Question

A student walk 60m on a bearing

Answer 1

Answer:

d = 234.6 m

Step-by-step explanation:

You can consider a system of coordinates with its origin at the beginning of the walk of the student.

When she start to walk, she is at (0,0)m. After her first walk, her coordinates are calculated by using the information about the incline and the distance that she traveled:

$x_1=60cos28\°=52.97m\\\\y_1=60sin28\°=28.16m$

she is at the coordinates (52.97 , 28.16)m.

Next, when she walks 180m to the east, her coordinates are:

(52.97+180 , 28.16)m = (232.97 , 28.16)m

To calculate the distance from the final point of the student to the starting point you use the Pythagoras generalization for the distance between two points:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}\\\\x=232.97\\\\x_o=0\\\\y=28.16\\\\y_o=0\\\\d=\sqrt{(232.97-0)^2+(28.16-0)^2}m=234.6m$

The displacement of the student on her complete trajectory was of 234.6m