Question

3. A lady walks 15.0 meters east and then 20.0 meters north.
a. Calculate her resultant displacement.
b. How far away is she from her starting point?
c. What distance did she travel?

Answer 1

Answer:

a. 25m

b. 25m

c. 35m

Explanation:

Given the distances walked as 15m east and 20m north:

Using Pythagoras Theorem, we calculate displacement as:

$a^2+b^2=c2, a=15,b=-20\\\\15^2+20^2=c^2\\\\c=\sqrt{15^2+20^2}=\sqrt{625}\\\\c=25m$

Hence, the resultant displacement is 25m

b. The distance from her start point is obtained by drawing a straight line from her final point to her initial point.

Given the distances walked as 15m east and 20m north:

Using Pythagoras Theorem, we calculate displacement as:

$a^2+b^2=c2, a=15,b=-20\\\\15^2+20^2=c^2\\\\c=\sqrt{15^2+20^2}=\sqrt{625}\\\\c=25m$

Hence, she is 25m away from her starting point.

c. The distance traveled is calculated by summing the total distance walked:

$d=d_1+d_2\\\\=15+20\\\\=35m$

Hence, the total distance traveled is 35m