# Question #25d4b

##### 1 Answer

#### Explanation:

All you have to do here is account for

thetotal distancecovered by the mamba as it strikes and as it retreatsthetotal timeit takes it to strike and to retreat

You know that the mamba strikes at a speed of **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*,

#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 **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 xx "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

#bar(v) = "25 m"/"14.5 s" = "1.724 m s"^(-1)#

Rounded to two **sig figs**, the answer will indeed be