Answer:
The distance is 2077.77m
Step-by-step explanation:
Given
Represent speed with s and time with t
[tex](s_1, t_1) = (2.1m/s,16min)[/tex] ---- Left
[tex](s_2, t_2) = (1.1m/s,12min)[/tex] --- South
[tex](s_3, t_3) = (3.4m/s,28sec)[/tex] ---- Right
Required
How far are you from your car?
[tex]distance=speed * time[/tex]
To solve this question, I will use the attached image to illustrate the movement.
First, we calculate distance AB
Using: [tex](s_1, t_1) = (2.1m/s,16min)[/tex] ---- Left
[tex]d_1 = s_1 * t_1[/tex]
[tex]d_1 = 2.1m/s * 16min[/tex]
Convert min to secs
[tex]d_1 = 2.1m/s * 16*60s[/tex]
[tex]d_1 = 2016m[/tex]
[tex]AB = 2016m[/tex]
Next, we calculate the distance BC
Using: [tex](s_2, t_2) = (1.1m/s,12min)[/tex] --- South
[tex]d_2 = 1.1m/s * 12min[/tex]
[tex]d_2 = 1.1m/s * 12*60s[/tex]
[tex]d_2 = 792m[/tex]
[tex]BC = 792m[/tex]
Next, we calculate distance CD
Using: [tex](s_3, t_3) = (3.4m/s,28sec)[/tex] ---- Right
[tex]d_3 = 3.4m/s * 28s[/tex]
[tex]d_3 = 95.2m[/tex]
[tex]CD = 95.2m[/tex]
The distance between you and the car is represented as AD.
Considering triangle AOD, we have:
[tex]AD^2 = AO^2 + OD^2[/tex]
Where:
[tex]AO = AB - BO[/tex]
and
[tex]BO = CD[/tex]
So:
[tex]AO = AB - CD = 2016m - 95.2m[/tex]
[tex]AO = 1920.8m[/tex]
[tex]OD = BC[/tex]
[tex]OD = 792m[/tex]
The equation becomes:
[tex]AD^2 = AO^2 + OD^2[/tex]
[tex]AD^2 = 1920.9^2 + 792^2[/tex]
[tex]AD^2 = 4317120.81m^2[/tex]
[tex]AD= \sqrt{4317120.81m^2[/tex]
[tex]AD= 2077.76822817 m[/tex]
[tex]AD= 2077.77 m[/tex] --- approximated