Two canoeists start paddling at the same time and head toward a small island in

Question

Two canoeists start paddling at the same time and head toward a small island in a lake, as shown in the figure (Figure 1) . Canoeist 1 paddles with a speed of 1.20m/s at an angle of 45 ? north of east. Canoeist 2 starts on the opposite shore of the lake, a distance of 1.5 km due east of canoeist 1.

Part A

In what direction relative to north must canoeist 2 paddle to reach the island?

Express your answer using two significant figures.

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Part B

What speed must canoeist 2 have if the two canoes are to arrive at the island at the same time?

Express your answer using two significant figures.

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Figure 1 of 1

Explanation / Answer

let:

A be point 1 (start canoe #1)
B be point 2 (start canoe #2)
C be point 3 (island)
and
D be point 4 (where arrow touches line AB)

assuming angle CDA = 90 degrees...
angle CAB = 45 degrees thus DA = DC = 1.0 km

DA + DB = 1.5 km so DB = 1.5 - 1.0 = 0.5 km


assuming angle CDA = 90 degrees...
angle CDB = 90 degrees too
so we can use Pythagoras-rule for the triangle CDB...

CB^2 = DC^2 + DB^2
CB^2 = 1.0^2 + 0.5^2 = 1.25 = 5/4
CB = 1/2 * sqr(5) = 0.5 * sqr(5)

assuming angle CDA = 90 degrees...
so we can use Pythagoras-rule for the triangle CDA too...

CA^2 = DC^2 + DA^2
CA^2 = 1.0^2 + 1.0^2 = 2

CA = sqr(2)

now
let:
V = velocity
S = distance
T = time

then

S = V*T
V = S/T
T = S/V

s1 = v1*t1
s2 = v2*t2

in this instance...
t1 = t2 (arrive at the same time)

v2 = s2/t2
t1 = s1/v1

v2 = s2/t1 (t1 = t2)
v2 = s2/(s1/v1)

v2 = v1 * (s2/s1)

v1 = 1.20
s2 = 0.5*sqr(5)
s1 = sqr(2)

Answer a) is...
arctan(DB/DC)
arctan(0.5/1.0) = 26.57 degrees ~= 27 degrees

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