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a man is walking across a 300 foot long field at the same time his daughter is walking towards him from the opposite end. the man is walking at 9 feet per second and the daughter is moving at 6 feet per second how many seconds will it take them to meet somewhere in the middle.

2 Answers

4 votes

Solution :

Given that, A man is walking across a 300 foot long field at the same time his daughter is walking towards him from the opposite end.

Speed of man = 9 feet per second.

Speed of his daughter = 6 feet per second.

To calculate after how many seconds both will, meet we first need to calculate their relative speed.

And as we know that, when two bodies move in opposite direction then

Relative speed = sum of their speed = 6 + 9 = 15 feet per second.

Time taken by both of them to meet some where in the middle of the 300 foot long field

distance =
(300)/(2) =150


\Rightarrow Distance\:= speed* time


\Rightarrow time =(distance)/(speed)


\Rightarrow time=(150)/(15) =10

Hence ,it will take 10 seconds to them to meet somewhere in the middle.

User Sreimer
by
5.5k points
2 votes

Let distance traveled by father be x feet

then distance traveled by daughter will be 300-x feet

speed = distance / time

and time = distance / speed

Since they both meet at the same time


(x)/(9)=(300-x)/(6)

cross multiplying gives


6x= 2700-9x


15x=2700

this gives x= 180 feet

So the distance traveled by father is 180 feet

Now time = distance/ speed

So time taken by father =
(180)/(9) = 20 seconds.

Hence it will take both of them 20 seconds to meet somewhere in the middle.

User Ramesh Dharan
by
5.1k points
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