Question

A dog running in an open field has components of velocity
*v**x* = 3.0 m/s and *v**y* = -1.7 m/s
at time *t*1= 10.0 s . For the time interval from
*t*1 = 10.0 s to *t*2= 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 *t*2 = 22.0 s , what is the *x*-component
of the dog's velocity?

At time *t*2 = 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

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