IIT JEE , AIEEE Forum - Physics , Mechanics

joyfrancis

A man's velocity wrt a river is always 5km/h. The shortest time in which he crosses the river is 15min. Find the velocity of the river (width=1km)

kavitha

Is the velocity of river 1 km/h?

pinnaka

I think u are problem concenns about going from a point A to a point B directly on the A
For it to happen we need travel at angle a to the vertical
now there is displacement only in the y direction
Now in the x direction the diplacements is caused by the current and the motion of the man
the below equation describes the toatl displacement of teh boat
d(total)=d(man)+d(current)=(-5sinat+vt)i+5cosat j
now since we know that the displacement in the x direction is zero
-5sina = v-----1
but we also know that
5cosat =1----2
now we know time taken
that t=1/4hr

now solve 2 for a and plug it in the equation 1 u will get the answer

---Hope that helps

vkvashistha

Man can cross river in short time only when it cross river by shortest river. Hence velocity = dis/time = 1*60/15 = 4

Now Apply pythagoras u'll get answer = 3Km/hr

nesita

it will be 1 km/hr

joyfrancis

hello pinnaka.

You have made a mistake, the problem says that he crosses the river in shortest time=15m, so it is GIVEN that he travels due north.

There will be some displacement in the x.direction also because of the river velocity.

You have solved for the case when the man reaches directly opposite to the starting point, but that is not asked.

thanks....

nigitha_17

the ans is 1kmph

magiclko

see the attached figure,

for the man to go in shortest time, he should row from frm P to Q, i.e. it shud go along the perpendicular way...

and for this thr shud be no net horizontal velocity of the boat... thus

V sin theta = U ,

whr V is the boat's velocity = 25/18 m/sec

now time taken is given, so we can calculate the velocity of man along PQ which is 20/18 m/sec

i.e V cos theta = 20/18

so we can easily calculate theta and hence V sin theta also, which is equal to U(river's velocity)

pmcsubir

the answer is 0

joyfrancis

u mean to say that the river is not flowing, so he'll reach the opposite end in 1/15hrs or 4min but it is given 15min. so your answer is incorrect..

@magiclko

you have made the same mistake as pinnaka.It is not mentioned anywhere that he will reach the directly opposite point..READ CAREFULLY.

To cover the distance in shortest time he will go "due north" and WILL have some displacement in the x dir.

V sin theta need not be equal to 0

KAB

The man will cross the river in shortest time when he moves perpendicular to the direction of flow of river.

Let v be the velocity of man in still water and v_r be the velocity of river.

Given,

v²+v_r²=25

Also v=1/(1/4)=4km/hr

So v_r=3km/hr

pinnaka

Well there seems to be something fundamentally wrong with the Question itself

@Kab

Now in a river the motion of the boat is due to velocity of the boat relative to the river not relative to the shore. Hence Ur solution is wrong
However the ans from my method is also 3km/hr
calculate it

@joy

In your first comment u have said that the man moves perpendicular to the flow of the river. Before trying out the solution I tried out the same idea that the man should move perpendicular to the river. But I found that there were several unknowns involved and I discovered that the idea was not going to work.

Here is the explanation for why ur suggestion is incorrect

Now as we now that the sum of v_br + v_rg = v_bg
now v_bg is inclined at angle theta to the horizontal

now the time needed to cross the river is
t = d/ v-bgsin theta---1
v_bg sin theta has to be equal to v_br
Plugging in the values and solving for v_bgsin theta u get v_bg sintheta = 4km/hr
now v_bgsin theta is not equal to v_br
so that is not a possible way to do trhe probelm

I also tried for plugging v_rg = 1km/hr
then u r t= 1hr

so that is not a possible way to do the problem using the given data

My previous correctly depicts the time needed from going A to B (on head)

---Hope This Helps
---Stuart Anderson

shobna

ans is 3 km\hr
for man to reach within shortest time he would have to reach the opposite pt. therefore, he'll make an angle theta with the vertical. now his vertical component vel=vcos theta=5cos theta.
now, 4 =5 cos theta. therefore cos theta=4\5. hence sin theta =3\5.
but he shud not hav horizontal vel . hence, vel of river is 3 km\hr.
correct me if my concept is wrong.

KAB

Hey pinnaka my soln is CORRECT!!!!!!!!!!!!!
I've considered the relative velocity of boat WRT RIVER AND NOT WRT SHORE.
Since both are perpendicular v^2+Vr^2=5^2
You are simply making the problem complicated.

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