A dog running in an open field has components of velocity v_x = 3.0 m/s and v_y

Question

A dog running in an open field has components of velocity v_x = 3.0 m/s and v_y = -1.2 m/s at time t_1 = 11.4 s. For the time interval from t_1 = 11.4 s to t_2 = 24.7 s, the average acceleration of the dog has magnitude 0.40 m/s^2 and direction 36.0 degree measured from the + x-axis toward the + y-axis .

a. At time t_2 = 24.7 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 + x - axis toward the + y - axis)?

Explanation / Answer

V2x = 5.832 m/s V2y = -.072 m/s T = -.707° Decompose the acceleration into (Ax, Ay) = (0.35 * cos(36.0), 0.35 * sin(36.0)) Vfinal = Vinitial + A * DeltaT (Vfx, Vfy) = (1.8, -1.1) + (Ax, Ay) * (20.8 - 11.5) | Vfinal | = sqrt((Vfx)^2 + (Vfy)^2) Direction = arctan(Vfy / Vfx) { note: Consider the signs of Vfx and Vfy to determine the quadrant where the final velocity vector is. You may need to add 180 degrees to get the final answer. }

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