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A dog running in an open field has components of velocity vₓ= 2.7 m/s and vᵧ = -1.1 m/s at time t1 = 11.2 s . For the time interval from t1 = 11.2 s to t2 = 24.0 s , the average acceleration of the dog has magnitude 0.49 m/s² and direction 26.5° measured from the +x-axis toward the +y-axis.
(a) At time t2 = 24.0 s, what are the x- and y-components of the dog's velocity?
(b) What is the magnitude of the dog's velocity?
(c) What is the direction of the dog's velocity (measured from the toward the + x-axis and +y-axis)?

Respuesta :

Answer:

a)  vₓ = 8.3128 m / s , v_{y} = 1,698 m / s , b)v = 8.484 m / s , c)   θ = 83.14°

Explanation:

Let's use the kinematic equations for two-dimensional movement

Let's look for the components of acceleration using trigonometry

             sin26.5 = ay / a

             cos 26..5 = ax / a

             ay = a sin26.5

             ax = a cos 26.5

             ay = 0.49 sin26.5

             ax = 0.49 cos 26.5

             ay = 0.2186 m / s2

             ax = 0.4385 m / s2

a) Having the accelerations we look for the speed in each axis

              vₓ = v₀ₓ + ax (t-t₀)

              vₓ = 2.7 + 0.4385 (24.0 - 11.2)

              vₓ = 8.3128 m / s

             [tex]v_{y}[/tex]  = v_{oy} + a_{y} (t-to)

              v_{y} = -1.1 + 0.2186 (24.0 - 11.2)

             

b) let's use Pythagoras' theorem to find the magnitude

              v = √ Vₓ² + v_{y}²

              v = Ra (8.3128² + 1,698²)

              v = 8.484 m / s

c) Let's use trigonometry to find the angles

            tan tea = v_{y} / vₓ

            θ = tan⁻¹ v_{y} / vₓ

            θ = tan⁻¹ 1,698 / 8.3128

             θ = 83.14°

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