A passenger in a boat approaching a dock attempts to step onto the dock before the boat has been tied up. The passenger, whose mass is 57 kg, accelerated at a rate of 3.2 m/s^2 [forward]. If the mass of the boat is 97 kg, what is the boat's acceleration relative to the passenger? Neglect the horizontal force of the water.
57kg x 3.2m/s2 forward = 182.4
182.4/97kg = 1.88 round to two significant figures so
1.9m/s2 backwards
1.9m/s2 backwards
To find the boat's acceleration relative to the passenger, we need to use Newton's second law of motion. This law states that the force acting on an object is equal to its mass multiplied by its acceleration:
Force = mass × acceleration
In this case, we want to find the boat's acceleration relative to the passenger. We can start by finding the force acting on the passenger, and then use that information to find the boat's acceleration.
We know the passenger's mass is 57 kg, and the passenger is accelerating at a rate of 3.2 m/s^2.
So, the force acting on the passenger can be calculated as:
Force = mass × acceleration
Force = 57 kg × 3.2 m/s^2
Force = 182.4 N [forward]
Now that we have the force acting on the passenger, we can find the boat's acceleration relative to the passenger by dividing this force by the boat's mass:
Acceleration = Force / mass
Acceleration = 182.4 N / 97 kg
Acceleration ≈ 1.8825 m/s^2 [forward]
Therefore, the boat's acceleration relative to the passenger is approximately 1.8825 m/s^2 [forward].
Sorry, I forgot to add the second part of the question.
What is the net acceleration of the passenger relative to the dock?