Question

3. A lady walks 15.0 meters east and then 20.0 meters north.

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