# A skier has an deceleration of 2.5 m/s^2. How long does it take her to come to a complete stop from a speed of 18 m/s?

May 28, 2018

It takes her 7.2 seconds to come to a complete stop.

#### Explanation:

The skier is slowing from a speed of $18 \text{ m/s}$ to $0 \text{ m/s} ,$ at a rate of –2.5 " m/s"^2. Her speed is decreasing by $2.5 \text{ m/s}$ every second.

Think of it like this: she starts with $18$ units of velocity, and every second, she loses $2.5$ units of velocity.

At this rate, how many seconds will it take for her to lose all $18$ units of velocity?

This we can write as an equation:

$\left[\left(2.5\right) , \left(\text{units/second")] xx [(t),("seconds")] = [(18),("units}\right)\right]$

or simply

$2.5 t = 18$

Solving for $t$:

$\frac{\cancel{2.5} t}{\cancel{2.5}} = \frac{18}{2.5}$

$t = 7.2$

So it will take 7.2 seconds to lose all 18 units of velocity.

This process is more commonly represented in Physics by the following equation:

${v}_{f} - {v}_{i} = a t$

where

• ${v}_{f}$ is the $f$inal $v$elocity,
• ${v}_{i}$ is the $i$nitial $v$elocity,
• $a$ is the $a$cceleration (negative when slowing down), and
• $t$ is the $t$ime taken.

Using this equation (with units) on this problem gives

0" m/s" - 18" m/s" = (–2.5 " m/s"^2)t

(–18" m/s")/(–2.5 " m/s"^2) = t

7.2" "cancel"(m/s)"/(cancel"(m/s)""/s") = t

$7.2 \text{ s} = t$

which is the same answer as before.