A model sailboat is slowly sailing west across a pond at 0.310 m/s. A gust of wind blowing at 28.3° south of west gives the sailboat a constant acceleration of magnitude 0.314 m/s2 during a time interval of 2.18 s. If the net force on the sailboat during the 2.18-s interval has magnitude 0.318 N, what is the sailboat's mass? What is the new velocity of the boat after the 2.18 s gust of wind? And what is the new direction of the boat after the 2.18 s gust of wind?
Question
A model sailboat is slowly sailing west across a pond at 0.310 m/s. A gust of wind blowing at 28.3° south of west gives the sailboat a constant acceleration of magnitude 0.314 m/s2 during a time interval of 2.18 s. If the net force on the sailboat during the 2.18-s interval has magnitude 0.318 N, what is the sailboat's mass? What is the new velocity of the boat after the 2.18 s gust of wind? And what is the new direction of the boat after the 2.18 s gust of wind?
Solution
To solve this problem, we will use the following physics formulas:
- F = ma (Newton's second law of motion)
- v = u + at (equation of motion)
- θ = tan^-1 (Vy/Vx) (to find the direction)
First, let's find the mass of the sailboat. We know the net force (F) and the acceleration (a), so we can rearrange the formula F = ma to solve for m:
m = F/a m = 0.318 N / 0.314 m/s² m = 1.01 kg
Next, let's find the new velocity of the boat. We know the initial velocity (u), the acceleration (a), and the time (t), so we can use the formula v = u + at:
v = u + at v = 0.310 m/s + (0.314 m/s² * 2.18 s) v = 0.310 m/s + 0.684 m/s v = 0.994 m/s
Finally, let's find the new direction of the boat. We know the angle of the wind (28.3° south of west), so we can use the formula θ = tan^-1 (Vy/Vx) to find the new direction. However, since the wind is blowing south of west, the vertical component of the velocity (Vy) will be negative:
Vy = v * sin(28.3°) Vy = 0.994 m/s * sin(28.3°) Vy = -0.467 m/s
Vx = v * cos(28.3°) Vx = 0.994 m/s * cos(28.3°) Vx = 0.879 m/s
θ = tan^-1 (Vy/Vx) θ = tan^-1 (-0.467 m/s / 0.879 m/s) θ = -28.3°
So, the new direction of the boat is 28.3° south of west.
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