Exercise 3.6 A dog running in an open field has components of velocity vx = 2.8

Question

Exercise 3.6 A dog running in an open field has components of velocity vx = 2.8 m/s and vy = -1.3 m/s at time t1 = 10.9 s . For the time interval from t1 = 10.9 s to t2 = 22.5 s , the average acceleration of the dog has magnitude 0.50 In/s^2 and direction 35.0 measured from the +x - axis toward the + y - axis. Part C What is the magnitude of the dog's velocity? Express your answer using two significant figures. Part D What is the direction of the dog' velocity (measured from the +x - axis toward the + y - axis)? Express your answer using two significant figures.

Explanation / Answer

acceleration along +ve x axis=0.5*cos(35)=0.4095 m/s^2
so if at t2=22.5 seconds, veloicty along x axis is vx, then

(vx-2.8)/(22.5-10.9)=0.4095

==> vx=7.55 m/s

acceleration along +ve y axis=0.5*sin(35)=0.2867 m/s^2
so if at t2=22.5 seconds, veloicty along y axis is vy, then

(vy-(-1.3))/(22.5-10.9)=0.2867

==> vy=2.0257 m/s

part C:

magnitude of velocity=sqrt(vx^2+vy^2)=7.817 m/s

part D:

angle with +ve x axis=arctan(vy/vx)=15 degrees

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