A dog running in an open field has components of velocity v x = 2.8 m/s and v y

Question

A dog running in an open field has components of velocity vx = 2.8 m/s and vy = -1.7 m/s at time t1 = 11.5 s . For the time interval from t1 = 11.5 s to t2 = 23.7 s , the average acceleration of the dog has magnitude 0.45 m/s2 and direction 35.0 measured from the +xaxis toward the +yaxis.

At time t2 = 23.7 s , what is the x-component of the dog's velocity?

At time t2 = 23.7 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)?

Explanation / Answer

Initial speed in x direction=2.8m/s

Initial speed in y direction=-1.7m/s

Acceleration in x direction=0.45cos35 =0.37 m/s²

Acceleration in y direction =0.45sin35=0.26 m/s²

X component of velocity at 23.7 seconds =2.8+(23.7-11.5)*0.37 =7.3 m/s

Y component of velocity at 23.7 seconds =-1.7+(23.7-11.5)*0.26=1.47 m/s

Magnitude of dogs velocity =sqrt(7.3²+1.47²)=7.45 m/s

Direction of dogs velocity =arctan(1.47/7.3)=11.4 degrees

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