Final answer:
The relative velocity of the car moving at 24 m/s NE and the bus moving at 7 m/s NW can be found using vector addition. The car's velocity is split into x and y components using trigonometric functions, and then the x and y components of the bus's velocity are also determined.
Step-by-step explanation:
To find the relative velocity between the car and the bus, we need to combine their velocities using vector addition. The car is moving at 24 m/s in the northeast (NE) direction, which can be split into horizontal (x) and vertical (y) components. The bus is moving at 7 m/s in the northwest (NW) direction, which also has x and y components.
The x component of the car's velocity is found using the cosine function:
Vcar,x = Vcar * cos(45°) = 24 m/s * 0.707 = 16.97 m/s
The y component of the car's velocity is found using the sine function:
Vcar,y = Vcar * sin(45°) = 24 m/s * 0.707 = 16.97 m/s
The x component of the bus's velocity is found using the cosine function:
Vbus,x = Vbus * cos(45°) = 7 m/s * 0.707 = 4.95 m/s
The y component of the bus's velocity is found using the sine function:
Vbus,y = Vbus * sin(45°) = 7 m/s * 0.707 = 4.95 m/s
To find the relative velocity, we subtract the x and y components of the bus's velocity from the x and y components of the car's velocity respectively:
Vrelative,x = Vcar,x - Vbus,x = 16.97 m/s - 4.95 m/s = 12.02 m/s
Vrelative,y = Vcar,y - Vbus,y = 16.97 m/s - 4.95 m/s = 12.02 m/s
The magnitude of the relative velocity is found using the Pythagorean theorem:
|Vrelative| = √[(Vrelative,x)² + (Vrelative,y)²] = √[(12.02 m/s)² + (12.02 m/s)²] = 17.0 m/s
Therefore, the car is moving at a speed of 17.0 m/s relative to the bus.