A dog running in an open field has components of velocity vx = 2.8 m/s and vy = -1.2 m/s at time t1 = 12.0 s. For the time interval from t1 = 12.0 s to t2 = 22.0 s, the average acceleration of the dog has magnitude 0.50 m/s2 and direction 36.0 ∘ measured from the +x-axis toward the +y-axis.

Question

A dog running in an open field has components of velocity vx = 2.8 m/s and vy = -1.2 m/s at time t1 = 12.0 s. For the time interval from t1 = 12.0 s to t2 = 22.0 s, the average acceleration of the dog has magnitude 0.50 m/s2 and direction 36.0 ∘ measured from the +x-axis toward the +y-axis.
Part A
At time  t2  = 22.0  s , what is the  x -component of the dog's velocity?
Express your answer with the appropriate units.
Part B
At time  t2  = 22.0  s , what is the  y -component of the dog's velocity?
Express your answer with the appropriate units.
Part C
What is the magnitude of the dog's velocity?
Express your answer with the appropriate units.
Part D
What is the direction of the dog's velocity (measured from the  +x -axis toward the  +y -axis)?
Express your answer in degrees.

Anonymous · Answer

Do an x problem and a y problem
x problem
at t = 12 Vxi = 2.8 m/s
Ax = 0.5 cos 36 = 0.405 m/s^2
so
Vx = 2.8 + 0.405 (t-12)
so at t = 22
Vx = 2.8 + 0.405 (22-12) = 6.85 m/s (answer Part A)
Get it ? Now do the Y problem the same way.
I think you can do sqrt(Vx^2+Vy^2) etc