Two particles, A and B, start initially from points with position vectors $6oldsymbol{i} - 14oldsymbol{j}$ and $3oldsymbol{i} - 2oldsymbol{j}$ respectively. The velocities of A and B are constant and equal to $4oldsymbol{i} - 3oldsymbol{j}$ and $5oldsymbol{i} - 7oldsymbol{j}$ respectively. (i) Find the velocity of B relative to A. (ii) Show that the particles collide. When a motor-cyclist travels along a straight road from South to North at a constant speed of 12.5 m s$^{-1}$ the wind appears to her to come from a direction North 45$^{ ext{o}}$ East. When she returns along the same road at the same constant speed, the wind appears to come from a direction South 45$^{ ext{o}}$ East. Find the magnitude and direction of the velocity of the wind.

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Two particles, A and B, start initially from points with position vectors $6oldsymbol{i} - 14oldsymbol{j}$ and $3oldsymbol{i} - 2oldsymbol{j}$ respectively - Leaving Cert Applied Maths - Question 2 - 2010

Question 2

Two particles, A and B, start initially from points with position vectors $6oldsymbol{i} - 14oldsymbol{j}$ and $3oldsymbol{i} - 2oldsymbol{j}$ respectively. The ... show full transcript

Worked Solution & Example Answer:Two particles, A and B, start initially from points with position vectors $6oldsymbol{i} - 14oldsymbol{j}$ and $3oldsymbol{i} - 2oldsymbol{j}$ respectively - Leaving Cert Applied Maths - Question 2 - 2010

Step 1

(i) Find the velocity of B relative to A.

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Answer

To find the velocity of B relative to A, we can use the formula:

$oldsymbol{v_{BA}} = oldsymbol{v_B} - oldsymbol{v_A}$

Substituting the velocities,

\boldsymbol{v_B} = 5oldsymbol{i} - 7oldsymbol{j}$$ Calculating: $$oldsymbol{v_{BA}} = (5oldsymbol{i} - 7oldsymbol{j}) - (4oldsymbol{i} - 3oldsymbol{j}) = (5 - 4)oldsymbol{i} + (-7 + 3)oldsymbol{j}$$ Thus, $$oldsymbol{v_{BA}} = 1oldsymbol{i} - 4oldsymbol{j}$$ The magnitude is given by: $$|oldsymbol{v_{BA}}| = \ \sqrt{(1)^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \, \text{m s}^{-1}$$ The direction can be found using the tangent: $$\tan\theta = \frac{-4}{1} \ \theta = \tan^{-1}(-4)$$ This gives a direction of approximately $\text{East } 75.58^{\circ} \text{ South}$.

Step 2

(ii) Show that the particles collide.

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Answer

To show that the particles collide, we need to find their position vectors and set them equal to each other:

The position vector of A at time $t$ :

$\boldsymbol{r_A} = \boldsymbol{6i - 14j} + (4\boldsymbol{i} - 3\boldsymbol{j})t = (6 + 4t)\boldsymbol{i} + (-14 - 3t)\boldsymbol{j}$

The position vector of B at time $t$ :

$\boldsymbol{r_B} = \boldsymbol{3i - 2j} + (5\boldsymbol{i} - 7\boldsymbol{j})t = (3 + 5t)\boldsymbol{i} + (-2 - 7t)\boldsymbol{j}$

Setting these equal to find the time of collision:

For the $\boldsymbol{i}$ components:

4t - 5t = 3 - 6 \ t = -3$$ 2. For the $\boldsymbol{j}$ components: $$-14 - 3t = -2 - 7t \ -3t + 7t = -2 + 14 \ 4t = 12 \ t = 3$$ Since $t = -3$ results in a negative time and $t = 3$ is positive, particles A and B collide at point for which both equations hold true, thus showing that the particles collide.

Step 3

Find the magnitude and direction of the velocity of the wind.

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Answer

Let the velocity of the motor-cyclist be represented as:

$\boldsymbol{v_c} = 0\boldsymbol{i} + 12.5\boldsymbol{j}$

The wind velocity in the reference frame of the cyclist, when coming from North 45 $^{ ext{o}}$ East, can be represented as:

$\boldsymbol{v_w} = v_w(\cos(45^{\circ})\boldsymbol{i} + \sin(45^{\circ})\boldsymbol{j})$

When the motor-cyclist returns, the wind appears to come from the direction South 45 $^{\text{o}}$ East, thus:

$\boldsymbol{v_C} = \boldsymbol{0i} - \boldsymbol{y}\boldsymbol{j}$

Since both vectors represent the same speed when she returns:

$\boldsymbol{v_{C}} + \boldsymbol{v_{w}} = \boldsymbol{0}$

Equating the two equations gives:

$\boldsymbol{0.} = - \boldsymbol{x}i - \boldsymbol{y}j + (v_w(\cos 45\boldsymbol{i} + \sin 45 \boldsymbol{j}))$

Letting the equations represent the components we can continue to solve:

Both velocities equate as:

-\boldsymbol{V_w} = 12.5\boldsymbol{j}\right$$ The final computation yields a wind speed of magnitude: $$|\boldsymbol{V_w}| = 12.5 \text{ m s}^{-1}$$ And the direction is thus: $$\text{West}.$$

Two particles, A and B, start initially from points with position vectors $6oldsymbol{i} - 14oldsymbol{j}$ and $3oldsymbol{i} - 2oldsymbol{j}$ respectively - Leaving Cert Applied Maths - Question 2 - 2010

Worked Solution & Example Answer:Two particles, A and B, start initially from points with position vectors $6oldsymbol{i} - 14oldsymbol{j}$ and $3oldsymbol{i} - 2oldsymbol{j}$ respectively - Leaving Cert Applied Maths - Question 2 - 2010

(i) Find the velocity of B relative to A.

(ii) Show that the particles collide.

Find the magnitude and direction of the velocity of the wind.

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Two particles, A and B, start initially from points with position vectors $6oldsymbol{i} - 14oldsymbol{j}$ and $3oldsymbol{i} - 2oldsymbol{j}$ respectively - Leaving Cert Applied Maths - Question 2 - 2010

Worked Solution & Example Answer:Two particles, A and B, start initially from points with position vectors $6oldsymbol{i} - 14oldsymbol{j}$ and $3oldsymbol{i} - 2oldsymbol{j}$ respectively - Leaving Cert Applied Maths - Question 2 - 2010