Question

A drone is launched with a velocity of 63 m/s, 29 . Three minutes after the drone is launched it suddenly changed its course to 22 m/s, 233 to avoid high trees on its way. Then, it finally goes back at a rate of 98 m/s, 321. Calculate for the total resultant vector of the drone.

Answer 1

Answer:

   v = 127.66 m / s      θ’= 337.59

Explanation:

For this exercise we must use the speed composition of the drone.

The first speed is 63 m / s in direction 29, three minutes later the speed reaches 22 m / s and direction 233, finally it returns to the launch point with 98 m / s in direction 321, in the attachment you can see a diagram of these speeds.

To find the resulting average velocity, the easiest thing is to decompose each velocity into the x and y coordinate system, then add each velocity

let's break down the speeds

             cos 29 = v₁ₓ / v₁

              sin 29 = $v_{1y}$ / v₁

             v₁ₓ = v₁ cos 29

             v_{1y} = v₁ sin 29

             v₁ₓ = 63 cos 29 = 55.10 m / s

             v_{1y} = 63 sin 29 = 30.54 m / s

speed 2

              cos 22 = v₂ₓ / v₂

              sin 22 = v_{2y} / v₂

               v₂ₓ = v₂ cos 233

               v_{2y} = v₂ sin 233

               v₂ₓ = 22 cos 233 = -13.24 m / s

                v_{2y} = 22 sin 233 = -17.57 m / s

speed 3

              cos 321 = v₃ₓ / v₃

              sin 321 = v_{3y} / v₃

               v₃ₓ = v₃ cos 321

               v_{1y} = vₐ sin 321

               v₃ₓ = 98 cos 321 = 76.16 m / s

               v_{3y} = 98 sin 321 = -61.67 m / s

We already have all the component of the speeds, the resulting speed is

               vₓ = v₁ₓ + v₂ₓ + v₃ₓ

               vₓ = 55.10 -13.24 +76.16

               vₓ = 118.02 m / s

               v_{y} = v_{1y} + v_{2y} + v_{3y}

                v_{y} = 30.54 -17.54 - 61.67

                v_{y} = -48.67 m / s

there are two ways to give the result

               v = (118.02 i -48.67 j) m / s

or in the form of magnitud and angle.

We use the Pythagorean theorem for the module

             v = √ (vₓ² + v_{y}²)

             v = RA (118.02² + 48.67²)

             v = 127.66 m / s

let's use trigonometry for the angle

             tan θ = v_{y} / vₓ

             θ = tan⁻¹ (v_{1} / vₓ)

             θ = tan⁻¹ (-48.67 / 118.02)

             θ = -22.41

if we want to measure the angles with respect to the positive side of the x axis

               θ’= 360 - 22.41

               θ’= 337.59