The black mamba is one of the world's most poisonous snakes, and with a maximum speed of 18.0 km/h, it is also the fastest. suppose a mamba waiting in a hide-out sees prey and begins slithering toward it with a velocity of +18.0 km/h. after 2.50 s, the mamba realizes that its prey can move faster than it can. the snake then turns around and slowly returns to its hide-out in 12.0 s. calculate a. the mambas average velocity during its return to the hideout. by. the mambas average velocity for the complete trip. c. the mamba's average speed for the complete trip.

Speed of the black Mamba = 18 km/h = 5m/s

now the distance that black mamba cover to catch prey

Now it returns to its hide in 12 s

so average speed of return is given as

Part b)

For the whole trip we know that displacement of the mamba is zero

so average velocity of whole trip will be zero

Part c)

For average speed we know that total distance of the black mamba

d = 12.5 + 12.5 = 25 m

total time taken = 2.5 + 12 = 14.5 s

now average speed is given as

as technology advances, new experiments often expose problems in accepted theories. c

hope i

-lavira: )

i don’t know what the answer is i wish i