A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.60 m/s2 for 11.0 s. 2. Maintain a constant velocity for the next 2.60 min. 3. Apply a constant negative acceleration of −9.38 m/s2 for 3.05 as. (a) What was the total displacement for the trip

x_total = 4662.5 m

Explanation:

This is a one-dimensional math problem, we must find the displacement in each section and then add them together.

1) let's use the equation

       x = v₀ t + ½ a t²

where it indicates that part of rest for which v₀ = 0

       x₁ = ½ a₁ t²

       x₁ = ½ 2.60 11²

       x₁ = 157.3 m

2) The second displacement is at constant speed,

   let's find the final speed of the previous displacement

       v = v₀ + a₁ t₁

       v = a₁ t₁

       v = 2.60 11

       v = 28.6 m / s

now we use the uniform speed

      v = x₂ / t₂

      x₂ = v t₂

let's reduce the time to SI units

      t₂ = 2.60 min (60 s / 1min) = 156 s

      x₂ = 28.6  156

      x₂ = 4461.6 m

3) it is braking

       x₃ = v t₃ - ½ a₃ t₃²

       x₃ = 28.6 3.05 - ½ 9.38 3.05²

        x₃ = 43.60m

a) total displacement is the sum of each displacement

     x_total = x₁ + x₂ + x₃

     x_total = 157.3 + 4461.6 + 43.60

     x_total = 4662.5 m

c

explanation:

gravity

explanation: