IIT JEE , AIEEE Forum - Physics , Mechanics - motion in 1-D

kislay

13. A river of width 'a' with straight parallel bank flows due north with speed 'u'. The points O and A are on opposite banks and A is due east of O. Co-ordinate axes O_x and O_y are taken in the east and north direction respectively. A boat, whose speed is v relative to water, starts from O and crosses the river. If the boat is steered due east and u varies with x as : u=x*(a - x)*(v/a²). Find :
(a) equation of trajectory of the boat.
(b) time taken to cross the river.
(c) absolute velocity of boatman when he reaches the opposite bank.
(d) the displacement of boatman when he reaches the opposite bank from the initial position.
Ans
(a)v=(x²/2a)-(x²/3a²)
(b)a/v
(c)v (due east)
(d) a i + (a/6) j

krishna.gopal

At any time t

x=vt

dy/dt = u = x(a-x)*(v/a²)=vt(a-vt)*(v/a²)

y = (at²/2-vt³/3)*(v/a)²

Putting t = x/v we get y = (x²/2a)-(x3/3a²) (there is a misprint in your answer)

b) River is crossed when x= a So t= a/v

c) At x = a u=0

So velocity is v i

d) At x= a, y= a/6 So displacement = a i + a/6 j

cvramana

Nice solution by krishna.

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