A train goes 25000 m East in 1500 sec and then turns around and travels 29,000 m West for 1740 sec. Determine the distance, speed, displacement, and velocity.

scalar d = 54000 m , v_average = 16,667 m / s

vector    d = -4000 m , moved 4000 to the west

                 v_average = 0                

Explanation:

his is a uniform motion exercise, but we must be careful with quantities that are scalars and vector quantities

The distance traveled is a scalar

       d = d₁ + d₂

       d = 25000 + 29000

       d = 54000 m

the speed is a scalar, in the exercise it is not specified if the speed of each trajectory or the average speed. Therefore we will calculate the two

       v₁ = d₁ / t₁

       v₁ = 25000/1500

       v₁ = 16,667 m / s

       v₂ = 29000/1740

        v₂ = 16,667 m / s

Since the two speeds are equal, the average speed is

       v = (v1 + v2) / 2

       v_average = 16,667 m / s

now let's calculate the displacement that is a vector, so it has direction in addition to modules

suppose the eastward direction is positive and the bold are vectors

      d = d₁ - d₂

       d = 25000 - 29000

        d = -4000 m

this means that it moved 4000 to the west

velocity is a vector, we assume positive eastward movement

        v₁ = 16,667 m / s

         v₂ = - 16,667 m / s

         v_average = (v1 -v2) / 2

         v_average = 0

truck y has a greater velocity than truck x.

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