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A black mamba snake has a length of 2.60 m and a top speed of 4.50 m/s. Suppose a mongoose and a black mamba find

themselves nose to nose. In an effort to escape, the snake accelerates past the mongoose at 8.37 m/s² from rest.

How much time ftop does it take the snake to reach its top

speed?

How far dsnake does the snake travel in that time?

How much time freact does the mongoose have to react before

the black mamba's tail passes the mongoose's nose?

ftop

dsnake =

freact=

S

m

S

Macmillan Learning A black mamba snake has a length of 2 60 m and a top speed of 4 50 m s Suppose a mongoose

Answer :

The black mamba travels a distance of 2.0828 m, which is less than its length of 4.50 m.

How far dsnake does the snake travel in that time?

The first equation of motion states:

vf = at + vi

A top speed is _f.

Initial or lowest speed is v i, and acceleration is a.

2.60 = 4.50 t

t +0 = 1.28571sec

The snake needs 1.28571 seconds to attain its peak speed.

Part B:

The second equation of motion states:

S=0.5(4.50) S=vit+at2 w i=0 (1.28571)

2 S = 2.0828m

In that period, the snake moves 2.0828 metres.

Part C:

The black mamba travels a distance of 2.0828 m, which is less than its length of 3.12 m.

2.0828 m<3.12 m

A potential exists for monkeys to capture a black mamba.

To learn more about black mamba refer to:

https://brainly.com/question/14748177

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