Relative to a landing stage (the origin), the position vectors in kilometres of two boats A and B at noon are 3i + j and i-2j  respectively. The velocities of A and B, which are constant and in kilometres per hour are 10i + 24j  and 24i +32j 
 

(a)What is the distance between boats A and B after t hours?

To find the distance between the two boats \( A \) and \( B \) after \( t \) hours, we first need to determine their position vectors at any time \( t \). The position vectors of the boats at noon (when \( t = 0 \)) and their velocity vectors are given as follows:

- Initial position vector of boat \( A \): \(\mathbf{r}_A(0) = 3\mathbf{i} + \mathbf{j}\)
- Velocity vector of boat \( A \): \(\mathbf{v}_A = 10\mathbf{i} + 24\mathbf{j}\)

- Initial position vector of boat \( B \): \(\mathbf{r}_B(0) = \mathbf{i} - 2\mathbf{j}\)
- Velocity vector of boat \( B \): \(\mathbf{v}_B = 24\mathbf{i} + 32\mathbf{j}\)

### 1. Position vectors after \( t \) hours:
The position vector of each boat after \( t \) hours is given by the initial position vector plus the velocity vector multiplied by \( t \):

\[
\mathbf{r}_A(t) = \mathbf{r}_A(0) + \mathbf{v}_A \cdot t = (3\mathbf{i} + \mathbf{j}) + (10\mathbf{i} + 24\mathbf{j})t = (3 + 10t)\mathbf{i} + (1 + 24t)\mathbf{j}
\]

\[
\mathbf{r}_B(t) = \mathbf{r}_B(0) + \mathbf{v}_B \cdot t = (\mathbf{i} - 2\mathbf{j}) + (24\mathbf{i} + 32\mathbf{j})t = (1 + 24t)\mathbf{i} + (-2 + 32t)\mathbf{j}
\]

### 2. Distance between the boats after \( t \) hours:
The distance between the two boats is the magnitude of the difference between their position vectors:

\[
\text{Distance} = |\mathbf{r}_A(t) - \mathbf{r}_B(t)|
\]

First, find the difference in their position vectors:

\[
\mathbf{r}_A(t) - \mathbf{r}_B(t) = \left[(3 + 10t)\mathbf{i} + (1 + 24t)\mathbf{j}\right] - \left[(1 + 24t)\mathbf{i} + (-2 + 32t)\mathbf{j}\right]
\]

Simplifying this:

\[
\mathbf{r}_A(t) - \mathbf{r}_B(t) = [(3 + 10t) - (1 + 24t)]\mathbf{i} + [(1 + 24t) - (-2 + 32t)]\mathbf{j}
\]

\[
= (2 - 14t)\mathbf{i} + (3 - 8t)\mathbf{j}
\]

Now, calculate the magnitude of this vector:

\[
\text{Distance} = \sqrt{(2 - 14t)^2 + (3 - 8t)^2}
\]

Expanding the squares:

\[
\text{Distance} = \sqrt{(2 - 14t)^2 + (3 - 8t)^2}
= \sqrt{(4 - 56t + 196t^2) + (9 - 48t + 64t^2)}
\]

\[
= \sqrt{(196t^2 - 56t + 4) + (64t^2 - 48t + 9)}
\]

\[
= \sqrt{260t^2 - 104t + 13}
\]

So, the distance between boats \( A \) and \( B \) after \( t \) hours is:

\[
\text{Distance} = \sqrt{260t^2 - 104t + 13} \text{ kilometres}
\]

Would you like any further details or have any questions?

### Follow-up Questions:

1. How would the distance change if the velocities of the boats were doubled?
2. What would be the distance between the boats at \( t = 2 \) hours?
3. How can you find the time at which the boats are closest to each other?
4. What if the velocity of boat B was in the opposite direction?
5. How would the equation change if the boats started at different positions?

### Tip:

When solving vector problems, always remember to carefully handle each component separately before combining them.

Relative to a landing stage (the origin), the position vectors in kilometres of two boats A and B at noon are 3i + j and i-2j respectively. The velocities of A and B, which are constant and in kilometres per hour are 10i + 24j and 24i +32j. (a) What is the distance between boats A and B after t hours?

Find out how to calculate the distance between two boats A and B using position vectors and velocities. This problem involves vector addition and the magnitude formula, ideal for students in Grades 10-12. A step-by-step guide explains how to derive the final distance formula: sqrt(260t^2 - 104t + 13).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation