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.
A) What happens to the time interval required for the package to stop?

Answer 1

Answer:

increases by a factor of $\sqrt{2}$

Explanation:

First we need to find the initial velocity for it to stop at the distance 2d using the following equation of motion:

$v^2 - v_0^2 = 2a\Delta s$

where v = 0 m/s is the final velocity of the package when it stops, $v_0$ is the initial velocity of the package when it, a is the deceleration, and $\Delta s = d$ is the distance traveled.

So the equation above can be simplified and plug in Δs = d, $v_0 = v_1$ for the 1st case

$-v_1^2 = 2ad$(1)

For the 2nd scenario where the ramp is changed and distance becomes 2d, $v_0 = v_2$

$-v_2^2 = 4ad$(2)

let equation (2) divided by (1) we have:

$\left(\frac{v_2}{v_1}\right)^2 = 4ad / 2ad = 2$

$v_2 / v_1 = \sqrt{2}$

$v_2 = \sqrt{2}v_1$

So the initial speed increases by $\sqrt{2}$. If the deceleration a stays the same and time is the ratio of speed over acceleration a

$t = v / a$

The time would increase by a factor of $\sqrt{2}$