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A car moves from X to Y with a uniform s...

A car moves from `X` to `Y` with a uniform speed `v_u` and returns to `X` with a uniform speed `v_d`. The average speed for this round trip is :
(a) `(2v_(d)v_(u))/(v_(d)+v_(u))` (b) `sqrt(v_(u)u_(d))` (c) `(v_(d)v_(u))/(v_(d)+v_(u))` (d) `(v_(u)+v_(d))/(2)`

A

`(2v_(d)v_(u))/(v_(d)+v_(u))`

B

`sqrt(v_(u)u_(d))`

C

`(v_(d)v_(u))/(v_(d)+v_(u))`

D

`(v_(u)+v_(d))/(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the average speed for the round trip of the car moving from point X to Y and back to X, we can follow these steps: ### Step 1: Define the Variables Let: - \( d \) = distance from X to Y - \( v_u \) = speed from X to Y (upward speed) - \( v_d \) = speed from Y to X (downward speed) ### Step 2: Calculate the Time Taken for Each Leg of the Trip 1. **Time taken to go from X to Y**: \[ t_1 = \frac{d}{v_u} \] 2. **Time taken to return from Y to X**: \[ t_2 = \frac{d}{v_d} \] ### Step 3: Calculate the Total Time for the Round Trip The total time \( T \) for the round trip is the sum of the time taken for both legs: \[ T = t_1 + t_2 = \frac{d}{v_u} + \frac{d}{v_d} \] ### Step 4: Simplify the Total Time Expression Factor out \( d \): \[ T = d \left( \frac{1}{v_u} + \frac{1}{v_d} \right) = d \left( \frac{v_d + v_u}{v_u v_d} \right) \] ### Step 5: Calculate the Total Distance Covered The total distance \( D \) for the round trip is: \[ D = d + d = 2d \] ### Step 6: Calculate the Average Speed Average speed \( V_{avg} \) is defined as total distance divided by total time: \[ V_{avg} = \frac{D}{T} = \frac{2d}{d \left( \frac{v_d + v_u}{v_u v_d} \right)} \] ### Step 7: Simplify the Average Speed Expression Cancel \( d \) in the numerator and denominator: \[ V_{avg} = \frac{2}{\frac{v_d + v_u}{v_u v_d}} = \frac{2 v_u v_d}{v_d + v_u} \] ### Final Answer Thus, the average speed for the round trip is: \[ \boxed{\frac{2 v_u v_d}{v_d + v_u}} \]
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