A car is traveling in a direction 38.0◦ from +x-axis with a speed of 5.50 m/s relative to the water. A
passenger is walking with a velocity of 2.50 m/s in the +x direction relative to the car. What is the
velocity (magnitude and direction) of the passenger with respect to the water?
a car travelling with velocity 5.50m/s at angle 38° to the +x axis relative to the water. a passenger walking with 2.50m/s to +x axis relative to the car. find velocity
Help me solving this problem
Well, well, well, looks like we've got a passenger going on a walk! Let's break it down, shall we?
First, we have the car traveling at 5.50 m/s at an angle of 38.0 degrees from the +x-axis. Then, we have our brave passenger strolling at 2.50 m/s in the +x direction.
To find the velocity of the passenger with respect to the water, we need to add these two velocities together. We can do this by considering the components of each velocity separately.
The car's velocity has two components: one in the x-direction and another in the y-direction (since it's making an angle with the x-axis). The passenger's velocity is only in the x-direction.
To find the x-component of the car's velocity, we use cosine: Vcar_x = Vcar * cos(angle).
Substituting the values, we get Vcar_x = 5.50 m/s * cos(38.0 degrees).
Now, since the passenger's velocity is only in the x-direction, the y-component of the car's velocity does not affect our final answer.
Finally, to find the passenger's velocity with respect to the water, we just add the x-component of the car's velocity and the passenger's velocity: Vpassenger_water = Vcar_x + Vpassenger.
And there you have it, the magnitude and direction of the passenger's velocity with respect to the water! You can calculate the final answer, and don't forget to double-check the math because I'm just a clown bot.
To find the velocity of the passenger with respect to the water, we need to add the velocities of the car and the passenger.
1. Convert the velocity of the car from polar coordinates to Cartesian coordinates.
Given:
Angle of car's direction from the +x-axis = 38.0°
Speed of the car relative to the water = 5.50 m/s
Using trigonometry, we can find the x- and y-components of the car's velocity:
x-component of car's velocity = (5.50 m/s) * cos(38.0°)
y-component of car's velocity = (5.50 m/s) * sin(38.0°)
2. Calculate the velocity of the passenger relative to the car in Cartesian coordinates.
Given:
Velocity of the passenger relative to the car = 2.50 m/s in the +x direction
Since the velocity of the car is also in the x direction, the velocity of the passenger relative to the water remains the same.
3. Add the velocities of the car and the passenger.
x-component of passenger's velocity relative to water = x-component of car's velocity + x-component of passenger's velocity
y-component of passenger's velocity relative to water = y-component of car's velocity + y-component of passenger's velocity
4. Calculate the magnitude and direction of the passenger's velocity relative to the water.
Magnitude = sqrt[(x-component)^2 + (y-component)^2]
Direction = arctan(y-component / x-component)
Now let's calculate the values step by step.
Using the given values:
Angle of car's direction (θ) = 38.0°
Speed of the car relative to the water (v_car) = 5.50 m/s
Velocity of the passenger relative to the car (v_passenger) = 2.50 m/s
1. Convert the velocity of the car from polar coordinates to Cartesian coordinates:
x-component of car's velocity (v_car_x) = (5.50 m/s) * cos(38.0°)
y-component of car's velocity (v_car_y) = (5.50 m/s) * sin(38.0°)
2. Velocity of the passenger relative to the water remains the same:
x-component of passenger's velocity relative to water (v_passenger_x) = 2.50 m/s
3. Add the velocities of the car and the passenger:
x-component of passenger's velocity relative to water = v_car_x + v_passenger_x
y-component of passenger's velocity relative to water = v_car_y
4. Calculate the magnitude and direction of the passenger's velocity relative to the water:
Magnitude = sqrt[(x-component)^2 + (y-component)^2]
Direction = arctan(y-component / x-component)
Now we can substitute the values and calculate the result.
To find the velocity of the passenger with respect to the water, we need to combine the velocity of the car and the velocity of the passenger relative to the car.
Let's break down the velocities into their x and y components:
Velocity of the car relative to the water:
V_car_x = car_speed * cos(car_angle)
V_car_y = car_speed * sin(car_angle)
Velocity of the passenger relative to the car:
V_passenger_x = passenger_speed
V_passenger_y = 0 (since the passenger is only moving in the x direction relative to the car)
Now, we can add the x and y components of the velocities to get the final velocity of the passenger with respect to the water:
V_passenger_x_wrt_water = V_car_x + V_passenger_x
V_passenger_y_wrt_water = V_car_y + V_passenger_y
To find the magnitude and direction of the velocity of the passenger with respect to the water, we can use these components:
Magnitude: |V_passenger_wrt_water| = sqrt(V_passenger_x_wrt_water^2 + V_passenger_y_wrt_water^2)
Direction: θ = arctan(V_passenger_y_wrt_water / V_passenger_x_wrt_water)
Now we can substitute the values and calculate the velocity:
V_car_x = 5.50 m/s * cos(38.0°)
V_car_y = 5.50 m/s * sin(38.0°)
V_passenger_x_wrt_water = V_car_x + 2.50 m/s
V_passenger_y_wrt_water = V_car_y + 0 m/s
|V_passenger_wrt_water| = sqrt((V_passenger_x_wrt_water^2) + (V_passenger_y_wrt_water^2))
θ = arctan(V_passenger_y_wrt_water / V_passenger_x_wrt_water)
Now, you can substitute the values and solve for the magnitude and direction of the velocity of the passenger with respect to the water.