A boat travels at 15 m/s in a direction 45° east of north. The boat then turns and travels at 18 m/s in a direction 5° north of east.
What is the magnitude of the boat’s resultant velocity? Round your answer to the nearest whole number.
What is the direction of the boat’s resultant velocity? Round your answer to the nearest whole degree.
31 m/s
23 ° north of east
To find the magnitude of the boat's resultant velocity, we can use the Pythagorean theorem.
Let's first find the north and east components of the boat's velocity after the turn.
North component = 15 m/s * sin(45°) = 10.61 m/s
East component = 15 m/s * cos(45°) = 10.61 m/s
Now let's find the north and east components of the boat's velocity before the turn.
North component = 18 m/s * sin(5°) = 1.56 m/s
East component = 18 m/s * cos(5°) = 17.33 m/s
To find the resultant velocity, we can add the north and east components of each stage together.
North component of resultant velocity = 10.61 m/s + 1.56 m/s = 12.17 m/s
East component of resultant velocity = 10.61 m/s + 17.33 m/s = 27.94 m/s
Now we can find the magnitude of the resultant velocity using the Pythagorean theorem.
Resultant velocity = sqrt((12.17 m/s)^2 + (27.94 m/s)^2)
Resultant velocity ≈ 30.78 m/s
Rounded to the nearest whole number, the magnitude of the boat's resultant velocity is 31 m/s.
To find the direction of the resultant velocity, we can use the inverse tangent function.
Direction = arctan(North component of resultant velocity / East component of resultant velocity)
Direction = arctan(12.17 m/s / 27.94 m/s)
Direction ≈ 24.2°
Rounded to the nearest whole degree, the direction of the boat's resultant velocity is 24°.
To find the magnitude of the boat's resultant velocity, we need to use vector addition. We can break down the boat's velocity into its north and east components.
First, let's find the north component of the first velocity:
North component = Velocity * sin(angle)
= 15 m/s * sin(45°)
≈ 10.61 m/s (rounded to two decimal places)
Next, let's find the east component of the first velocity:
East component = Velocity * cos(angle)
= 15 m/s * cos(45°)
≈ 10.61 m/s (rounded to two decimal places)
Now, let's find the north component of the second velocity:
North component = Velocity * sin(angle)
= 18 m/s * sin(5°)
≈ 1.57 m/s (rounded to two decimal places)
Next, let's find the east component of the second velocity:
East component = Velocity * cos(angle)
= 18 m/s * cos(5°)
≈ 17.38 m/s (rounded to two decimal places)
Now, let's add the north components and east components separately:
North component total = 10.61 m/s + 1.57 m/s
≈ 12.18 m/s (rounded to two decimal places)
East component total = 10.61 m/s + 17.38 m/s
≈ 27.99 m/s (rounded to two decimal places)
Finally, let's find the magnitude of the resultant velocity using the Pythagorean theorem:
Magnitude of resultant velocity = √(North component total)^2 + (East component total)^2
= √(12.18 m/s)^2 + (27.99 m/s)^2
≈ 30.66 m/s (rounded to the nearest whole number)
Therefore, the magnitude of the boat's resultant velocity is approximately 31 m/s.
To find the direction of the boat's resultant velocity, we can use inverse tangent (tan^(-1)) to find the angle.
Direction = tan^(-1)(North component total / East component total)
= tan^(-1)(12.18 m/s / 27.99 m/s)
≈ 23.30° (rounded to the nearest whole degree)
Therefore, the direction of the boat's resultant velocity is approximately 23°.
V1 = 15m/s[45o E of N.] = 45o N. of E.
V2 = 18m/s[5o N. of E.].
Vr = 15m/s[45o] + 18m/s[5o].
Vr = 15*Cos45+15*sin45 + 18*Cos5+18*sin5.
Vr = (10.61+10.61i) + (17.93+1.57i) = 28.54 + 12.18i.
Magnitude = sqrt(28.54^2 + 12.18^2) =
Tan A = Y/X = 12.18/28.54 = 0.42887. A =