# Answer to Question #26502 in Mechanics | Relativity for Bhawani

Question #26502

A hiker sets out on a trek heading [N35degreesE] at a pace of 5.0km/h for 48.0 min. He then heads west at 4.5km/h for 40.0 min. Finally, he heads [N30degreesW] for 6.0km/h, until he reaches the camground 1.5 hours later. What is his total displacement?

Expert's answer

In terms or vectors {x, y}, the displacement is

{(5.0km/h)*(48.0 min)*sin(35 deg), (5.0km/h)*(48.0 min)*cos(35 deg)} + {-(4.5km/h)*(40.0 min), 0} + {-(6.0km/h)*(1.5 hours)*sin(30 deg), (6.0km/h)*(1.5 hours)*cos(30 deg)}

The displacement is sqrt( ( (5.0km/h)*(48.0 min)*sin(35 deg) - (4.5km/h)*(40.0 min) -(6.0km/h)*(1.5 hours)*sin(30 deg) )^2 + ( (5.0km/h)*(48.0 min)*cos(35 deg) + (6.0km/h)*(1.5 hours)*cos(30 deg) )^2 ) = 12.2 km.

Answer: 12.2 km

{(5.0km/h)*(48.0 min)*sin(35 deg), (5.0km/h)*(48.0 min)*cos(35 deg)} + {-(4.5km/h)*(40.0 min), 0} + {-(6.0km/h)*(1.5 hours)*sin(30 deg), (6.0km/h)*(1.5 hours)*cos(30 deg)}

The displacement is sqrt( ( (5.0km/h)*(48.0 min)*sin(35 deg) - (4.5km/h)*(40.0 min) -(6.0km/h)*(1.5 hours)*sin(30 deg) )^2 + ( (5.0km/h)*(48.0 min)*cos(35 deg) + (6.0km/h)*(1.5 hours)*cos(30 deg) )^2 ) = 12.2 km.

Answer: 12.2 km

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