SOLUTION: respected sir, 1] A hare sees a dog 100 metres away from her,& scuds off in the opposite direction at a speed of 12 km an hour.A minute later the dog perceives her & gives chase

Question 195768: respected sir,
1] A hare sees a dog 100 metres away from her,& scuds off in the opposite direction at a speed of 12 km an hour.A minute later the dog perceives her & gives chase at a speed of 16 km per hour.How soon will the dog overtake the hare ,& at what distance from the spot whence the hare took flight ?
2] A carriage driving in a fog passed a man who was walking at the rate of 3 km an hour in the same direction.He could see the carriage for 4 min.& it was visible to him upto a distance of 100 m. what was the speed of the carriage ?
plz explain me this sums...waiting for yurs reply.
thank u.
Answer by josmiceli(6786) (Show Source):
You can put this solution on YOUR website!
The equations for both hare and dog are:

and

These are laws of constant speed motion, and
the key to not getting confused is to only
apply them when the hare and dog are moving.
I would like to have them both moving from
start to finish, but how do I do that?
I have a stopwatch and I will start it
when the dog starts running. But where is the hare
then? The hare has a 1 minute headstart, and
in that time, she goes

km
I start my stopwatch and stop it when the dog
catches the hare

The dog has to make up 100 m plus 200 m
km

The dog's time and the hare's time will now be the
same on my stopwatch, so I'll say they both are

(1)
and
(2)
Subtract (2) from (1)
(1)
(2)

hr
min
min
It takes the dog 4.5 min to overtake the hare
How far does the hare go in that time?
(1)

km
But the hare is already 200 m from her starting point, so
The dog overtakes the hare at
km
1100 m from the hare's starting point
check answer:
The dog has to go
km in .075 hr to catch the hare

OK
I don't have time for the other question, sorry