Question

A dog running in an open field has components of velocity vx = 3.0 m/s and vy = -1.2 m/s at time t1 = 11.7 s . For the time interval from t1 = 11.7 s to t2 = 23.8 s , the average acceleration of the dog has magnitude 0.54 m/s2 and direction 25.0 ? measured from the +x?axis toward the +y?axis.

A)At time t2 = 23.8 s , what is the x-component of the dog's velocity?

B)At time t2 = 23.8 s , what is the y-component of the dog's velocity?

C)What is the magnitude of the dog's velocity?

D)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= 23.8 -11.7 = 12.1s; ax = 0.54 * cos 25 = 0.4894

vy= -1.2 m/s; t= 12.1s; ay = 0.54 * sin 25 = 0.2282

At time t2= 23.8s , what are the x- and y-components of the dog's velocity

v= u+at

A) vx( t2= 23.8s) = 3 + 0.4894 * 12.1 = 8.92 m/s ---answer

B) vy( t2= 23.8s) = -1.2 + 0.2282 * 12.1 = 1.56 m/s ---answer

C) the magnitude of the dog's velocity

= sq root [ 8.92^2 + 1.56^2 ]

= 9.06 m/s -----answer

D) the direction of the dog's velocity (measured from the +x-axis toward the +y-axis

= tan (-1) (1.56 / 8.92) = 9.92 deg --answer