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A dog running in an open field has components of velocity vx = 2.2m/s and vy = -2.3m/s at time t1 = 11.1s . For the time interval from t1 = 11.1s to t2 = 23.6s , the average acceleration of the dog has magnitude 0.55m/s2 and direction 30.0❝ measured from the +x❝axis toward the +y❝axis.

Part A

At time t2 = 23.6s , what are the x-component of the dog's velocity?

Express your answer using two significant figures.

Part B

At time t2 = 23.6s , what are the y-component of the dog's velocity?

Express your answer using two significant figures.

Part C

What is the magnitude of the dog's velocity?

Express your answer using two significant figures.

Part D

What is the direction of the dog's velocity (measured from the +x?axis toward the +y?axis)?

Express your answer using two significant figures.

Added Wed, 01 Jul '15

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Solutions

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Net accleration towards X-axis=0.55cos(30) =0.476 m/sec^2

a)Vx=Ux + a(dt) =2.2+0.476*(23.6-11.1)=8.15 m/sec (ans)

b)Net accleration towards Y-axis=0.55sin(30) =0.275 m/sec^2

so )Vy=Uy + a(dt) = -2.3+0.275*(23.6-11.1)=1.1375 m/sec (ans)

c)magnitude,V=sqrt (8.15^2 +1.1375^2) =8.228 m/sec (ans)

d) direction =arctan(1.1375/8.15) =7.945 degree measured from the +x axis toward the +y axis. (ans)

Added Wed, 01 Jul '15

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