Question

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.

Answer 1

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°.