A fast moving vehicle travelling at a speed of 25.4 m/s comes up behind another vehicle which is travelling at a slower constant speed of 13.6 m/s. If the faster vehicle does not begin braking until it
is 11.4 meters away from the car in front of it, what is the minimum acceleration that the faster car
must exhibit if it is to avoid colliding with the car in front? Assume that both cars are travelling in the
positive direction

     a = 6.1 m / s²

Explanation:

For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle

Let's find the relative initial velocity of the two vehicles

          v₀ = v₀₂ - v₀₁

          v₀ = 25.4 - 13.6

          v₀ = 11.8 m / s

the fastest vehicle

          x = v₀ t + ½ a t²

The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is

          x = 11.4 m

let's use the expression

           v² = v₀² - 2 a x

how the vehicle stops v = 0

          a = v₀² / 2x

          a =

          a = 6.1 m / s²

this velocity is directed to the left

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2