river problem
i tried doing this question but i ended up getting a hard equation in terms of sin0 and cos0 and on proceeding further i got stuck again
136 Replies
ans is 120 degree wrt to stream
@Gyro Gearloose
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the ans i get will be 180 - 0
0 = theta
A motor boat is to reach at a point 30∘ upstream on other side of a river flowing with the velocity 5 m/s. Velocity of motor boat with respect to water is 53√ m/sec. The driver should steer the boat at an angle
wrt to water speed is 5root3
what is the given ans
120 degree wrt to the direction of stream
💀
alr
got nowhere near that
By taking components.
I'm sorry, I made a mistake
isnt 30 deg the angle between perpendicular and vel vector
Tantheta = -1/root3
I did have a confusion with this due to the questions wordings.
vB-vR = 5√3 na?
Vector diff rather
ye
so vb = 5(root3 + 1)
thats what i did
u cant do that ig
bruh why
these are magnitude of vectors
dirn mein alag alag hoga so maybe actual vector ka magnitude nikale toh maybe its diff
not sure tho
it says Velocity of Boat wrt to river we wont take angle its just velocity
the direction is introduced later
wait i am confused
oh wait right
its wrong
then what do we do
The whole question is at the end to find the angle right ? It's actually very different sort of screenshot
yes i also wrote the question
A motor boat is to reach at a point 30∘ upstream on other side of a river flowing with the velocity 5 m/s. Velocity of motor boat with respect to water is 53√ m/sec. The driver should steer the boat at an angle
wrt to water speed is 5root3
Okay lemme try
Oh my god im getting 60°
The hell
can u send the solution
maybe its a minor mistake
Wait lemme go through again because I did half of it in mind
@Gamertug I got stuck in this math identity can you solve it?
i just want the basic of what to do
i will do calcn no problem
Rest is solved
Resolution in x and y helped me
i dont understand what u did under boat wrt to ground
Solving in x and y axis
With respect to ground
The speed of boat
i still dont get it
thats the speed of Vbw
See the speed of boat with respect to water and ground are different and I just resolved it
it wont give speed of boat and we also dont know which direction boat is going initally
An made a equation for it
Well I assumed it to be upstream hence the - Vr
It's not fully on a axis but has a angle
Because it has to reach the 30° point
Vbw will be 30 wrt to stream
i get it
no wait
ok
i think i understand
waht u did
u first took components then applied the concept of relative velocity
Yep
also that cosTheta
isnt that for the Vbw
thats how much angle Vbw makes with stream
not Vb
Huh where
if u broke in components
u took angle of Vbw right?
Nah I've used theata for whatever is the angle that boat will need to move
then how can u break Vbw using that
I don't know I just smh did I am no physics master but I've solved relative velocity first time and I'm happy
either i dont understand it completlely or its wrong idk
Dunno myself
the more i think abt this question the more confused i become
But I'm stuck in this equation for theta
3sintheta +1 = √3 costheta
So I don't even know if I am right
doesnt give the right ans
Shit
Yeah I'm physics dumb
My suggestion would be, to make equations in x and y by taking components. Simple, yet effective.
Tell me what did you not understand here, I'll help you.
how would we do that
like take components of what
I got answer as arctan(3(2+√3))
.....
No wait nvm
Crap
@Gamertug what was the answer?
its 120 degree wrt to the direction of the stream
Getting 120
Yup
With x axis
ok lemme see
I think this method was overkill.
There's gotta be a simpler way
what exactly is u
which at 30 degree?
speed of boat?
u is velocity along the path AB
Ye
oh boats speed component along AB?
wait but shouldnt the Veloicty at 30degree be the net velocity so (Vb + Vw)^1/2 like we vectoraly add them
Haan. It needs to have at least 5 along X because wrt to A and B, it needs to be only moving along the path between.
No, that's where you're wrong.
what why
if net veloicty is anywhere else it wont reach the 30 degree mark
?
I didn't get this.
it says
the boat has to land at shore
like at 30 degree
Yeah. And it is 30°
so the NET Velocity of The BOAT and River
has to be along the 30 degre
and no where else
Because the points A and B are always at that angular separation 30°
yes
and net the velocity should point there , i dont get what u did
My brain ain't braining wait.
@Nimboi Pls explain
do i send the soln
i will just send
That's exactly what I did.....
what
and i dont understand niether
Yeah, look at the bottom half of what I sent.
i mean i will do the question i just want to know how to start
i still dont get what u did in starting
Y'know what, I think I don't get it either. I'll sleep and try again in the morning.
lol ok
Try explaining rather
Not the question
lol shall we close this out?
oh its still open?: my b
this is chill no kya hua?
what opt did seem right too
https://www.toppr.com/ask/question/a-motor-boat-is-to-reach-at-a-point-30o-upstream-on-the-other-side/
does this help?
Toppr Ask
A motor boat is to reach a point 30^o upstream on the other side of...
