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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 :

Answer:

Explanation:

Δt = Δv/a = (4.5 m/s) / (8.37 m/s²) = 0.538 s = time to reach top speed

d = Vi + 1/2at²

d = 0 + 1/2(8.37 m/s²)(0.538 s)² = 1.21 m = distance snake traveled while accelerating

t = d/v = (2.60 m - 1.21 m) / (4.5 m/s) = 0.31 s = reaction time before snake tail passes mongoose nose

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Rewritten by : Barada

According to the question of speed, ftop is 0.9578 s, dsnake is 2.14 m, freact is 0.4773 s.

What is speed?
Speed is the rate at which an object moves from one place to another. It is expressed in units of distance per unit of time, such as miles per hour or kilometers per hour. Speed is a scalar quantity, meaning that it has magnitude but no direction. Acceleration, on the other hand, is a vector quantity that has both magnitude and direction. Speed is directly related to the amount of time it takes an object to move from one point to another, and is inversely proportional to the time it takes to cover a given distance. Speed can also be affected by external factors such as wind, terrain, and friction. Speed is an important concept in physics and is often used to calculate the kinetic energy of an object.

ftop = 0.5 × (8.37 m/s²) / (4.50 m/s) = 0.9578 s

dsnake = 0.5 × (4.50 m/s) × (0.9578 s) = 2.14 m

freact = 2.14 m / (4.50 m/s) = 0.4773 s

To learn more about speed
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