Ilya and also Anya each deserve to run in ~ a speed of 8.30mph and walk in ~ a speed of 3.90mph . They set off together on a path of length 5.00miles . Anya walks fifty percent of the distance and runs the other half, when Ilya walks half of the time and also runs the various other half.

You are watching: How long does it take anya to cover the distance of 5.00 miles ?

How lengthy does it take Anya come cover the street of 5.00miles ?

Express her answer numerically, in minutes.

Find Anya"s average speed.

Express Anya"s mean speed save,Anya numerically, in miles per hour.

How lengthy does it take Ilya to cover the distance?

Express the time tIlya taken by Ilya numerically, in minutes.

Now discover Ilya"s average speed.

Express Ilya"s typical speed save,Ilya numerically, in miles per hour.



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added Tue, 29 Mar "16

Treebeard
31978
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Given , running speed(RS) = 8.3 mph

Walking speed (WS) = 3.9 mph

Total distance = 5 miles

Time take away by Anya come cover the distance = (2.5/3.9) +(2.5/8.3) hours

= < (2.5/3.9) + (2.5/8.3) > * 60 mins

= 56.53 minutes

Anya"s mean speed = (5/56.53)*60 mph = 5.30 mph

Time bring away by Ilya come cover the street =

let time take away by Ilya to cover the distance be t hours.

then Ilya walks for t/2 hours and runs because that t/2 hours

so , RS*t/2 + WS*t/2 = 5 miles

8.3*t/2 + 3.9*t/2 = 5

or t/2*(8.3+3.9) = 5

or t = 0.81 hours

or t= 0.81*60 mins = 49.1 mins

Ilyas average speed = 5/0.81 mph = 6.17 mph



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included Tue, 29 Mar "16

Treebeard
31978
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anya runs half of the track

t = d / s

t = 2.5 (Half the total distance) / 8.5 (speed of running)

This is .294 hrs which is around 1058s...

And because that the walking part...

t = d / s

t = 2.5 / 3.9

t = 2307s...

So girlfriend then add the two..

2307 + 1058 = 3365s = 56.09 minutes.

Therefore to work out the median speed, the is simply distance / time...

See more: How Are The Pacific’S Low Islands Distinct From The High Islands?

(5 / 56.09)x60 = 5.3 mph

for llya

let it takes t time for she to sheathe the track

so because that t/2 time she runs and for various other t/2 time she walks.

for run

d1 = 8.3 x t/2

for walk

d2 = 3.9 x t/2

so d1+d2 = d = 5 miles

s0 6.1 x t = 5

t = 5/6.1 = 0.819 hrs = 49.18 minutes (llya"s time)

for llya"s typical speed

5/0.819 = 6.10 mph


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added Tue, 29 Mar "16
Treebeard
31978
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