A dog running in an open field has components of velocity vx = 2.5 m/s and vy =

Question

A dog running in an open field has components of velocity vx = 2.5 m/s and vy = -2.4 m/s at t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration of the dog has magnitude 0.41 m/s2 and direction 25.0° measured from the +x-axis toward the +y-axis.

(a) At t2 = 20.0 s, what are the x and y components of the dog's velocity? vx = m/s vy = m/s

(b) At t2 = 20.0 s, what are the magnitude and direction of the dog's velocity? magnitude m/s direction ° clockwise from the +x-axis

Explanation / Answer

ANSWER

vx= 2.5 m/s ; t= 10s; ax = 0.41 * cos 25 = 0.37
vy= -2.4 m/s; t= 10s; ay = 0.41 * sin 25 = 0.17

A)At time t2= 20s , what are the x- and y-components of the dog's velocity
v= u+at

vx( t2= 20s) = 2.5 + 0.37*10 = 6.2 m/s ANS.
vy( t2= 20s) = -2.4 + 0.17*10 = -0.7 m/s ANS.

B) the magnitude of the dog's velocity
= sq root [ 6.2^2 + (-0.7)^2 ] =6.24 m/s ANS.

C) the direction of the dog's velocity (measured from the +x-axis toward the +y-axis
= tan -1( -0.7/6.2) = -6.44° ANS.

Regards!!!

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