Question:
Two trains A and B are moving in the same direction with velocities \(V_A\) and \(V_B\) respectively and a third train C is moving in opposite direction. If velocity of C with respect to B is twice the velocity of A with respect to B, then velocity of A with respect to C is:
Two trains A and B are moving in the same direction with velocities \(V_A\) and \(V_B\) respectively and a third train C is moving in opposite direction. If velocity of C with respect to B is twice the velocity of A with respect to B, then velocity of A with respect to C is:
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Always choose one direction positive and assign signs before applying relative velocity formulas.
Updated On: Jun 18, 2026
- \(3(V_A-V_B)\)
- \(2(V_A-V_B)\)
- \((V_A-V_B)/3\)
- \((V_A-V_B)/2\)
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The Correct Option is A
Solution and Explanation
Concept:
Relative velocity is defined as:
\[
V_{PQ}=V_P-V_Q.
\]
Carefully assign signs according to directions.
Step 1: Write the given condition.
Velocity of A relative to B: \[ V_{AB}=V_A-V_B. \] Since C moves opposite to A and B, \[ V_C=-V_C. \] Velocity of C relative to B: \[ V_{CB}=V_C-V_B. \] Magnitude given: \[ |V_{CB}|=2|V_{AB}|. \]
Step 2: Express using directions.
Since C moves opposite, \[ V_{CB}=-(V_C+V_B). \] Hence \[ V_C+V_B=2(V_A-V_B). \] \[ V_C=2V_A-3V_B. \]
Step 3: Find velocity of A relative to C.
\[ V_{AC}=V_A-(-V_C). \] \[ =V_A+V_C. \] Substituting, \[ V_{AC} = V_A+2V_A-3V_B. \] \[ = 3(V_A-V_B). \] \[ \boxed{V_{AC}=3(V_A-V_B)}. \]
Step 1: Write the given condition.
Velocity of A relative to B: \[ V_{AB}=V_A-V_B. \] Since C moves opposite to A and B, \[ V_C=-V_C. \] Velocity of C relative to B: \[ V_{CB}=V_C-V_B. \] Magnitude given: \[ |V_{CB}|=2|V_{AB}|. \]
Step 2: Express using directions.
Since C moves opposite, \[ V_{CB}=-(V_C+V_B). \] Hence \[ V_C+V_B=2(V_A-V_B). \] \[ V_C=2V_A-3V_B. \]
Step 3: Find velocity of A relative to C.
\[ V_{AC}=V_A-(-V_C). \] \[ =V_A+V_C. \] Substituting, \[ V_{AC} = V_A+2V_A-3V_B. \] \[ = 3(V_A-V_B). \] \[ \boxed{V_{AC}=3(V_A-V_B)}. \]
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