A dog running in an open field has components of velocity vx= 2.8 m/s and vy= -2.3 m/s at time t1= 10.3s . For the time interval from t1= 10.3 to t2= 22.7 , the average acceleration of the dog has magnitude 0.43 m/s2 and direction 33.5 degree measured from the +x-axis toward the +y-axis .
A)At time t2= 22.7s , what are the x- and y-components of the dog’s velocity?
B)What is the magnitude of the dog’s velocity?
C)What is the direction of the dog’s velocity (measured from the +x-axis toward the +y-axis… (answer in θ)
1 Answer
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vx= 2.8 m/s ; t= 12.4s; ax = 0.43 * cos 33.5 = 0.3585
vy= -2.3 m/s; t= 12.4s; ay = 0.43 * sin 33.5 = 0.2373
A)At time t2= 22.7s , what are the x- and y-components of the dog’s velocity
v= u+at
vx( t2= 22.7s) = 2.8 + 0.3585 * 12.4 = 7.25 m/s —answer
vy( t2= 22.7s) = -2.3 + 0.2373 * 12.4 = 0.64 m/s —answer
B) the magnitude of the dog’s velocity
= sq root [ 7.25^2 + 0.64^2 ]
answer
C) the direction of the dog’s velocity (measured from the +x-axis toward the +y-axis
= tan (-1) 0.64 / 7.25 = 5.04 deg –answer
Source(s): my brain (Prof TBT)
