Question

a man walks on a sraight road from his home to the market2.5km away with the speed 5km/h.he instantly walks back home withspeed of 7.5km/h.the average speed of the man over the interval of time0 to 40 min,is equal to

Answer 1

Answer:

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Answer 2

Question :-

$⟶$A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he immediately turns and walks back home with a speed of 7.5 km/h. The average speed of the man over the interval of time 0 to 40 min is equal to

➭ To find :-

→ average speed in 0 to 40 minutes

➭ Formula required :-

$\orange{\bold{\boxed{average speed = total\: distance / total \:time }}}$

➭ Solution :-

$⟹$Time taken to go to market = $\frac{distance}{speed}$

$→ \frac{2.5}{5}hrs$

$→ \frac{1}{2}hrs$

$→ 30 minutes$

$⟹$Time taken to return back = $\frac{distance}{speed}$

$→ \frac{2.5}{7}$

$→ \frac{1}{3}hrs$

$→ 20 minutes$

Distance travelled in 10 minutes while

$⟹$Returning back = $\frac{10}{60} \times 7.5 = <strong>1.25km</strong>$

( Speed is constant at 7.5 km/hr )

Hence Distance traveled in 40 minutes

(30 mins towards the market and 10 mins towards home)

$⟶$2.5+1.25=3.75km

So average speed in 0 to 40 minutes =

$\frac{3.75km}{ \frac{40}{60}hrs }$

$⟹ \frac{45}{8} km/hr$

Hence, average speed in 0 to 40 minutes is $\frac{45}{8} km/hr$