Answer:
a) Ant B was 173.2 cm away from ant M along the northern line/in the northern direction.
b) Ant B was 257.8 cm away from ant M along the northeastern line/in the northeastern direction.
Explanation:
Let the speed of ant M be x
The speed of ant B = 2x
Let the distance between the two ants be d
The diagram of the situation described in the first part of the question is drawn in the first image attached to the question.
The position of ant B, ant M and the crumb of bread form a right angled triangle as shown in the first image attached.
Using Pythagoras theorem, the distance between ant B and the crumb of bread is given as
√(d² + 100²]
Speed = (distance)/(time)
Time = (distance)/(speed)
It is given in the question that the time taken for the two ants to reach the crumb of bread is exactly the same.
For ant M
Time = (100/x)
For ant B
Time = [√(d² + 100²)]/2x
Time = Time
[√(d² + 100²)]/2x = 100/x
√(d² + 100²) = 200
(d² + 100²) = 200²
d² = 40000 - 10000 = 30000
d = 173.2 cm
b) Determine the distance between the two ants if ant B is directly northeast from ant M
The image of the description is shown in the second attached image to this solution.
The distance between ant B and the crumb of bread can be obtained using cosine rule
Distance between ant B and the crumb of bread = √[d² + 100² - (2×d×100×cos 45°)]
The distance gives √[d² + 100² - 141.42d]
Using the speed, distance, time relation used in (a)
Time for ant M to reach crumb of bread
Time = (100/x)
Time for any B to reach crumb of bread
Time = (√[d² + 100² - 141.42d])/2x
They both reach the crumb at the same time
Time = Time
(√[d² + 100² - 141.42d])/2x = 100/x
√[d² + 100² - 141.42d] = 200
d² + 100² - 141.42d = 200²
d² - 141.42d + 10000 - 40000 = 0
d² - 141.42d - 30000 = 0
Solving the quadratic equation
d = 257.8 cm or - 116.4 cm
Since distance cannot be negative,
d = 257.8 cm