Click here👆to get an answer to your question ✍️ a motor boat is to reach at a point 30o upstream on the other side
Lets break it down (take time to be 1 sec, similar triangles, should work):
1. Upstream, ie direction of water flow is opp that of boat, hence the x axis distance on the triangle is 5 m (b).
2. The boat covers a distance greater than the distance at the 30 deg angle, that distance is 5 rt 3 (a).
3. the 120 deg angle (A) comes from sum of interior angles = exterior angle
4. Now we need to find the angle opposite to b, ie B
Then sine rule got used to solve. Makes sense?
a/sin A = b/sin B
bas :)
so theta = 30 degrees.
We good?
@Gamertug
shouldnt the situation be more like this
theta is what we have to find
realistically speaking to reach at a point on the other bank at 30deg , Vbr should be along the line joining those very two point
the angle (wrt normal ) of Vb should be def more than 30 cuz as the boat moves water will exert force so as to move the boat along that line
Vr is vel of river and Vb vel of boat
the theta we get from this should tell us that -
driver must steer the boat at an angle theta wrt line joining initial and final points
driver must steer the boat at an angle theta + 30 wrt normal on river
driver must steer the boat at an angle 120+theta wrt stream velocity
It has asked the angle with the stream, not with the velocity of man w.r.t river.
why Vbr has to be along that line , shouldnt V net be along that line
i dont understand that diagramn
thats kinda what i wanna convey , theta se sab aa jayega
velocity of boat perpendicular to line joining initial and final points should be zero
Vb is moving at an angle , water exerts a force on the boat which makes it shift the dirn
thus net dirn being diff
what is going on here
what part of the question is tricky
wording?
I got the answer, but not the words to explain my solution
lemme try
lmao fair, lemme try on my own rq
did you assume here that the velocity of the river and the destination was along the same direction for the boat?
cuz it says 30 deg upstream, not downstream
It's upstream, yeah.
you wrote 5m/s towards left and the boat component seems to be going, well, along the river
what did you do there
No, look at the very first sentence. This is the river frame of reference
is my approach wrong??
@Gamertug Basically this and then just apply sine rule to find theta.
the problem is ques doesnt mention "still"
it should be velocity of boat wrt "still" water is 5root3
In my opinion, velocity w.r.t water or velocity in still water, have the same meaning, which is what you are trying to tell.
absence of "still" does make the ques ambiguous tbh
why does still water make any diffrence
wont that just be the normal velocity of boat then
wait , i am confused about why is vBR at 30degree
shouldnt it be Net velocity of boat and river combined
Oh right, sorry. Just switch vbr and vb there ig.
what wil taht do
I mean, the labeling is sorta wrong but:
We need to have the boat go 30 degree from perpendicular to the bank so,
Velocity of river is 5m/s, and that of the boat in still water is 5sqrt(3). For the boat to go 30 degree to perpendicular, the vector sum triangle must be as shown in the figure with v_br (as labeled in figure) the net velocity of the boat
Similarly v_b must be the velocity of boat w.r.t the river, which is the direction where he needs to be steering
exactly
then expression will turn very ugly very quick
and ans will be in arctan smthing , def not a good value
Velocity of river is horizontal i got it , but how do u know that Vbr Is 30 degree to the perpn
Becase the question states "30 degrees upstream"
ye so if Velocity of boat relative to water is 30degrees to perpendicualt
how will it reach
wait
ohhhh
i finally get it
...So is this resolved?
i will try to solve it tmrw
correct
Well if we take this diagram
And find theta
Since we have to fins it wrt to flow of river it will be 90 + 30+ theta
And ans is 120...
Theta comes out to be in this case 30 degree, which means the third angle of the triangle is 60 degree. So, the angle between the flow of river and direction of steering when joined tail to tail is 180-60=120 I think
if theta is 30
30+30 = 60 and flow of river is in +x direction
so that should be 60 + 90
150
We have to find the angle between (as ;abeled in the figure) v_b and v_r.
ye
that will be 150 if theta is 30
@SirLancelotDuLac
Oh right...
Wait in this triangle the required angle absolutely can't be 60 can it?
which triangle
The one with the vector arrows
wait a se
c
in this diagrma
they took Vb
at 30degree
and Vbr as hypotenus
Yeah
and in this diagram its opp
Again, the labelling was wrong in that one
Precisely, its sorta wrong
but if Vb is at 30degree it will never reach shore at that point
Yeah the answer should be 150 degrees.
You can check assuming we are steering the boat at 150 degree to river and then finding what angle the boat goes to
@Gamertug
The last line is sort of checking the answer. But yeah it should be 150 degrees not 120
ic
i will try again
exactly , the wording is very vague. search up the ques online and ull see it as a multi correct ques with options which actually make sense
Toppr Ask
A motor boat is to reach a point 30^o upstream on the other side of...
Click here👆to get an answer to your question ✍️ a motor boat is to reach at a point 30o upstream on the other side
same link which sir shared
Ah yes, the direction of boat steering would be 30 degree, but as shared above the answer given, i.e., 120 degrees is absolutely wrong.
broskis
please :()
dont use this to chat :(
interesting!
So you are saying 150 to horizontal, and the phi being found out is basically 30, ie 60 to the normal, right?
haan this
@Gamertug , close this out? :D
I will close it after today
samajh me aaya kya karna hai right?
ye
+solved @SirLancelotDuLac @hardcoreisdead
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