Question #87882

From a point A, a body travels for 2s at a constant velocity of 12m/s due north. It then travels for 5s due east whilst its velocity decreases uniformly from 12m/s to 6m/s. What is the resultant displacement from A? Also determine the average velocity, in magnitude and direction, of the body for it to return along a straight path to point A in 3s.

Expert's answer

vx,avg=12+62=9msv_{x,avg}=\frac{12+6}{2}=9 \frac{m}{s}

sy=12(2)=24ms_y=12(2)=24 m

sx=9(5)=45ms_x=9(5)=45 m

s=242+452=51m.s=\sqrt{24^2+45^2}=51 m.

The direction is


a=arctan2445=28a=\arctan{\frac{24}{45}}=28^\circ

north of east.

The direction, of the body for it to return along a straight path to point A in 3s is 28 degree south of west.

The magnitude is


V=513=17msV=\frac{51}{3}=17 \frac{m}{s}



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