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

The distance is 2077.77m

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