Question

a person moves such a way that he travels first half time in speed 36 km per hour and in remaining half time he covers first half distance with speed 20 km per hour and remaining half distance with speed 30 kilometre per hour find average speed​​

Answer 1

average speed in 2nd half time=2v1v2/v1+V2

so average speed is=2×20×30/50=24km/h

average speed when times are eqaul is =v1+V2/2

so, average speed =24+36/2=30km/h

Answer 2

Answer:

Average Speed is 30 km/h .

Explanation:

Let the person have total time = t for his journey.

For first half time:

Speed (v1) = 36 km/hour

So distance traveled in first half time ;

$s1 = v1 \times (\frac{t}{2}) \: km$

So the distance is:

$s1 = v1 \times (\frac{t}{2}) \: km \\ = 36 \times (\frac{t}{2})\: km$

For remaining half time:

Let total distance traveled in second half time is

= s2 km

So according to the question;

Half of s2 is traveled at a speed (v2) = 20 km/h

time taken = (t')1

$(t')1 = \frac{(\frac{s2}{2} )}{20} \\ = \frac{s2}{40}$

And other half distance is traveled at a speed (v3) = 30 km/h

time taken = (t')2

$(t')2 = \frac{(\frac{s2}{2} )}{30} \\ = \frac{s2}{60}$

So average speed in this remaining half time;

$(v2)avg = \frac{total \: distance}{total \: time} \\ = \frac{s2}{(t')1 + (t')2 } \\ = \frac{s2}{ \frac{s2}{40} + \frac{s2}{60} } \\ = \frac{1}{ \frac{1}{40} + \frac{1}{60} } \: km \:. {h}^{ - 1} \\ = 24 \: km \:. {h}^{ - 1}$

We know (t')1 + (t')2 = t/2

Total distance traveled in this remaining half time;

$s = (v)avg \times \frac{t}{2} \: km \\ = 24 \times ( \frac{t}{2} ) \: km$

So now we can calculate the average speed (v)avg of total time t;

$(v)avg = \frac{s1 + s}{t} \\ = \frac{(36 \times \frac{t}{2}) + (24 \times \frac{t}{2})}{t} \: km. {h}^{ - 1} \\ = \frac{36 + 24}{2} \: km. {h}^{ - 1} \\ = 30 \: km. {h}^{ - 1}$

Hence Average Speed is 30 km/h (Answer).