contestada

The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 14.0 ft/s at point A and 19.0 ft/s at point C. The cart takes 3.50 s to go from point A to point C, and the cart takes 1.40 s to go from point B to point C. What is the cart's speed at point B

Respuesta :

Answer:

    [tex]v_b[/tex] = 17.98 m / s

Explanation:

To solve this exercise we will use the kinematics relations in one dimension, let's start by looking for the acceleration between points A and C

         v = v₀ + a t

         a = [tex]\frac{v - v_o}{t}[/tex]

         a = [tex]\frac{19 - 14}{3.50}[/tex]

         a = 1.43 m / s²

this acceleration is uniform throughout the path.

Let's find the velocity at point B

          [tex]v_c = v_b + a t[/tex]

          [tex]v_b = v_c - a t[/tex]

 

let's calculate

          [tex]v_b[/tex] = 19 - 1.43 1.40

          [tex]v_b[/tex] = 17.98 m / s

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