As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward
the truck, 15 m back to retrieve his wind-blown hat, and another 20 m to reach the truck. He walked a total
time of 75 s.
What is Justin’s average velocity over the 75 s period? What is Justin’s average speed over the 75 s period?
Round answers to two significant digits.
Answer using coordinate system where rightward is positive.

Respuesta :

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is   [tex]v =0.333 \  m/s [/tex] in positive x -direction

The speed is [tex]s = 0.733 \ m/s[/tex]

Explanation:

From the question we are told that

The distance from the house to truck is  D =  20 m

  The distance traveled back to retrieve  wind-blown hat is  d =  15

  The distance from the wind-blown hat position too the truck is  k =  20  m

  The total time taken is  t  =  75 s

Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

      [tex]L = 20 - 15 + 20[/tex]

=>    [tex]L  =  25 \ m [/tex]

Generally the average velocity is mathematically represented as

          [tex]v = \frac{L}{t}[/tex]

=>      [tex]v = \frac{25}{75}[/tex]

=>      [tex]v =0.333 \  m/s [/tex]

Generally the distance covered by Justin is mathematically represented as  

         [tex]R = D+ d + k[/tex]

=>      [tex]R = 20 + 15 +20[/tex]

=>     [tex]R =  55 \  m [/tex]

Generally Justin's average speed over a 75 s period is mathematically represented as

            [tex]s = \frac{R}{ t}[/tex]

=>         [tex]s = \frac{55}{ 75}[/tex]

=>        [tex]s = 0.733 \ m/s[/tex]

Ver imagen okpalawalter8