I have the following problem: A dog running in an open field has components of v

Question

I have the following problem:

A dog running in an open field has components of velocity vx = 3.3 m/s and vy = -1.3 m/s at time t1 = 11.0 s . For the time interval from t1 = 11.0 s to t2 = 24.4 s , the average acceleration of the dog has magnitude 0.55 m/s2 and direction 36.0 measured from the +xaxis toward the +yaxis.

At time t2 = 24.4 s , what is the x-component of the dog's velocity?

At time t2 = 24.4 s , what is the y-component of the dog's velocity?

What is the magnitude of the dog's velocity?

What is the direction of the dog's velocity (measured from the +xaxis toward the +yaxis)?

Thanks.

Explanation / Answer

Given dog's velocity at t = 11 s is

   Vx= 3.3 m/s, vy = -1.3 m/s

for the time interval 24.4-11 = 13.4 s

angle theta = arc tan (-1.3/3.3)= -21.50 degrees

the components of acceleration is ax = 0.55cos(-21.50) = 0.512 m/s2

                   ay = 0.55 sin (21.50) = -0.20 m/s2

at time t=24.4 s

   Vx = ux +ax*t = 3.3 +0.512*13.4 = 10.1608 m/s

   Vy = uy + ayt =-1.3 +(-0.20)*13.4 = -3.98 m/s

magnitude is V = sqrt(vx^2vy^2) = sqrt(10.1608^2+(-3.98)^2) = 10.91 m/s
()

direction is theta = arc tan (-3.98/10.1608)==> theta = -21.39 degrees

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