A particle take the following the following successive displacement . 10km south west, 50km 60° north of East . Find the magnitude and the direction of resistant.
To find the magnitude and direction of the resultant displacement, we can add the given displacements using vector addition.
First, let's convert the second displacement from polar to cartesian coordinates:
Magnitude = 50 km
Direction = 60° north of East
To convert the direction to Cartesian coordinates, we need to split it into its north and east components.
North component = Magnitude * sin(direction) = 50 km * sin(60°)
East component = Magnitude * cos(direction) = 50 km * cos(60°)
North component = 50 km * sqrt(3)/2 = 25√3 km
East component = 50 km * 1/2 = 25 km
Now, we can add the displacements together:
Total north component = -10 km + 25√3 km
Total east component = 25 km
Total north component = 25√3 km - 10 km ≈ 43.3 km
Total east component = 25 km
To find the magnitude of the resultant displacement, we use the Pythagorean theorem:
Magnitude = sqrt((Total north component)^2 + (Total east component)^2)
Magnitude = sqrt((43.3 km)^2 + (25 km)^2)
Magnitude ≈ sqrt(1878.49 km^2 + 625 km^2)
Magnitude ≈ sqrt(2503.49 km^2)
Magnitude ≈ 50.03 km
To find the direction of the resultant displacement, we use the inverse tangent function:
Direction = atan(Total north component / Total east component)
Direction = atan(43.3 km / 25 km)
Direction ≈ 59.69°
Therefore, the magnitude of the resultant displacement is approximately 50.03 km and the direction is approximately 59.69°.
To find the magnitude and direction of the resultant displacement, we can apply vector addition.
First, let's break down the given displacements into their respective x and y components.
1. 10km south-west:
This displacement can be split into two components: 10km south and 10km west.
The south component has a magnitude of 10km and a direction of 180° (in the negative y-axis direction).
The west component has a magnitude of 10km and a direction of 270° (in the positive x-axis direction).
2. 50km 60° north of East:
To find the x and y components, we need to convert the given angle to a standard reference angle.
The reference angle is 180° - 60° = 120°.
The x component has a magnitude of 50km * cos(120°) and a direction of 0° (in the positive x-axis direction).
The y component has a magnitude of 50km * sin(120°) and a direction of 90° (in the positive y-axis direction).
Now, let's add up the x and y components separately:
X-component:
10km (west) + 50km * cos(120°) = 10km + (-25km) = -15km
Y-component:
10km (south) + 50km * sin(120°) = 10km + 43.30km = 53.30km
The resultant displacement can be found by combining these x and y components using the Pythagorean theorem:
Resultant magnitude (R) = √((-15km)^2 + (53.30km)^2) ≈ 55.51km
To find the direction of the resultant displacement, we use the inverse tangent function (tan^(-1)):
Resultant direction (θ) = tan^(-1)((y-component)/(x-component))
= tan^(-1)(53.30km/(-15km))
≈ tan^(-1)(-3.55)
≈ -74.68°
Since the resultant displacement is in the negative x-axis direction, we need to add 180° to get the direction with respect to the positive x-axis:
Resultant direction (θ) ≈ -74.68° + 180° ≈ 105.32°
Therefore, the magnitude of the resultant displacement is approximately 55.51km, and the direction is approximately 105.32° north of the east.