A dog running in an open field has components of velocity vx = 3.0m/s and vy = -1.1m/s at time t1 = 11.7s . For the time interval from t1 = 11.7s to t2 = 22.2s , the average acceleration of the dog has magnitude 0.55m/s2 and direction 34.5∘ measured from the +x−axis toward the +y−axis. at t=22.2, what are the x and y components of the dogs velocity?

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67 To find the x and y components of the dog's velocity at t = 22.2 s, we need to use the information given about the average acceleration and the initial velocity components.

First, let's calculate the change in time (Δt) from t1 to t2:
Δt = t2 - t1
Δt = 22.2 s - 11.7 s
Δt = 10.5 s

Next, let's find the change in velocity (Δv) using the average acceleration:
Δv = a * Δt
Δv = 0.55 m/s² * 10.5 s

Now, let's determine the change in velocity for the x and y components separately. Since the acceleration is given in magnitude and direction, we need to split it into x and y components.

The x component of the acceleration is given by:
ax = a * cos(θ)
ax = 0.55 m/s² * cos(34.5°)

The y component of the acceleration is given by:
ay = a * sin(θ)
ay = 0.55 m/s² * sin(34.5°)

Once we have the x and y components of the acceleration, we can calculate the change in velocity for each component:

Δvx = ax * Δt
Δvx = (0.55 m/s² * cos(34.5°)) * 10.5 s

Δvy = ay * Δt
Δvy = (0.55 m/s² * sin(34.5°)) * 10.5 s

To find the final velocity components, we add the changes in velocity to the initial velocity components:

vxf = vx + Δvx
vyf = vy + Δvy

Therefore, the x and y components of the dog's velocity at t = 22.2 s are given by:
vxf = 3.0 m/s + ((0.55 m/s² * cos(34.5°)) * 10.5 s)
vyf = -1.1 m/s + ((0.55 m/s² * sin(34.5°)) * 10.5 s)