Question

In reaching her destination, a backpacker walks with an average velocity of 1.19 m/s, due west. This average velocity results because she hikes for 5.96 km with an average velocity of 3.10 m/s, due west, turns around, and hikes with an average velocity of 0.744 m/s, due east. How far east did she walk

Answer 1

Answer:

 x₂ = 4.455 m

Explanation:

The average speed is defined as the displacement traveled between the time interval

         v_average =  $\frac{\Delta x}{\Delta t}$

in this case we have a first stop walking west at speed v = 3.10 m / s a ​​distance of x = 5.96 km = 5.96 10³ m

Let's find the time it takes on this tour

          v = x / t

          t = x / v

          t₁ = 5.96 10³ / 3.10

          t₁ = 1.9226 10³ s

in a second to walk east at a speed of v₂ = 0.744 m / s

           v₂ = x₂ / t₂

           t₂ = x₂ / v₂

for full movement

let's assume that the eastward movement is positive

         v =$\frac{- x_1 + x_2}{t_1 +t_2}$

         v = $\frac{-x_1 +x_2}{t_1 + \frac{x_2}{t_2} }$

the only unknown term is the distance to the east. We replace and resolve

         1.19 = $\frac{-5.96 \ 10^3 + x_2}{1.92 \ 10^3 + \frac{x_2}{0.744} }$

         1.19 (1.92 10³ + $\frac{x_2}{0.744}$) = x₂ 1.92 10³ - 5.96 10³

             

          2.2848 10³ + 1.599 x₂ = 1.92 10³ x₂ - 5.96 10³

          x₂ (1.92 10³ - 1.599) = 2.2848 10³ + 5.96 10³

          x₂ = 8.5448 10³ / 1.918 10³

          x₂ = 4.455 m