Three runners are competing in a 120-m race. Each begins when the starting signal sounds and all three runners cross the finish line at the same time. The first runner moves at a constant speed of 6 m/s. The second runner sprints at a constant speed of 7.5 m/s, stops after 60 m to rest for 4 s, then runs the second half of the race at a constant speed of 7.5 m/s. The third runner dashes 60 m at a constant speed of 12 m/s but realizes they forgot their keys, so runs back to the starting line at a constant speed of 12 m/s, then turns around and runs straight to the finish line at a constant speed of 12 m/s. Which of the following statements about this situation is NOT true?
Assume all average velocities and average speeds are for the period from the starting signal to the moment the runners cross the finish line. You may select more than one option.
a. The average speed of the third runner is not equal to his average velocity
b. The average velocity of the third runner is zero.
c. The first runner's average velocity is equal to the second runner's average velocity
d. The second runner's average speed is less than 7.5 m/s.
e. All three runners have the same average velocity.

Respuesta :

Answer:

The correct answer is B is false

Explanation:

This problem highlights the difference between vectors (speed) and scalars (speed)

average speed is defined by

      v = Δx / Δt

bold indicates vectors

the average speed is

        v= Δx / Δt

Let's calculate

first runner

         Δt = Δx / v₁

         Δt = 120/6

         Δt = 20 s

Second runner

          Δt₁ = dx1 / v2

          Δt₁ = 60 / 7.5

          Δt₁ = 8 s

          v₂ = 0

          Δt₂ = 4 s

          Δt₃ = dx₃ / v₂

          Δt₃ = 60 / 7.5

          Δt₃ = 8 s

the displacement is Dx = 60 + 60 = 120 m

time is t_total = Δt₁ + Δt₂ + Δt₃

 

the average speed is

                    v = Dx / Dt

                    v = 120 / (8 + 4 + 8)

                    v = 6 m / s

speed has the same value

third runner

           Δt₁ = 60/12

           Δt₁ = 5 s

       

Get back to the finish line

          Δt₂ = (-60) / (- 12)

          Δt₂ = 5 s

        Δt₃ = 120/12

        Δt₃ = 10 s

The displacement (vector) is

          dx = 60 -60 + 120

          dx = 120

Time is t_total =  Δt₁ + Δt₂ + Δt₃

         

veloicity is

           v = dx / t_total

            v = 120 / (5 + 5+ 10)

            v = 6 m / s

the radidity is

 the distance (scalar) traveled

          dx = 60 +60 + 120

          dx = 240 m

time t_total = dt1 + dt2 + dt3

         the speed

          v = 240 / (20)

          v = 12 m / s

with these results we can review the final statements

a) True The speed is twice the speed

b) False The velocity is 6 m / s

c) True the speeds from the three runners are equal

d) True the speeds of all runners are equal

The correct answer is B is false