Question

A ship is travelling due east at 30 km/hr and a boy runs across a deck in a south west direction at 10 km/hr. find the velocity of the boy relative to sea

Answer 1

Answer on Question #42198 - Physics - Mechanics | Kinematics | Dynamics

A ship is travelling due east at $30\mathrm{km / hr}$ and a boy runs across a deck in a south west direction at $10\mathrm{km / hr}$ . find the velocity of the boy relative to sea

Solution:

$\mathrm{V_s} = 30\frac{\mathrm{km}}{\mathrm{hr}} -$ velocity of the ship;

$\mathrm{V_b} = 10\frac{\mathrm{km}}{\mathrm{hr}} -$ velocity of the boy;

Relative velocity of the boy with respect to the sea is the difference between vectors of the boy's and the ship's velocities:

$$\overline {{V}} _ {\mathrm {b}, \mathrm {s}} = \overline {{V}} _ {\mathrm {b}} - \overline {{V}} _ {\mathrm {s}} = \overline {{V}} _ {\mathrm {s}} + (- \overline {{V}} _ {\mathrm {s}})$$

Triangle ABC:

$$\alpha = 1 8 0 {}^ {\circ} - 4 5 {}^ {\circ} = 1 3 5 {}^ {\circ}$$

We can use law of cosines to find the unknown velocity $V_{b,c}$ :

$$\begin{array}{l} V _ {b, c} ^ {2} = V _ {b} ^ {2} + V _ {s} ^ {2} - 2 V _ {b} V _ {c} \cos \alpha \\ V _ {b, c} = \sqrt {V _ {b} ^ {2} + V _ {s} ^ {2} - 2 V _ {b} V _ {c} \cos \alpha} = \\ = \sqrt {\left(3 0 \frac {\mathrm {k m}}{\mathrm {h r}}\right) ^ {2} + \left(1 0 \frac {\mathrm {k m}}{\mathrm {h r}}\right) ^ {2} - 2 \cdot 3 0 \frac {\mathrm {k m}}{\mathrm {h r}} \cdot 1 0 \frac {\mathrm {k m}}{\mathrm {h r}} \cos 1 3 5 {}^ {\circ}} = 3 7. 7 \frac {\mathrm {k m}}{\mathrm {h r}} \\ \end{array}$$

Answer: relative velocity of the boy with respect to the sea is 37.7 $\frac{\mathrm{km}}{\mathrm{hr}}$

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