How Do You Calculate Average Speed with Variable Speeds and Distances?

AI Thread Summary
To calculate average speed for a journey with variable speeds, the average speed Va for a journey where the first half of the time is spent at speed V1 and the second half at speed V2 can be expressed as Va = 2V1V2/(V1 + V2). For a journey where half the distance is traveled at speed V1 and the other half at speed V2, the average speed Vb is given by Vb = 2V1V2/(V1 + V2), similar to the time-based calculation. It is also established that traveling equal distances at two different speeds will take longer than traveling for equal times at those speeds. The discussions emphasize the importance of understanding the relationship between distance, speed, and time in calculating average speeds. This knowledge is crucial for solving related problems in physics and mathematics.
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Homework Statement



a) A journey is completed by travailing for the first half of the time at speed V1 and the second half at speed V2. Find the average speed Va for the journey in terms of V1 and V2 .

b) A journey is completed by traveling at speed V1 for half the distance and at speed V2 for the second half. Find the average speed Vb for the journey in terms of V1 and V2.

c) Deduce that a journey completed by traveling at two different speeds for equal distance will take longer than the same journey completed at the same two speeds for equal times.


Homework Equations



Va = d/t, where d = distance covered , t = time taken.



The Attempt at a Solution



a) d1 = distance covered by t1 and d2 distance covered by t2,
therefore 2d = V1t1+V2t2

I have no clue how to convert this in terms of V1 and V2

b) t1 = d/2v1 and t2 = d/2v2,

Va = d/t1 + d/t2

This is as far as I could go.
 
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On a), you've assumed that d_1=d_2. Since d_1=\frac{v_1t}{2} and d_2=\frac{v_2t}{2}, take note of the fact that d_1+d_2=d_{total} and v_{avg}=\frac{d_{total}}{t_{total}}, where t_{total}=t.

Try a similar approach on b), noting that t_1+t_2=t_{total}.
 
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Thank you so much, I solved them.
 
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