Question
Asked Sep 13, 2019
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Jesse takes a 3-day kayak trip and travels 27 km south the city to a camp area. The trip to the camp with a 3 km/hour current takes 6 hours less time than the return against the current.  Find the speed that Jesse travels in still water. 

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Expert Answer

Step 1

Let, speed of Jesse in still water is v km/h.

Given that speed of current = 3 km/h

While going to the camp, Jesse travels with the current. So, her speed = (v+3) km/h

And, while returning from the camp, Jesse travels against the current. So, her speed = (v-3) km/h.

Step 2

Let, the time taken by Jesse while going to the camp = t hours

According to the question,

The time taken by Jesse while returning from the camp = (t+6) hours.

We know that, for Jesse,
Distance travelled while going to camp
Distance travelled while returning from camp = 27 km
=
(v 3) t6)
27
(v3)t
=>
By using the first two conditions, we get
(v3)t
(v 3)(t6)
=
vt - 3t + 6v - 18
expanding the brackets]
vt +3t
3t 6v 18[subtracting vt from both sides]
3t
6t
бу — 18
[adding 3t to both sides]
=
[dividing both sides by 6]
v - 3
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Image Transcriptionclose

We know that, for Jesse, Distance travelled while going to camp Distance travelled while returning from camp = 27 km = (v 3) t6) 27 (v3)t => By using the first two conditions, we get (v3)t (v 3)(t6) = vt - 3t + 6v - 18 expanding the brackets] vt +3t 3t 6v 18[subtracting vt from both sides] 3t 6t бу — 18 [adding 3t to both sides] = [dividing both sides by 6] v - 3

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Step 3

Now we proceed fu...

By using the first and third conditions, we get
(v3)t
27
Using t v3 as obtained in Step - 2
(v3)v3)
27
v2 32 =27
a2 - b2
[as (ab)a b)
v2
36
±6 km/h
12 =
But speed v can't be negative, hence we will eliminate v = -6 km/h
6 km/h
help_outline

Image Transcriptionclose

By using the first and third conditions, we get (v3)t 27 Using t v3 as obtained in Step - 2 (v3)v3) 27 v2 32 =27 a2 - b2 [as (ab)a b) v2 36 ±6 km/h 12 = But speed v can't be negative, hence we will eliminate v = -6 km/h 6 km/h

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