When the pad is replaced by one that requires a distance of 4d to stop the same package, what happens to the length of time it takes for the package to stop?

t = 4 t₁ , We see that time is increased by a factor of 4

Explanation:

For this exercise let's use the concepts of kinematics

       v² = v₀² + 2 a d

With the first pad when the package stopped zero speed, let's look for acceleration

       0 = v₀² - 2 a₁ d

       a₁ = v₀² / 2 d

Let's use the equation to enter time

       v = v₀ - a₁ t₁

        t₁ = v₀ / a₁

When we replace the pad with another one where the distance is 4d

        a = v₀² / 2 (4d)

        a = a₁ / 4

We see that the acceleration is 4 times lower.

The equation  kinematics

       v = v₀ - at

       t = v₀ / a

We replace the acceleration

       t = v₀  4/a1

       t = t₁ 4

We see that time is increased by a factor of 4

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