Below is the solution:
Let us say that the disk goes through a vertical elevation change of one meter.
<span>The change in potential energy will equal the change in kinetic energy </span>
<span>PE = KEt + KEr </span>
<span>mgh = ½mv² + ½Iω² </span>
<span>for a uniform disk, the moment of inertia is </span>
<span>I = ½mr² </span>
<span>and </span>
<span>ω = v/r </span>
<span>mgh = ½mv² + ½(½mr²)(v/r)² </span>
<span>mgh = ½mv² + ¼mv² </span>
<span>gh = ¾v² </span>
<span>v² = 4gh/3 </span>
<span>v² = u² + 2as </span>
<span>if we assume initial velocity is zero </span>
<span>v² = 2as </span>
<span>a = v² / 2s </span>
<span>s(sinθ) = h </span>
<span>s = h/sinθ </span>
<span>a = 4gh/3 / 2(h/sinθ) </span>
<span>a = ⅔gsinθ </span>
<span>a = ⅔(9.8)sin25 </span>
<span>a = 2.8 m/s² </span>
Answer:
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