A sailboat is traveling east at 8.0 m/s. A sudden gust of wind gives the boat an acceleration a = (1.30 m/s2 35° north of east). What is the boat's speed 10.0 seconds later when the gust subsides?

What is the boat's direction (in deg) 10.0 seconds later when the gust subsides?

I somehow got the first question right but don't fully understand how and I am completely confused on how to do the last question.

Ah, sailing and physics, two things that go together like a ship in the night. Let's break it down.

For the first question, we need to find the boat's final speed 10.0 seconds later. We know the initial speed is 8.0 m/s and the acceleration from the wind gust is 1.30 m/s² 35° north of east. So, we can add the two velocities together to find the boat's final velocity.

Now, without going into too much mathematical detail, we can use the concept of vector addition. The initial velocity can be represented by a vector pointing east with a magnitude of 8.0 m/s. The acceleration due to the wind gust can be represented by a vector pointing 35° north of east with a magnitude of 1.30 m/s². Adding these two vectors together using vector addition, we get the boat's final velocity.

As for the second question, we need to find the boat's direction, which can be represented by an angle. We can use some trigonometry here. To find the boat's direction, we need to find the angle between the boat's final velocity vector and the east direction.

Now, the final velocity vector will have both an eastward and a northward component due to the wind gust. So, we can use trigonometry to find the angle. We can find the northward component of the velocity using the magnitude of the acceleration and the angle of 35°. Then, we can use this northward component and the eastward velocity to find the angle between the boat's velocity vector and the east direction.

And voila! You've got the boat's speed and direction 10.0 seconds later. Just remember, if you ever get lost at sea, just follow the North Star and hope it's not a rock band giving directions. Keep sailing and keep calculating, my friend!

To find the boat's speed 10.0 seconds later when the gust subsides, we can use the equation of motion:

v = u + at

Where:
v = final velocity (unknown)
u = initial velocity (8.0 m/s eastwards)
a = acceleration (1.30 m/s^2 35° north of east)
t = time (10.0 seconds)

First, we need to resolve the acceleration vector into its eastward and northward components:
a_east = a * cos(35°)
a_north = a * sin(35°)

Now, we can apply the equation of motion to both the eastward and northward components separately:

v_east = u_east + a_east * t
v_north = u_north + a_north * t

Given that the boat is initially traveling east, its initial velocity in the eastward direction (u_east) is 8.0 m/s, and the initial velocity in the northward direction (u_north) is 0 m/s.

So, substituting the values into the equations:

v_east = 8.0 m/s + (1.30 m/s^2 * cos(35°)) * 10.0 s
v_north = 0 m/s + (1.30 m/s^2 * sin(35°)) * 10.0 s

Calculate the values:

v_east = 8.0 m/s + (1.30 m/s^2 * 0.819) * 10.0 s
v_north = 0 + (1.30 * 0.573) * 10.0

v_east = 17.71 m/s
v_north = 7.46 m/s

Finally, we can find the boat's speed (magnitude of velocity) using the Pythagorean theorem:

v = √(v_east^2 + v_north^2)

v = √((17.71 m/s)^2 + (7.46 m/s)^2)

v ≈ 19.2 m/s (rounded to one decimal place)

Now, let's move on to the second question:

To find the boat's direction in degrees, we can use trigonometry and the inverse tangent function:

θ = tan^(-1)(v_north / v_east)

θ = tan^(-1)(7.46 m/s / 17.71 m/s)

θ ≈ 23.2°

Therefore, the boat's direction 10.0 seconds later when the gust subsides is approximately 23.2° north of east.

To solve this problem, we can break it down into two parts: finding the boat's velocity after 10.0 seconds and determining its direction.

1. Finding the boat's velocity after 10.0 seconds:
The boat's initial velocity, given as 8.0 m/s east, remains constant throughout the problem. However, the question introduces a gust of wind that gives the boat an additional acceleration of (1.30m/s^2, 35° north of east).

To find the boat's velocity after 10.0 seconds, we need to add the change in velocity caused by the gust to the initial velocity.

First, we calculate the change in velocity caused by the gust in terms of components:
a = 1.30 m/s^2
θ = 35°

Using trigonometry, we can find the components of the gust's acceleration along the east and north directions:
a_east = a * cos(θ)
a_north = a * sin(θ)

Since the gust acts for 10 seconds, we can find the change in velocity, Δv, in each direction by multiplying the components of acceleration by the time:
Δv_east = a_east * t
Δv_north = a_north * t

Now, we add the change in velocity caused by the gust (Δv) to the boat's initial velocity to find the boat's final velocity after 10.0 seconds:
v_final = v_initial + Δv

2. Determining the boat's direction:
To find the boat's direction after 10.0 seconds, we need to calculate the angle it makes with the east direction.

Using trigonometry, we can find the angle as follows:
θ_final = arctan(v_final_north / v_final_east)

Now let's use the given values to solve the problem step by step:

Given:
v_initial = 8.0 m/s (east)
a = 1.30 m/s^2 (35° north of east)
t = 10.0 seconds

Step 1: Finding the boat's velocity after 10.0 seconds:
a_east = a * cos(θ) = 1.30 m/s^2 * cos(35°)
a_north = a * sin(θ) = 1.30 m/s^2 * sin(35°)

Δv_east = a_east * t
Δv_north = a_north * t

v_final_east = v_initial_east + Δv_east
v_final_north = v_initial_north + Δv_north

Step 2: Determining the boat's direction:
θ_final = arctan(v_final_north / v_final_east)

Now you can plug in the values and calculate the final velocity and direction of the boat after 10.0 seconds.