11) arthur and betty start walking toward each other when they are 100 m apart. arthur has a speed of 3.0 m/s and betty has a speed of 2.0 m/s. their dog, spot, starts by arthur's side at the same time and runs back and forth between them at 5.0 m/s. by the time arthur and betty meet, what distance has spot run?

100 m

Explanation:

Arthur and Betty should be walking the same amount of time if they start walking at the same time and stop when they meet = t

Speed of Arthur = 3 m/s

Speed of Betty = 2 m/s

Distance = Speed × time

Distance covered by Arthur = 3t

Distance covered by Betty = 2t

The distance covered by both of them will be 100 m

The speed of dog is 5 m/s

Spot is running back and forth at 5 m/s for 20 seconds

In 20 seconds the distance covered by the dog is

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y