help please
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?
To solve this problem, we can apply vector addition to find the resulting velocity of the sailboat after 10.0 seconds.
First, let's break down the given acceleration into its components. The acceleration given is a = 1.30 m/s^2 at an angle of 35° north of east. So, the acceleration in the x-direction (east) is given by ax = a * cos(35°) and in the y-direction (north) by ay = a * sin(35°).
Next, we need to find the change in velocity in both the x and y directions after 10.0 seconds. Since the sailboat has an initial velocity of 8.0 m/s east, the change in velocity in the x-direction is Δvx = ax * t and in the y-direction is Δvy = ay * t, where t is the time interval of 10.0 seconds.
To find the new velocity of the sailboat after 10.0 seconds, we add the changes in x and y velocities to the initial velocity vector. The resulting velocity vector is given by V = √((vx + Δvx)^2 + (vy + Δvy)^2), where vx is the initial velocity in the x-direction.
Using this formula, we can find the speed of the boat after 10.0 seconds. The direction can be determined using trigonometry.
Let's calculate it step by step:
1. Find the components of acceleration:
ax = a * cos(35°)
ay = a * sin(35°)
2. Calculate the change in velocity in the x and y directions:
Δvx = ax * t
Δvy = ay * t
3. Find the new velocity after 10.0 seconds:
V = √((vx + Δvx)^2 + (vy + Δvy)^2)
4. To find the speed, take the magnitude of the velocity vector:
speed = |V|
5. To find the direction in degrees, use trigonometry:
direction = atan(Δvy / (vx + Δvx)) * (180 / π)
Now, let's plug in the values and calculate the speed and direction.