a) A young man measuring 1.89 m. walks at a velocity of 1m/s towards a lamppost which its light is at 4m from the ground. At what velocity does the young man's shadow decrease when he's 10 m away from the lamppost?
b) A young girl, standing at the edge of a dock, pulls a boat towards herself with a cable attached to the front of a boat at 0.3m from sea level. If the girl pulls the cable at a velocity of 1.6 m/s, her hands being at 2.5m from sea level, at what velocity does the boat approach the dock when it's 3m away?
a) To find the velocity at which the young man's shadow decreases when he is 10m away from the lamppost, we can use similar triangles.
We can establish the following relationship between the height of the man, the height of the lamppost, the distance from the man to the lamppost, and the length of the shadow:
(h1 - h2) / L = h2 / (L + d)
where:
h1 = height of the man (1.89m)
h2 = height of the lamppost (4m)
L = length of the shadow
d = distance from the man to the lamppost
We can rewrite this equation to solve for L:
L = (h2 * d) / (h1 - h2)
Substituting the given values:
L = (4 * 10) / (1.89 - 4)
L ≈ 27.96m
Now, to find the velocity at which the shadow decreases, we can use the concept of similar triangles. The ratio of the side lengths of similar triangles is equal to the ratio of their corresponding velocities.
So, we can set up the following relationship between the shadow length (L) and the velocity of the shadow (v):
L / v = (L + d) / v1
where:
v = velocity of the shadow
v1 = velocity of the man (1m/s)
d = distance from the man to the lamppost (10m)
Rearranging the equation, we can solve for v:
v = (L + d) * v1 / L
Substituting the values:
v = (27.96 + 10) * 1 / 27.96
v ≈ 1.53m/s
Therefore, the velocity at which the young man's shadow decreases when he is 10m away from the lamppost is approximately 1.53m/s.
b) To find the velocity at which the boat approaches the dock when it is 3m away, we can again use similar triangles.
We can establish the following relationship between the heights and distances of the girl, the boat, and the dock:
(h1 - h2) / L = h2 / d
where:
h1 = height of the girl's hands (2.5m)
h2 = height of the front of the boat (0.3m)
L = length of the cable
d = distance from the girl to the boat
We can rewrite this equation to solve for L:
L = (h2 * d) / (h1 - h2)
Substituting the given values:
L = (0.3 * 3) / (2.5 - 0.3)
L ≈ 0.39m
Now, to find the velocity at which the boat approaches the dock, we can use the concept of similar triangles. The ratio of the side lengths of similar triangles is equal to the ratio of their corresponding velocities.
So, we can set up the following relationship between the cable length (L) and the velocity of the boat (v):
L / v = d / v1
where:
v = velocity of the boat
v1 = velocity of the girl pulling the cable (1.6m/s)
d = distance from the girl to the boat (3m)
Rearranging the equation, we can solve for v:
v = (L * v1) / d
Substituting the values:
v = (0.39 * 1.6) / 3
v ≈ 0.208m/s
Therefore, the velocity at which the boat approaches the dock when it is 3m away is approximately 0.208m/s.