Question

You park your car, and start walking west on Robinson Street at a speed of 2.1 m/s. After 16 minutes, you turn left. It really is a beautiful day out, and so you slow your speed to 1.1 m/s and continue walking to the south on Rosalind Avenue for another 12 minutes. At this point, you realize that you have a library book on hold at the downtown branch, and decide to make a quick detour to pick it up. You turn right and jog west at a speed of 3.4 m/s for 28 seconds to cross the street before the light changes. At this point, how far (in meters) are you from your car?

Answer 1

Answer:

The distance is 2077.77m

Step-by-step explanation:

Given

Represent speed with s and time with t

$(s_1, t_1) = (2.1m/s,16min)$ ---- Left

$(s_2, t_2) = (1.1m/s,12min)$ --- South

$(s_3, t_3) = (3.4m/s,28sec)$ ---- Right

Required

How far are you from your car?

$distance=speed * time$

To solve this question, I will use the attached image to illustrate the movement.

First, we calculate distance AB

Using: $(s_1, t_1) = (2.1m/s,16min)$ ---- Left

$d_1 = s_1 * t_1$

$d_1 = 2.1m/s * 16min$

Convert min to secs

$d_1 = 2.1m/s * 16*60s$

$d_1 = 2016m$

$AB = 2016m$

Next, we calculate the distance BC

Using: $(s_2, t_2) = (1.1m/s,12min)$ --- South

$d_2 = 1.1m/s * 12min$

$d_2 = 1.1m/s * 12*60s$

$d_2 = 792m$

$BC = 792m$

Next, we calculate distance CD

Using: $(s_3, t_3) = (3.4m/s,28sec)$ ---- Right

$d_3 = 3.4m/s * 28s$

$d_3 = 95.2m$

$CD = 95.2m$

The distance between you and the car is represented as AD.

Considering triangle AOD, we have:

$AD^2 = AO^2 + OD^2$

Where:

$AO = AB - BO$

and

$BO = CD$

So:

$AO = AB - CD = 2016m - 95.2m$

$AO = 1920.8m$

$OD = BC$

$OD = 792m$

The equation becomes:

$AD^2 = AO^2 + OD^2$

$AD^2 = 1920.9^2 + 792^2$

$AD^2 = 4317120.81m^2$

$AD= \sqrt{4317120.81m^2$

$AD= 2077.76822817 m$

$AD= 2077.77 m$ --- approximated