Question

An object which starts at the origin is traveling in the positive direction with a constant velocity V. After traveling a distance D with this constant velocity, the object begins to slow down at a constant rate until it stops after traveling an addition distance of 2D. In terms of D and V, determine the average velocity of the entire trip.

Answer 1

Answer:

Explanation:

time to travel first leg

D = Vt₁

t₁ = D/V

time to travel second leg.

at constant deceleration, the average velocity will be half of the original.

2D = (V/2)t₂

t₂ = 4D/V

t = t₁ + t₂ = 5D/V

Vavg = d/t = (D + 2D)/(5D/V) = (3D/5D)V = 0.6V

Answer 2

With the concept of kinematics we find the average speed is 0.6V

given parameters

   * initial velocity V

   * the distances in each part of the movements x₁ = D and x₂ = 2D

to find

  The average speed of the entire trajectory

In kinematics we analyze the motion of bodies, the average velocity is defined by

        $V_{avg} = \frac{\Delta x}{\Delta t}$           1

where $V_{avg}$ is the average velocity, Δx and Δt are the variation of displacement and time in the interval

For this exercise we have two types of movement in the first part a uniform movement and in the second part an accelerated movement  let's solve each onaccelerated movement,e separately

1 Part. uniform motion

we look for the time

          V = $\frac{x_1}{t_1}$

          t₁ = x₁ / V

          t₁ = $\frac{D}{V}$

2 Part. Accelerated movement, let's start by looking for acceleration, they tell us that the body stops at the end of the interval so its velocity is zero v_f=0, the initial velocity is v₀ = V

           $v_f^2 = v_o^2 - 2 \ a \ x_2$

           0 = v₀² - 2 a x₂

           a = $\frac{v_o^2}{2x_2}$

we substitute

           a = $\frac{V^2}{D}$

now we look for the time it takes to stop

          v = v₀ - a t₂

          0 = v₀ - a t₂

          t₂ = $\frac{v_o}{a}$

we substitute

          t₂ = $\frac{V}{\frac{V^2}{4D} }$

          t₂ = $\frac{4D}{V}$4D / V

for the entire movement, the total displacement is

          Δx = x₁ + x₂

          Δx = D + 2D

          Δx = 3D

the total time is  

          Δt = t₁ + t₂

          Δt = $\frac{D}{V} + \frac{4D}{V}$

          Δt = $\frac{5D}{V}$

we substitute in equation 1 of average velocity

         $V_{avg} = \frac{3D}{\frac{5D}{V} }$

         $V_{avg} = \frac{3}{5} \ V$

         $V_{avg } = 0.6 V$

With the use of the kinematics equations with constant acceleration we can find the average velocity of the body throughout the journey is 0.6V

                                                               

learn more about average speed here:

brainly.com/question/11265533