Two ants, Ant M and Ant B, often compete with each other to see which can be the first to grab a sweet morsel of food. Ant B, the faster runner of the two, always runs at twice the speed of Ant M. One day they both saw a particularly inviting crumb of bread 100 cm directly to the east of Ant M. At that moment Ant B happened to be some distance straight north of Ant M's position. At the signal "1, 2, 3, GO!" they both ran to the crumb of bread, arriving at the same moment. A tie! So they decided to share it. Your task for this problem is to determine just how far north of Ant M Ant B was at the start of this race. Bonus: What if Ant B started directly northeast of Ant M? Then how far apart were the two ants

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.

Step-by-step 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