Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs downfield at 45° with respect to the 50-yard line and covers 8.0 m in 1 s. He then runs straight down the field at 90° with respect to the 50-yard line for 12 m, with an elapsed time of 1.2 s. (a) What is Matthews' final displacement from the start of the play? (b) What is his average velocity?

a)   D_ total = 18.54 m,   b)        v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

           sin 45 = y₁ / d

           cos 45 = x₁ / d

           y₁ = d sin 45

           x₁ = d sin 45

           y₁ = 8 sin 45 = 5,657 m

           x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

            y₂ = 12 m

            x₂ = 0

total displacement

          y_total = y₁ + y₂

          y_total = 5,657 + 12

          y_total = 17,657 m

          x_total = x₁ + x₂

          x_total = 5,657 + 0

          x_total = 5,657 m

          D_total =   17.657 i^+ 5.657 j^  m

          D_total = Ra (17.657 2 + 5.657 2)

          D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

           v = Δx /Δt

the distance traveled using the pythagorean theorem is

         r = √ (d1² + d2²)

          r = √ (8² + 12²)

          r = 14.42 m

The time used for this shredding is

         t = t1 + t2

         t = 1 + 1.2

         t = 2.2 s

let's calculate the average speed

         v = 14.42 / 2.2

         v = 6.55 m / s

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.