53. Physics | Kinematics | A Floating Raft and Boat Going Downstream | by Ashish Arora (GA)

53. Physics | Kinematics | A Floating Raft and Boat Going Downstream | by Ashish Arora (GA)


in this illustration we’ll discuss about
a floating raft and a boat going downstream. and the situation stated as, a boat going
downstream in a river. overcame a raft at a point ay at t equal to zero. and at t equal
to 60 minutes that is after 1 hour that boat turn back. and after sometime passed the raft
at a distance 6 kilometer from point ay. and we are required to find the flow velocity
of the river. here we can analyze this situation and let us first draw, the situation. pictorially
to understand it better. here we are having a boat. and if river is flowing, at a speed
u and the boat speed with respect to river is v then, the speed of boat would be v plus
u. here i can take, u and v in kilometers per hour to, analyze the situation easier,
because we are given time as 1 hour, and distances in kilometers. so if at t equal to zero it
is overcoming, a wooden raft which is floating and obviously the speed of raft would be,
u only that is a speed with which, river current is flowing. so this is, the location, where
the boat crosses or overtakes, a raft at t equal to zero. and after 60 minute that is
after 1 hour the boat will travel a distance which is equal to v plus u only. because if
we are taking the speeds in kilometers per hour, that means in 1 hour it will travel
a distance v plus u. and in that duration the raft will travel a distance, u. and now,
the boat return. when it return the speed of boat would be v minus u. and after time
t, somewhere here the boat and raft will meet again. so if this is time t after returning.
then this distance, travelled would be. v minus u multiplied by t, and this distance
travelled by the raft would be u t here i am taking t as, time from, the returning point.
so, in this situation. this distance is v minus u t. and this distance is u t and this
distance is u. here if this point is taken as ay and this point is taken as b here, we
can write. the distance. ay b in, figure is. this distance ay b we can write it as, v plus
u which is the total distance minus. this should be v minus u multiplied by t and that
should be equal to 6 kilometer which is already given here, that it passes the raft at a distance
6 kilometer from point ay that is the point b which we are denoting. so in this situation
here, if we further simplify it here you can see what we are getting. in this situation.
this will give us t, is the time in which it will overtake, so. here we can write, this
as, v, minus v t. plus u. plus u t is equal to, 6 kilometers. and we also have 1 thing.
this distance, which is u plus u t. so here i can write u plus u t also equal to. 6 kilometer
which gets cancelled out. and here we are getting v is equal to v t. which further simplifying.
simplify is to t is equal to 1 hour. this implies if we substitute the value, of 1 hour
in, this, distance. because, again we can write, ay b is equal to u plus, u t. which
is equal to 6 kilometers. if we substitute t as 1 hour this will be 2 u is 6 kilometer.
this implies the value of u is equal to 6 by 2, so this will be 3 kilometers per hour.
this would be the answer to our problem.

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