In: Physics
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 1.40m/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.
a-
In what direction relative to north must canoeist 2 paddle to reach the island?
Express your answer using two significant figures.
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.
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 = 1.40
s2 = 0.5*sqr(5)
s1 = sqr(2)
Answer a) is...
arctan(DB/DC)
arctan(0.5/1.0) = 26.57 degrees