A car is driven east for a distance of 49 km, then north for 30 km, and then in a direction 35° east of north for 26 km. Determine (a) the magnitude (in km) of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.
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Hello
all angles are measured as north of east ( = above + x-axis)
the components of the single sections are
(49, 0)
(0, 30)
(26*cos55°, 26*sin55°) (35° east of north = 55° north of east).
Now add all x-components, and all y- components.
The final location is
(49+26*cos55, 30+26*sin55) = (63.912, 51.3).
The magnitude of the displacement vector is √(63.912^2+51.3^2) = 81.95 km.
The angle is tan^-1 (51.3/63.912) = 38.75° north of east ( above + x-axis) .
Regards
all angles are measured as north of east ( = above + x-axis)
the components of the single sections are
(49, 0)
(0, 30)
(26*cos55°, 26*sin55°) (35° east of north = 55° north of east).
Now add all x-components, and all y- components.
The final location is
(49+26*cos55, 30+26*sin55) = (63.912, 51.3).
The magnitude of the displacement vector is √(63.912^2+51.3^2) = 81.95 km.
The angle is tan^-1 (51.3/63.912) = 38.75° north of east ( above + x-axis) .
Regards