Photo AI

A ship P is moving north at a constant speed of 20 km/h - Leaving Cert Applied Maths - Question 2 - 2009

Question icon

Question 2

A-ship-P-is-moving-north-at-a-constant-speed-of-20-km/h-Leaving Cert Applied Maths-Question 2-2009.png

A ship P is moving north at a constant speed of 20 km/h. Another ship Q is moving south-west at a constant speed of $10 \sqrt{2}$ km/h. At a certain instant, P is ... show full transcript

Worked Solution & Example Answer:A ship P is moving north at a constant speed of 20 km/h - Leaving Cert Applied Maths - Question 2 - 2009

Step 1

the velocity of P in terms of $\hat{i}$ and $\hat{j}$

96%

114 rated

Answer

vP=0i^+20j^\mathbf{v}_P = 0 \hat{i} + 20 \hat{j}

Step 2

the velocity of Q in terms of $\hat{i}$ and $\hat{j}$

99%

104 rated

Answer

vQ=−10/2cos⁡45∘i^−10/2sin⁡45∘j^=−10i^−10j^\mathbf{v}_Q = -10/\sqrt{2} \cos 45^\circ \hat{i} - 10/\sqrt{2} \sin 45^\circ \hat{j} = -10 \hat{i} - 10 \hat{j}

Step 3

the velocity of P relative to Q in terms of $\hat{i}$ and $\hat{j}$

96%

101 rated

Answer

vPQ=vP−vQ={0i^+20j^}−{−10i^−10j^}=10i^+30j^\mathbf{v}_{PQ} = \mathbf{v}_P - \mathbf{v}_Q = \{0 \hat{i} + 20 \hat{j}\} - \{-10 \hat{i} - 10 \hat{j}\} = 10 \hat{i} + 30 \hat{j}

Step 4

the shortest distance between P and Q in the subsequent motion

98%

120 rated

Answer

Using the triangle formed, let tan⁡(α)=3010\tan(\alpha) = \frac{30}{10}, thus α=tan⁡−1(3)\alpha = \tan^{-1}(3).

The shortest distance is computed as: shortest distance=50sin⁡(tan⁡−1(3))=50sin⁡(71.5651)≈47.43 km\text{shortest distance} = 50 \sin(\tan^{-1}(3)) = 50 \sin(71.5651) \approx 47.43\text{ km}

Join the Leaving Cert students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;