Question

At the end of a delivery ramp, a skid pad exerts a constant force on a package so that the package comes to rest in a distance d. The ramp is changed so that the same package arrives at the skid pad at a higher speed and the stopping distance is 2d. What happens to the time interval required for the package to stop?

Answer 1

t₂> t₁

Step-by-step explanation:

When the package reaches the braking ramp it gives a constant acceleration, which slows it down in distance d, let's use the kinematic equations to find this acceleration ration, when the package stops or speed is zero

v²= v₀₁² + 2 a₁ d

0 = v₀₁² + 2 a₁ d

a1 = v₀₁² / 2d

This constant acceleration since it depends on the characteristics of the braking section,

Let's look for the time it takes to stop

v = vo - a₁ t

t = v₀₁ / a₁

Now let's calculate the time for the second package

t₂ = v₀₂ / a₁ (1)

As the initial velocity in the second case is greater and the acceleration is constant, the time must increase

t₂> t₁

We can calculate this value, write the equation for the two cases

v² = v₀₁² + 2 a₁ d

v² = v₀₂² + 2 a₁ 2d

v₀₁² + 2 a₁ d = v₀₂² + 2 a₁ 2d

v₀₂² - v₀₁² = 2 a₁ (2d - d)

v₀₂² = v₀₁² + 2 a₁ d

We substitute in 1

t₂ =1 / a₁ RA (v₀₁² + 2 a₁ d)

t₂ = RA (v₀₁₂ / a₁² + 2d / a₁)

t₂ = Ra (t₁² + 2 d / a₁)

t₂> t₁