A dog running in an open field has components of velocity v x = 2.9 m / s and v

ID: 1325124 • Letter: A

Question

A dog running in an open field has components of velocity vx = 2.9m/s and vy = -1.6m/s at time t1 = 12.0s . For the time interval from t1 = 12.0s to t2 = 23.5s , the average acceleration of the dog has magnitude 0.52m/s2 and direction 34.0? measured from the +x?axis toward the +y?axis. A dog running in an open field has components of velocity vx = 2.9m/s and vy = -1.6m/s at time t1 = 12.0s . For the time interval from t1 = 12.0s to t2 = 23.5s , the average acceleration of the dog has magnitude 0.52m/s2 and direction 34.0? measured from the +x?axis toward the +y?axis.

Explanation / Answer

Here ,

v = 2.9 i - 1.6 j m/s

time = 23.5 - 12

t = 11.5 s

a = 0.52 * cos(34) i + j 0.52 *sin(34)

a = 0.431 i + 0.29 j m/s^2

Now , Using first equation of motion ,

vf = v + at

vf = (2.9 + .431 * 11.5) i +(-1.6 * 0.29*11.5) j

vf = 7.66 i + 1.735 j m/s

the final velocity is 7.66 i + 1.735 j m/s

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A dog running in an open field has components of velocity v x = 2.8 m/s and v y

A dog running in an open field has components of velocity v x = 2.9 m / s and v

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A dog running in an open field has components of velocity v x = 2.9 m / s and v

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