A car travels 20km due north and then 35km in a direction 60 degree west of north. Find the magnitude and direction of the resultant.
is so diffecult
A car travels 3km due north, then 5km east.
48.14
To find the magnitude and direction of the resultant, we can use vector addition.
First, we need to convert the 35km distance traveled in a direction 60 degrees west of north into its x and y components.
The 35km can be broken down into two components: one component in the north direction and another component in the west direction. To find the north component, we multiply 35km by the cosine of 60 degrees. To find the west component, we multiply 35km by the sine of 60 degrees.
North component = 35km * cos(60°)
West component = 35km * sin(60°)
North component = 35 * 0.5
North component = 17.5km
West component = 35 * 0.866
West component ≈ 30.31km
Now, we can add the x and y components of the two vectors (20km north and 17.5km north, 30.31km west) to find the resultant vector.
X component = (20km + 17.5km)
X component = 37.5km
Y component = 30.31km
Now, we can use Pythagoras' theorem to find the magnitude of the resultant vector.
Magnitude = sqrt((X component)^2 + (Y component)^2)
Magnitude = sqrt((37.5km)^2 + (30.31km)^2)
Magnitude ≈ sqrt(1406.25km^2 + 918.1161km^2)
Magnitude ≈ sqrt(2324.3661km^2)
Magnitude ≈ 48.21km
To find the direction of the resultant vector, we can use the tangent function.
Direction = arctan(Y component / X component)
Direction = arctan(30.31km / 37.5km)
Using a calculator, we find:
Direction ≈ 39.3°
Therefore, the magnitude of the resultant vector is approximately 48.21km and its direction is approximately 39.3°.
Call north y and east x
draw axis system with the vectors
x = -35 sin 60
y = 20 + 35 cos 60
sin of resultant angle east of north = x/y
(it will be negative of course since going more west than east)
magnitude = sqrt (x^2+y^2)