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P is a point on the southern bank of a river - Leaving Cert Applied Maths - Question 2 - 2016
Question 2
P is a point on the southern bank of a river. Q is a point directly opposite P on the northern bank. Ship A departs from P at a constant speed of 52 km h⁻¹ and trav... show full transcript
Worked Solution & Example Answer:P is a point on the southern bank of a river - Leaving Cert Applied Maths - Question 2 - 2016
Step 1
(i) the velocity of A in terms of \( \hat{i} \) and \( \hat{j} \)
Answer
To find the velocity of Ship A, we first calculate the angle ( \alpha ) using the given tangent value.
[ \tan(\alpha) = \frac{1}{2} \Rightarrow \alpha = \tan^{-1}(\frac{1}{2}) ]
The components of velocity are given by:
[ \vec{v}_A = 52 \cos(\alpha) \hat{i} + 52 \sin(\alpha) \hat{j} ]
Using trigonometric values:
[ \vec{v}_A \approx 20 \hat{i} + 48 \hat{j} ]
So, the final answer is ( \vec{v}_A = 20 \hat{i} + 48 \hat{j} ) km h⁻¹.
Step 2
(ii) the velocity of B in terms of \( \hat{i} \) and \( \hat{j} \)
Answer
For Ship B, the angle ( \beta ) is determined from:
[ \tan(\beta) = \frac{1}{3} \Rightarrow \beta = \tan^{-1}(\frac{1}{3}) ]
The velocity is given by:
[ \vec{v}_B = 51 \cos(\beta) \hat{i} - 51 \sin(\beta) \hat{j} ]
Substituting the values:
[ \vec{v}_B \approx 45 \hat{i} - 24 \hat{j} ]
Thus, the final result is ( \vec{v}_B = 45 \hat{i} - 24 \hat{j} ) km h⁻¹.
Step 3
(iii) the velocity of A relative to B in terms of \( \hat{i} \) and \( \hat{j} \)
Answer
To find the relative velocity of A with respect to B, we calculate:
[ \vec{v}_{AB} = \vec{v}_A - \vec{v}_B ]
Substituting the velocities:
[ \vec{v}_{AB} = (20 \hat{i} + 48 \hat{j}) - (45 \hat{i} - 24 \hat{j}) ]
After performing the subtraction:
[ \vec{v}_{AB} = -25 \hat{i} + 72 \hat{j} ]
Hence, ( \vec{v}_{AB} = -25 \hat{i} + 72 \hat{j} ) km h⁻¹.
Step 4
(iv) the time it takes B to reach R and how much longer it takes A to reach R
Answer
The distance to point R is 9 km downstream.
For Ship B:
[ t_B = \frac{9}{45} = 0.2 \text{ h} ]
For Ship A:
[ t_A = \frac{9}{\sqrt{20^2 + 48^2}} = \frac{9}{\sqrt{400 + 2304}} = 0.45 \text{ h} ]
The difference in time is:
[ t_A - t_B = 0.45 - 0.2 = 0.25 \text{ h} ]
Step 5
(v) the width of the river, assuming its banks are parallel
Answer
To find the width of the river, we calculate the distance covered by Ship A in the j-direction:
[ d = 24 \times 0.2 + 48 \times 0.45 ]
After calculating:
[ d \approx 26.4 \text{ km} ]
Thus, the width of the river is approximately 26.4 km.