Answer:
The average speed of the second ant is 2 cm/s
Step-by-step explanation:
The details of the ants moving on the sides of the pentagon are;
The length of the sides of the pentagon = 12 cm
The path of the first ant = A to B, B to C, then C to D
The average speed of the first ant = 3 cm/s
The path of the second ant = A to E, and E to D
The time of arrival of the second ant at the point D = The time of arrival of the first ant at the point D
Given that both ants departed at the same time, from point A, we have;
The time it takes the first ant to arrive at the point D = The time it took the second ant to arrive at the point D
The distance from A to B = 12 cm
The distance from B to C = 12 cm
The distance from C to D = 12 cm
∴ The distance traveled by the first ant from A to D = 12 cm + 12 cm 12 cm = 36 cm
Average speed = (Total distance)/(Total Time)
Total time = (Total distance)/(Average speed)
Let 't' represent the time it takes the first ant to move from point A to point D, we have;
t = 36 cm/(3 cm/s) = 12 seconds
Therefore, the time it takes the first ant to move from point A to point D, t = 12 seconds = The time it takes the second ant to move from point A to point D through point E
The total distance traveled by the second ant from A to point E, then from point E to point D, = 12 cm + 12 cm = 24 cm
The average speed of the second ant, v₂ = /t
∴ v₂ = 24 cm/(12 seconds) = 2 cm/s
The average speed of the second ant, v₂ = 2 cm/s.
