Question

A dog running in an open field has components of velocity vx = 3.0 m/s and vy = -1.7 m/s at time t1= 10.0 s . For the time interval from t1 = 10.0 s to t2= 22.0 s , the average acceleration of the dog has magnitude 0.45 m/s2 and direction 31.5 ∘measured from the +x−axis toward the +y−axis.

At time t2 = 22.0 s , what is the x-component of the dog's velocity?

At time t2 = 22.0 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 +x−axis toward the +y−axis)?

Answer 1

vx= 3 m/s ; t= 12s; ax = 0.45 * cos 31.5 = 0.3836
vy= -1.7 m/s; t= 12.4s; ay = 0.43 * sin 33.5 = 0.2351

A)At time t2= 22.0s , what are the x- and y-components of the dog's velocity
v= u+at

vx( t2= 22.0s) = 3.0 + 0.3836 * 12 = 7.30 m/s ---answer
vy( t2= 22.0s) = -1.7 + 0.2351 * 12 = 1.12 m/s ---answer

B) the magnitude of the dog's velocity
= sq root [ 7.30^2 + 1.12^2 ]

7.38
answer

C) the direction of the dog's velocity (measured from the +x-axis toward the +y-axis
= tan (-1) 1.12 / 7.30 = 6.60 deg --answer