Ilya and Anya each have the right to run at a rate of 8.30mph and also walk in ~ a rate of 3.90mph . They set off with each other on a path of size 5.00miles . Anya walks half of the distance and runs the other half, when Ilya walks fifty percent of the time and runs the various other half.

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

How long does it take it Anya to cover the distance of 5.00miles ?

Find Anya"s median speed.

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

How lengthy does it take it Ilya to cover the distance?

Express the time tIlya bring away by Ilya numerically, in minutes.

Now find Ilya"s median speed.

Express Ilya"s average speed save,Ilya numerically, in miles every hour.  edit Treebeard
31978
Comment  solutions
2

Given , running speed(RS) = 8.3 mph

Walking speed (WS) = 3.9 mph

Total street = 5 miles

Time taken by Anya to 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 typical 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 street be t hours.

then Ilya walks for t/2 hours and also runs for 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 mean speed = 5/0.81 mph = 6.17 mph  edit Treebeard
31978
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include a Comment
1

anya runs half of the track

t = d / s

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

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

And because that the go part...

t = d / s

t = 2.5 / 3.9

t = 2307s...

So you then include the two..

2307 + 1058 = 3365s = 56.09 minutes.

Therefore to occupational out the average speed, the is just distance / time...

See more: How Many Days Are In 72 Hours ? How Long Is 72 Hours In Days

(5 / 56.09)x60 = 5.3 mph

for llya

let the takes t time for she to covering the track

so because that t/2 time she runs and for 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 minute (llya"s time)

for llya"s average speed

5/0.819 = 6.10 mph

modify
Treebeard
31978
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