# Question 25d4b

Sep 2, 2016

${\text{1.72 m s}}^{- 1}$

#### Explanation:

All you have to do here is account for

• the total distance covered by the mamba as it strikes and as it retreats
• the total time it takes it to strike and to retreat

You know that the mamba strikes at a speed of ${\text{18.0 km h}}^{- 1}$ for a total of $2$ seconds. Notice that the time it takes the mamba to strike is expressed in seconds, but that its maximum speed in this interval is given in kilometers per hour.

This means that you're going to have to convert the speed to more appropriate units such as meters per second, ${\text{m s}}^{- 1}$. To do that, use

color(purple)(bar(ul(|color(white)(a/a)color(black)("1 km" = 10^3"m")color(white)(a/a)|)))" " and " "color(purple)(bar(ul(|color(white)(a/a)color(black)("1 h" = "60 min" = 60 * 60" s")color(white)(a/a)|)))

You will have

18.0 color(red)(cancel(color(black)("km")))/color(red)(cancel(color(black)("h"))) * (10^3"m")/(1color(red)(cancel(color(black)("km")))) * (1color(red)(cancel(color(black)("h"))))/(60 color(red)(cancel(color(black)("min")))) * (1color(red)(cancel(color(black)("min"))))/(60" s") = "5 m s"^(-1)

Now, use the speed of the mamba and the time of the strike to find the distance it covers as it strikes

2.5 color(red)(cancel(color(black)("s"))) * "5.0 m"/(1color(red)(cancel(color(black)("s")))) = "12.5 m"

The problem tells you that the mamba turns around and retreats for a total of $12$ seconds. This means that the total time, which includes the time it takes it to strike and the time it takes it to retreat, is equal to

overbrace("2.5 s")^(color(blue)("strike time")) + overbrace("12 s")^(color(purple)("retreat time")) = "14.5 s"#

Now, the mamba covers the same distance as it strikes and as it returns, so the total distance covered will be

$2 \times \text{12.5 m" = "25 m}$

The average speed of the mamba will be equal to the total distance it covers divided by the total time it takes for it to do so

$\overline{v} = {\text{25 m"/"14.5 s" = "1.724 m s}}^{- 1}$

Rounded to two sig figs, the answer will indeed be ${\text{1.72 m s}}^{- 1}$.