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Question on Relative Velocity

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

My Attempt :
Velocity of ship $V_s=30km/hr$
Velocity of boy $V_b=10km/hr$
Now,
Velocity of boy relative to ship $V_bs=V_b - V_s=V_b + (-V_s)$

asked Dec 23, 2017 in Physics Problems by Albert Einstein (160 points)
The velocity of the boy relative to the ship $V_{bs}$ is already given.  You have to find $V_b$, the velocity of the boy relative to the sea. Your equation is correct. Remember that it is a vector equation.
@sammy gerbil,  How do I do that?  Could you please give some hints?
Write the vectors in terms of unit vectors i and j.
@sammy gerbil,  How? Could you please show?
I do not mind giving hints, but I do not want to teach you. Go back to your textbook and do some studying. Do some simple examples using vectors.

1 Answer

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Here is the vector triangle. The unknown velocity is $V_b$

To Find $V_b$ using Components

Suppose $\hat{i}$ is the unit vector in E direction and $\hat{j}$ is unit vector in S direction. Then
$V_s = 30 \hat{i}$
$V_{bs} = -10 \hat{i} \cos 45^{\circ} + 10 \hat{j} \sin 45^{\circ} = \frac{10}{\sqrt{2}}(-\hat{i}+\hat{j})$
So
$V_b=V_s+V_{bs} = (30-\frac{10}{\sqrt2})\hat{i} +\frac{10}{\sqrt2} \hat{j}$

The question asks for the magnitude of $V_b$ which is
$|V_b| = \sqrt{ (30-\frac{10}{\sqrt2})^2 + (\frac{10}{\sqrt2} )^2}$

To Find $V_b$ using the Cosine Rule

$V_b^2 = V_s^2+V_{bs}^2-2V_sV_{bs}\cos45^{\circ}$
$ = 30^2+10^2 - 2\times 30\times 10\times \frac{1}{\sqrt2}$

answered Dec 27, 2017 by sammy gerbil (28,448 points)
edited Dec 29, 2017 by sammy gerbil
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