Question

A car travels 13 km in south east direction and then 16 km 40° north of east. Find the cars resultant direction. PLEASE HELP! ALSO LOOK ON MY PROFILE AND TRY MY OTHER QUESTIONS PLEASE

Answer 1

Answer:

  about 2.9° north of east

Step-by-step explanation:

If we orient the directions so north is in the +y direction and east is in the +x direction, the car is traveling 13 km at an angle of -45°, then 16 km at an angle of +40°. We can add these vectors by adding their components in the x- and y-directions.

  13(cos(-45°), sin(-45°)) ≈ (9.19239, -9.19239)

  16(cos(40°), sin(40°)) ≈ (12.25671, 10.28460)

The sum of these vectors is then ...

  = (21.44910, 1.09221)

and the resultant angle is ...

  arctan(1.09221/21.44910) ≈ 2.915° . . . . measured north of east

The resultant direction is about 2.9° north of east.

Answer 2

Answer:

Magnitude of resultant direction = 3.253 Km

Direction of resultant motion = 19.62° North of East

Step-by-step explanation:

Here we have

13 km in SE direction and

16 km 40 ° North of East

Therefore

13 cos 45, 13 sin 45 = (9.192, 9.9192)

16 cos 40, 16 sin 40 = (12.2567, 10.245)

x₂ - x₁ = 3.064

y₂ - y₁ = 1.092

The Magnitude = $\sqrt{(x_2 - x_1)^2 + (y_2 -y_1)^2}$ = 3.253

The direction = Arctan (y₂ - y₁)/(x₂ - x₁) =  Arctan 1.092/3.064 = 19.62° North of East