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

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