# A dog running in an open field (physics)?

0

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?

Also Check This  What is the closest beach to Binghamton, NY?

C)What is the direction of the dog’s velocity (measured from the +x-axis toward the +y-axis… (answer in θ)

• 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 ]