 3) $. We assume these coordinates are in meters, with the z-coordinate representing the height from the ground. So, the initial position vector is $ \mathbf{r}_0 = 1\hat{i} + 2\hat{j} + 3\hat{k} $ meters.

The target is to find the time ($t$) when the bird reaches a height of 13 meters from the ground. This means the z-component of the position vector should be 13 meters.

Physics Principles: Motion in Three Dimensions

For an object moving with constant velocity $ \mathbf{v} $, its position vector $ \mathbf{r}(t) $ at time $ t $ is given by the formula:

$ \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}t $

Where:

• $ \mathbf{r}(t) $ is the position vector at time $ t $.
• $ \mathbf{r}_0 $ is the initial position vector.
• $ \mathbf{v} $ is the constant velocity vector.
• $ t $ is the time elapsed.

We are interested in the z-component of the position, which represents the height. Let $ z(t) $ be the height at time $ t $. The initial height is $ z_0 = 3 $ meters. The z-component of the velocity vector is $ v_z $.

Unit Conversion for Velocity

The velocity is given in km/hr, but the height is in meters. We need to convert the velocity components to meters per second (m/s) for consistency.

Conversion factor: $ 1 \text{ km/hr} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s} $.

Let's convert each component of the velocity vector:

• $ v_x = 10 \text{ km/hr} = 10 \times \frac{5}{18} \text{ m/s} = \frac{50}{18} \text{ m/s} = \frac{25}{9} \text{ m/s} $
• $ v_y = 6 \text{ km/hr} = 6 \times \frac{5}{18} \text{ m/s} = \frac{30}{18} \text{ m/s} = \frac{5}{3} \text{ m/s} $
• $ v_z = 1 \text{ km/hr} = 1 \times \frac{5}{18} \text{ m/s} = \frac{5}{18} \text{ m/s} $

So, the velocity vector in m/s is $ \mathbf{v} = \frac{25}{9}\hat{i} + \frac{5}{3}\hat{j} + \frac{5}{18}\hat{k} $ m/s.

Step-by-Step Calculation

We need to find the time $ t $ when the height $ z(t) $ reaches 13 meters. The height changes according to the z-component of the motion:

$ z(t) = z_0 + v_z t $

Substitute the known values:

• $ z(t) = 13 $ meters (target height)
• $ z_0 = 3 $ meters (initial height)
• $ v_z = \frac{5}{18} $ m/s (z-component of velocity)

The equation becomes:

$ 13 = 3 + \left(\frac{5}{18}\right) t $

Now, solve for $ t $:

1. Subtract 3 from both sides: $ 13 - 3 = \left(\frac{5}{18}\right) t $ $ 10 = \left(\frac{5}{18}\right) t $
2. Multiply both sides by $ \frac{18}{5} $ to isolate $ t $: $ t = 10 \times \frac{18}{5} $
3. Calculate the final value: $ t = \frac{10}{5} \times 18 $ $ t = 2 \times 18 $ $ t = 36 $

The time taken is 36 seconds.

Final Answer Explanation

The bird starts at a height of 3 meters and needs to reach 13 meters, meaning it needs to ascend an additional $ 13 - 3 = 10 $ meters. The vertical component of its velocity ($ v_z $) is $ \frac{5}{18} $ m/s. Using the formula distance = velocity × time for the vertical motion, we get $ 10 \text{ m} = \left(\frac{5}{18} \text{ m/s}\right) \times t $. Solving for $ t $ gives $ t = 36 $ seconds.