Question

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.

Answer 1

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

now the distance that black mamba cover to catch prey

$d = v*t$

$d = 5 * 2.5 = 12.5 m$

Now it returns to its hide in 12 s

so average speed of return is given as

$v_{avg} = \frac{12.5}{12} = 1.04 m/s$

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

$v = \frac{25}{14.5}$

$v = 1.72 m/s$

Answer 2

Answer:

A. 1.04 m/s

B. 0

C. 1.72 m/s

Explanation:

A. Displacement by Mamba in 2.50 s moving with the velocity 18 km/h (= 5 m/s) is

D = v t

D = (5 m/s) (2.5 s) = 12.5 m

On the return journey to the hideout, Mamba will cover the same displacement in 12.0 s. Thus, average velocity during it's return journey is

$v_{avg} =\frac{12.5 m}{12s}= 1.04 m/s$

B. For the complete trip, since initial and final position are same, the total displacement is zero, the average velocity for the round trip would be zero.

C. Average speed of the complete trip can be calculated by dividing the total distance by total time taken.

Total distance = 12.5+12.5 = 25 m.

Total time taken = 12 +2.5 = 14.5 s

Therefore, average speed of the complete trip of mamba is (25 m) /(14.5 s) = 1.72 m/s.