IIT JEE , AIEEE Forum - Physics , Mechanics

coolkrazy007

can u plzz explain river boat prbolems of relative velocity....

also plzz explain rel velocity b4 that....

plzz help....

a_JOSHI

all u have to do is find the vertical and horizontal components or velocity and then resultant and find the distand by multiplying velocity and time..
thats all...try hcv...its shows all...

a_JOSHI

i meant distance..

coolkrazy007

any more explanation

varun.tinkle

SEE SINCE A BOAT MOVES IN THE RIVER THE CURRENT OF THE RIVER WILL GIVE THE BOAT A HORIZONTAL VELOCITY
AND THE VERTICAL VELOCITY WILL BE DUE TO ITS OWN MOTION.WHENEVER A BOAT MOVES IN A STRAIGHT DIRECTION THE CURRENT AND THE VELOCITY OF THE BOAT R IN OPPOSITE DIRECTION SO THE RESULTANT VELOCITY WILL BE THE HYPOTHENUSE OF A RIGHT ANGLED TRIANGLE BY THE TRIANGLE LAW OF ADDITION OF VECTORS.
WHENEVER A BOAT CROSSES A RIVER THE HORIZONTAL COMPONENT OF THE RESULTANT VELOCITY OR THE VELOCITY OF THE BOAT BE RESPONSIBLE FOR THE CROSSING OF THE RIVER
PLZ RATE ME IF U FIND ME USEFUL
!!!!!!!!!CHEERS!!!!!!!!!!!!

paulparthapratim2

Let us take an example :

Q.

A river 400m wide is flowing at the rate of 2 m/s.A boat is sailing at the rate of 10m/s with repect to the water in a direction perpendicular to the river. a) Find the time taken by the boat to reach the opposite bank.

b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?

solution :

The component of the resultant velocity which is responsible for causing the boat to cross the river has magnitude 10m/s.

So time taken to cross the river is 400/10=40s.

Now the horizontal displacement of the boat is caused by the velocity of river

Distance from point directly opposite to the point from where boat started

= 40 x 2 =80m

u can use vector method also. For example :

if u be the velocity of river,& u' be the velocity of boat,

then the net velocity of boat = u i + u' j = 2 i + 10 j

now the time taken for boat to cross the river,

= ditance / velocity in y-direction

= 400/ u'

= 40 sec

distance travelled by boat in horizontal direction, will be purely due to the river's velocity, thrfore drift

= u X t

= 2 X 40

= 80 m

biki

let me try my hand at explaining.

So the figure will give u the necessary info. 'bout the river and the boat

The river having a velocity u and the boat a velocity v making an angle $\theta$ with the vertical.

So the velocity of the boat in the vertical direction becomes vcos $\theta$ and that in the horizontal direction becomes

u + vsin $\theta$ as the boat acquires the velocity of the river.

So time for crossing the river t = (width of river) / (relative velocity of boat w.r.t. river in vertical direction)

i hope this is clear.

So t = d / v.cos $\theta$

and x = (rel. vel. of boat w.r.t. river in horizontal dir.) x t

= (u + v.sin $\theta$ ) x (d / vcos $\theta$ ) = {(u + v.sin $\theta$ ) / (v.cos $\theta$ )} x d

and now u can thereby easily get the conditions for crossing the river in minimum time and to reach the point exactly opposite to point of starting.

For minimum time, v.cos $\theta$ should be maximum, so cos $\theta$ = 1 i,e, $\theta$ = 0⁰.

and for reaching the point exactly opposite to point of starting, x = 0. So, u + v.sin $\theta$ = 0 i,e sin $\theta$ = -u/v ... which is possible iff u < v.

biki

and for rel. velocity
its nothing..
its just the velocity of one body w.r.t. another moving body.
the velocity of the body is calculated taking that body at rest w.r.t. which it is calculated.

Suppose a car moves at 50mps and another at 30mps.
So for a man sitting in the 30mps car, the other car moves with a spd. 50 - 30 = 20mps.
and for a man in the 50mps var, the other car will appear to go back. So its vel. relative the the car at 50mps will be -20mps

elessar_iitkgp

Consider the problem:

A river flows west to east at a speed of 5m per minute. A man on the south bank can swim at the rate of 10m per minute in still water. In which direction should he swim so as to cross the river

1) in the smallest possible time

2) along the shortest path

Solution

Take the normal NSEW directions (N up, E right, etc). Assume XY axes such that+ve X is along east and +Y along north. Let the origin of the coordinate system be at a point on the south shore from where the man starts.

Speed of man wrt water, v_MW = 10 m/min

Velocity of water wrt to earth, v_WE= 5 i m/min

Let velocity of man wrt eath be v_ME.

Let the width of the river be d.

(a) Let the man try to swim in a direction making

with the X axis.

v_MW= 10(cos

i + sin

j)

v_ME = v_MW + v_WE= (10cos

+ 5) i + 10sin

j

Time taken by the man to cross, T = d/10sin

T is minimum if sin

is maximum, ie,

=

/2

Hence the man should swim perpendicular to the current if he is to reach in shortest possible time.

(b) The shortets possible path is pependicular to the river. For the man to follow this path, v_ME must not have any velocity component along X axis.

10cos

+ 5 = 0

= 120⁰

Hence the man should swim in a direction making 120⁰ the +ve X direction

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