IIT JEE , AIEEE Forum - Physics , Mechanics

ramyani

I can't solve this problem:

An aeroplane has to go from point A to point B which is 500 km away due 30 degree east of north. A wind is blowing due north at a speed of 20 m/s. The air speed of the plane is 150 m//s. Find the direction in which the pilot should head the plane to reach point B. Find also the time taken by the plane to go from A to B.

elessar_iitkgp

Let the plane head in a direction

North of East.
Let the velocity of the plane w.r.t be v_PW, the velocity of the wind wrt earth be v_WE & the velocity of the plane wrt earth be v_PEAlso, let the East direction be +X axis & the North direction be +Y axis.
Then,
v_{PE =}v_PW+v_WE= v_PW(sin

i + cos

j) + v_WE j = v_PWsin

i + (v_PWcos

+ v_WE )j
The plane should head along AB. So,
v_PWsin

/ (v_PWcos

+ v_WE )= tan30
150sin

/ (150cos

+ 20 ) = 1/

3
150(

3sin

- cos

) = 20
300sin(

-

/6 ) = 20

= sin^-1 (1/15) +

/6
Hence the angle made by the plane with the line AB ,

= (

-

/6 ) = sin^-1 (1/15)

Speed along AB = |v_PE| = v_PW cos

+ v_WEcos30 = 150

(1-1/15²) + 20 (

3/2)
= 40

14 + 10

3 m/s
Time taken to travel from A to B
=50000/(40

14 + 10

3) s

phyana

elessar has worked a bit hard to explain his steps...........i think we can do it in a simple manner.
this is a twisted river - boat problem explained in a different maner.

since there is wind-20m/s the plane should mave in such an angle such that the resultant is in the direction of B

tan( theta ) = [ Psin(aplha) ] / [ pcos(alpha) + q ]
p-150
q=20
alpha---angle bet two velocity.
theta ---angle between wind velocity and resultant==30

thus u get an alpha
now the time taken would be== 500*1000/150
becauz the verticall component(20m/s) only takes it vertically.it does not accelat=rates it horzontally.so the time taken to reach any vertical point along B the time would be the same

if u dont mind plz rate me

elessar_iitkgp

@phyana
Your calculation of time is incorrect for speed along the line AB isn't 150 m/s

nitin62225

altenate method:

to reach point B the net velocity should be in directin A-->B i.e the component of individual velocities perpendicular to AB should be zero.

so let the plane moves at an angle

further east with AB

so V_planesin

=V_windsin(30)

=arc sin(1/15)

so the plane moves at angle +30^o east of north.

and time taken can b calculated by distance/(velocity along AB)

i.e t=500000/(20 cos(30)+150 cos(arc(sin(1/15))) in seconds

punterjack

let us use vectors to indicate the velocities.

v=xi+yj

root(x2+y2)=150--------------------1

resultant velocity vector=xi+(y+20)j

as $=60degs

(y+20)/x=root(3)---------------------2

solve for x and y magnitude of initial velocity=root(x2+y2)

initial direction($)=tan inverse (y/x)

time taken=500/(root(x2+(20+y)2))

phyana

hey ramani what is the time taken to reach there
is nithin,elessar,punterjack correct??????

i think the time taken should be---500*1000/150.

elessar i shall tel u y i told so

IN RIVER BOAT PROBLEM WE CAN SEE THAT THE TIME TAKEN BY THE BOAT TO REACH THE OPPOSITE BANK IS ALWAY A CONSTANT=DISTANCE/VELOCITY OF THE SHIP.
OF COURSE DUE TO THE VELOCITY OF THE RIVER IT MOVES A HORIZONTAL DISTANCE BUT IT DIDNT EFFECT THE TIME TAKEN.

SO ONLY I TOLD THAT WAY

ELESSAR & OTHERS ...........IF U FIND IT WRONG PLZ CLARIFY...................

elessar_iitkgp

In a river boat problem, the boat tries to move perpendicular to the stream current. that analogy isn't true here.

omkarnet

nitin62225 find directions accurately i wfind time
sin^{-1 (1/15)
=3hr 48min

30+3-48=33-48
r=sq.root of 150+20+2(150)20cos33-48
sq.rt of 27886=167
time=s/v=500000/167
=2994 sec=49 min aprox=50 min
this answer is accurate because i have crakce problem after 10 times practise
please rate me}

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