The Question is from neet-physics motion-in-a-straight-line relative-velocity, and it was created by the author Mr. ANIL KUMAR.G

Question

Two cars A and B are moving with velocities $\vec{v_A} = 40\hat{i}$ km/h and $\vec{v_B} = -30\hat{j}$ km/h relative to the ground. A third car C is moving such that its velocity relative to car A is $\vec{v_{CA}} = 30\hat{j}$ km/h and relative to car B is $\vec{v_{CB}} = 40\hat{i}$ km/h. What is the speed of car C relative to the ground?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [Option C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Relative Velocity
	**Related Topics**: Vector Addition, Kinematics
	**Prerequisites**: Understanding of relative velocity, vector addition, and speed calculation.
## SOLUTION PATHWAY
**Given Information**:
	*  $\vec{v_A} = 40\hat{i}$ km/h
	*  $\vec{v_B} = -30\hat{j}$ km/h
	*  $\vec{v_{CA}} = 30\hat{j}$ km/h
	*  $\vec{v_{CB}} = 40\hat{i}$ km/h
**Method & Steps**:
	1.  Express the velocity of car C relative to the ground using the relative velocity formula for car A.
		Reasoning: The velocity of car C relative to car A is given by $\vec{v_{CA}} = \vec{v_C} - \vec{v_A}$.  We can rearrange this to find $\vec{v_C}$.
		Formula/Rule:  $\vec{v_{CA}} = \vec{v_C} - \vec{v_A}$  => $\vec{v_C} = \vec{v_{CA}} + \vec{v_A}$
	2.  Calculate $\vec{v_C}$ using the given values.
		Working: $\vec{v_C} = 30\hat{j} + 40\hat{i}$ km/h
		Key Point: This gives us one expression for $\vec{v_C}$
	3.  Express the velocity of car C relative to the ground using the relative velocity formula for car B.
		Reasoning: The velocity of car C relative to car B is given by $\vec{v_{CB}} = \vec{v_C} - \vec{v_B}$. We can rearrange this to find $\vec{v_C}$.
		Formula/Rule: $\vec{v_{CB}} = \vec{v_C} - \vec{v_B}$  => $\vec{v_C} = \vec{v_{CB}} + \vec{v_B}$
    4. Calculate $\vec{v_C}$ using the given values.
       Working: $\vec{v_C} = 40\hat{i} + (-30\hat{j})$ km/h
       Key Point: This gives us another expression for $\vec{v_C}$, and both should match.
	5.  Calculate the speed of car C relative to the ground by finding the magnitude of the velocity vector.
		Working:  $\vec{v_C} = 40\hat{i} + 30\hat{j}$. Speed =  $|\vec{v_C}| = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50$ km/h
		Result:  Speed of car C is 50 km/h
		Verification: Both expressions for $\vec{v_C}$ are identical.
 ## OPTION ANALYSIS
[A] :x: Incorrect. This answer seems to add the magnitudes of the given velocities without considering vector addition.
[B] :x: Incorrect. This answer is incorrect, it does not align with the calculated speed.
[C] ✓: Correct. The calculated speed of car C relative to the ground is 50 km/h.
[D] :x: Incorrect. The car is moving, so its speed cannot be 0.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Adding magnitudes of velocities directly instead of using vector addition. [How to avoid]: Always treat velocities as vectors, considering both magnitude and direction.
		2.[Error 2]: Incorrectly applying the relative velocity formula. [How to avoid]: Ensure proper arrangement of the formula and the correct use of signs.
	**Quick Tips**:
		1.[Helpful shortcut]:  When dealing with relative velocities, remember that $\vec{v_{AB}} = \vec{v_A} - \vec{v_B}$ which means the velocity of A as observed from B is the velocity of A minus the velocity of B, all relative to the same frame of reference.
		2.[Memory aid]: Draw a vector diagram to visualize the velocities and their directions. This will help in understanding the vector addition process.

Mr. ANIL KUMAR.G · Answer

70 km/h

Mr. ANIL KUMAR.G · Answer

10 km/h

Mr. ANIL KUMAR.G · Answer

The correct Option is "50 km/h"

Mr. ANIL KUMAR.G · Answer

0 km/h

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