A car is driven east for a distance of 88 km, then north for 28 km, then in a direction 30 degrees east of north for 25 km. Determine the magnitude of the displacement (in kilometers) of the car from its starting point.
thanks!
add them as vectors:
displacement=88km E + 28km N + 25cos30 N + 25sin30 E.
add the n, and the e components, and add.
To determine the magnitude of the displacement of the car, we need to find the straight-line distance and direction from the starting point to the ending point.
First, let's break down the motion of the car into its components.
1. The car is driven east for a distance of 88 km. This is a purely eastward movement, so the x-component of the displacement is +88 km, and the y-component is 0 km.
2. Then, the car is driven north for a distance of 28 km. This is a purely northward movement, so the y-component of the displacement is +28 km.
3. Finally, the car is driven in a direction 30 degrees east of north for 25 km. We can break down this movement into its x and y components. To do this, we can use trigonometry.
The x-component is given by 25 km * cos(30°), which is approximately 21.65 km.
The y-component is given by 25 km * sin(30°), which is approximately 12.5 km.
Now, we can add up the x and y components of the displacements:
x-component = +88 km + 21.65 km = 109.65 km
y-component = +28 km + 12.5 km = 40.5 km
To find the displacement, we can use the Pythagorean theorem:
displacement = √(x^2 + y^2) = √(109.65^2 + 40.5^2) ≈ 117.47 km
Therefore, the magnitude of the displacement of the car from its starting point is approximately 117.47 km.