Final answer:
Using the conservation of momentum, it is calculated that the canoe's final velocity is 0.12 m/s, thus the correct answer is d) 0.12 m/s.
Step-by-step explanation:
To solve for the velocity of the canoe after the dog jumps out, we need to use the conservation of momentum. The initial momentum of the system (canoe and dog) is zero because they are at rest. When the 15 kg dog jumps out of the canoe with a velocity of -3.2 m/s, the dog's momentum is the product of its mass and velocity (momentum = mass × velocity).
The dog's momentum is therefore 15 kg × -3.2 m/s = -48 kg·m/s. To conserve momentum, the canoe must have equal and opposite momentum. Let the mass of the canoe be 40 kg and its velocity be v.
Conservation of momentum: (mass of dog × dog's velocity) + (mass of canoe × canoe's velocity) = 0
-48 kg·m/s + (40 kg × v) = 0
40 kg × v = 48 kg·m/s
v = 48 kg·m/s / 40 kg
v = 1.2 m/s
Since the direction is opposite to the dog's direction, the canoe's final velocity is 0.12 m/s in the positive direction. Therefore, the correct answer is d) 0.12 m/s.