Question

Please answer it guys ​

Answer 1

Explanation:Let the one side distance travelled be "s"

speed = u

Time taken for side1 (t1) = s/u

[speed = distance/time ; Time = distance/speed]

Similarly, time taken for returning (t2) =s/v

Average speed = Total distance/Total time

Speed(avg.) = (s + s)/(s/u + s/v)

Speed (avg.) = 2s/[(sv + su)/uv]

Speed(avg.) = 2s/[s(u + v)/uv]

Speed(avg.) = 2uv/(u + v)PROVED

Answer 2

Given:

• A body covers a certain distance with speed 'u' and returns to the same point with speed 'v'. Then prove that the average speed is 2uv/(u + v)  

Proof:  

• Let the certain distance be covered in one side be d.  We know that,

♦ Time = Distance/Speed  Speed = u  Time = d/u  Let the distance covered while returning be d ( as he returns to the same point )  Speed = v  Time = d/v  Finally, using formula.

♦ Average Speed = Total distance covered/Total time taken  Putting the values,

• Average Speed = (d + d)/(d/u + d/v)
• Average Speed = 2d/((dv + du)/uv)
• Average Speed = 2d/(d(v + u)/uv)
• Average Speed = 2/((v + u)/uv)
• Average Speed = 2 × uv/(v + u)
• Average Speed = 2uv/(v + u)  Therefore,
• The average speed = 2uv/(u + v)

Hence Proved!!  ━━━━━━━━━━━━━━━━━━━━━