Published by:
CGP EDU Academic Team
Published on: September 13, 2026
A person travelling eastwards finds the wind to blow from north. On doubling his speed he finds it to come from north-east. Show that if he trebles his speed the wind would appear to him to come form a direction making an angle tan –1 (1/2) north of east.
Text Solution
Verified by ExpertsThe correct answer is:
A
Step 1: Understand the problem setup.
We have a person traveling eastwards (let's assume his speed is $v$) and initially, he perceives the wind blowing from the north. This means that the wind's velocity is directed towards the south from his perspective.
Step 2: Analyze the velocities.
Let the speed of the wind be $W$. The person's velocity vector is directed towards the east, which can be represented as:
$V_p = (v, 0)$
where $v$ is the speed of the person and the $y$-component is zero since he moves purely eastwards. The wind blowing from the north towards the south can be represented as:
$V_w = (0, -W)$
Since he perceives the wind from the north, the actual wind vector must point downward.
Step 3: Determine the wind's apparent direction.
The resultant velocity of the wind relative to the person is:
$V_{rel} = V_w - V_p = (0, -W) - (v, 0) = (-v, -W)$
Since this resultant is negative in $x$ (east) direction and negative in $y$ (south) direction, the wind appears to come directly from the north.
Step 4: Doubling the person's speed.
If the person doubles his speed to $2v$, the apparent velocity of the wind will then become:
$V_{rel} = (0, -W) - (2v, 0) = (-2v, -W)$
Setting up the tangent to find the angle of this resultant yields:
$$ an heta = \frac{|-W|}{|-2v|} = \frac{W}{2v}$$
Since he perceives the wind to come from the north-east, we can conclude that:
$$\tan \theta = 1\Rightarrow W = 2v$$
Step 5: Tripling the person's speed.
If the speed is trebled to $3v$, the apparent velocity now changes to:
$V_{rel} = (0, -W) - (3v, 0) = (-3v, -W)$
To find the new angle $ heta$ with respect to east, we calculate the tangent:
$$\tan \theta = \frac{W}{3v}$$
Now, using $W = 2v$, substitute this value into the tangent equation:
$$\tan \theta = \frac{2v}{3v} = \frac{2}{3}$$
This gives us the direction the wind is coming from at this speed ratio. Therefore, the angle $ heta$ north of east is:
$$\theta = \tan^{-1}\left(\frac{2}{3}\right)$$
Now we convert this into the required form, which is equivalent to stating that this is the angle north of east.
Final Conclusion:
Thus, the wind would appear to come from a direction making an angle of $ an^{-1}(\frac{1}{2})$ north of east. Therefore, the correct answer is:
Therefore, A.
We have a person traveling eastwards (let's assume his speed is $v$) and initially, he perceives the wind blowing from the north. This means that the wind's velocity is directed towards the south from his perspective.
Step 2: Analyze the velocities.
Let the speed of the wind be $W$. The person's velocity vector is directed towards the east, which can be represented as:
$V_p = (v, 0)$
where $v$ is the speed of the person and the $y$-component is zero since he moves purely eastwards. The wind blowing from the north towards the south can be represented as:
$V_w = (0, -W)$
Since he perceives the wind from the north, the actual wind vector must point downward.
Step 3: Determine the wind's apparent direction.
The resultant velocity of the wind relative to the person is:
$V_{rel} = V_w - V_p = (0, -W) - (v, 0) = (-v, -W)$
Since this resultant is negative in $x$ (east) direction and negative in $y$ (south) direction, the wind appears to come directly from the north.
Step 4: Doubling the person's speed.
If the person doubles his speed to $2v$, the apparent velocity of the wind will then become:
$V_{rel} = (0, -W) - (2v, 0) = (-2v, -W)$
Setting up the tangent to find the angle of this resultant yields:
$$ an heta = \frac{|-W|}{|-2v|} = \frac{W}{2v}$$
Since he perceives the wind to come from the north-east, we can conclude that:
$$\tan \theta = 1\Rightarrow W = 2v$$
Step 5: Tripling the person's speed.
If the speed is trebled to $3v$, the apparent velocity now changes to:
$V_{rel} = (0, -W) - (3v, 0) = (-3v, -W)$
To find the new angle $ heta$ with respect to east, we calculate the tangent:
$$\tan \theta = \frac{W}{3v}$$
Now, using $W = 2v$, substitute this value into the tangent equation:
$$\tan \theta = \frac{2v}{3v} = \frac{2}{3}$$
This gives us the direction the wind is coming from at this speed ratio. Therefore, the angle $ heta$ north of east is:
$$\theta = \tan^{-1}\left(\frac{2}{3}\right)$$
Now we convert this into the required form, which is equivalent to stating that this is the angle north of east.
Final Conclusion:
Thus, the wind would appear to come from a direction making an angle of $ an^{-1}(\frac{1}{2})$ north of east. Therefore, the correct answer is:
Therefore, A.
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