Please help
In a park, a man and a woman are walking toward one another each at a velocity of 1.2m/s. Their dog is running back and forth between the two incessantly . It is running at a velocity of 6.8m/s. At time t=0s, the distance between the man and woman is 150m and the dog is beside the woman. When the man and the woman reach one another, they both stop walking and their dog stops running.
a) what is the woman's displacement?
b) what is the distance travelled by the woman?
c) what is the dogs displacement?
d) what is the distance travelled by the dog?
In a park, a man and a woman are walking toward one another each at a velocity of 1.2m/s. Their dog is running back and forth between the two incessantly . It is running at a velocity of 6.8m/s. At time t=0s, the distance between the man and woman is 150m and the dog is beside the woman. When the man and the woman reach one another, they both stop walking and their dog stops running.
a) what is the woman's displacement?
b) what is the distance travelled by the woman?
c) what is the dogs displacement?
d) what is the distance travelled by the dog?
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Since the problem asks for the distance and the displacement, let's make sure we understand the difference. Say the women starts at the origin and the man is at 150 m mark on the x-axis. Since they walk at the same velocity, they will meet at the point x = + 75 m
a) The woman's displacement = + 75 m (this is a vector from the starting point to the end point)
b) The women traveled 75 m (this is a scalar)
c) The dogs displacement = +75 m [vector from the start point(dog beside the women) to the end point (dog beside the woman again)]
d) calculating the distance traveled by the dog is a bit tricky. The woman walked 75 m at the speed of 1.2 m/s, therefore it took her 62.5 s to walk that distance. At this time the dog was running constantly between the woman and the man at the speed of 6.8 m/s, therefore the distance covered by the dog is 425 m (d = vt)
a) The woman's displacement = + 75 m (this is a vector from the starting point to the end point)
b) The women traveled 75 m (this is a scalar)
c) The dogs displacement = +75 m [vector from the start point(dog beside the women) to the end point (dog beside the woman again)]
d) calculating the distance traveled by the dog is a bit tricky. The woman walked 75 m at the speed of 1.2 m/s, therefore it took her 62.5 s to walk that distance. At this time the dog was running constantly between the woman and the man at the speed of 6.8 m/s, therefore the distance covered by the dog is 425 m (d = vt)
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the man and women are both walking at 1.2 m/s . since they are walking at the same speed they will meet half way which is 150/2 = 75 m which is the displacement of both the man and women from their initial spot. the dog however keeps moving back and forth therefore cancelling the displacement .
the distance can be calculated simply by finding out how long it took for the man and women to meet which is 75/1.2 = 62.5 sec and multiplying that by the dogs velocity which would result in 62.5*6.8 = 425 m
the distance can be calculated simply by finding out how long it took for the man and women to meet which is 75/1.2 = 62.5 sec and multiplying that by the dogs velocity which would result in 62.5*6.8 = 425 m