0A dog running in an open field has components of velocity v x = 2.1 m/s and v y

Question

0A dog running in an open field has components of velocity vx = 2.1 m/s and vy = -1.7 m/s at time t1 = 10.4 s . For the time interval from t1 = 10.4 sto t2 = 23.3 s , the average acceleration of the dog has magnitude 0.51 m/s2 and direction 26.0 measured from the +xaxis toward the +yaxis.

a)At time t2 = 23.3 s , what is the x-component of the dog's velocity?

b)At time t2 = 23.3 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 +xaxis toward the +yaxis)?

Explanation / Answer

The acceleration along the x-axis is given by x =0.51cos26 =0.458m/s2

The valocity at t2 =23.3s along the x-axis is taken as =v

The time interval between the two times is (t) =t1-t2 =23.3-10.4=12.9s

Now the final velocity is given by vx =u+at =2.1+0.458*12.9 =8.0082m/s

Now the acceleration along the y-axis is given by a =0.51sin26 =0.223m/s2

Time difference is (t) =12.9s

Now the final velocity along the y -axis is vy =u+at =-1.7+(0.223)(12.9) =1.176m/s

Now the magnitude of dog's velocity is given by v =Sqrt( 64.032 +1.384)=8.087m/s

The direction of dog's velocity is given by

theta =tan-1(vy/vx) =ta-1(1.176/8.002) =8.360degrees (measured from the +xaxis toward the +yaxis)

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