Question

A motorboat going downstream overtook a raft at a point A. 30 minutes later it turned back and after some time passed the raft a distance e from point A. If the flow velocity is 4 km/h, then the value of e in km is​

Answer 1

Here,

• The boat crosses the raft at A.

• It turns back at a point B, which is at $\sf{d}$ kilometres away from A, after 30 minutes.

• Then after a time $\sf{t}$ hours, it meets the raft again at point C, which is at a distance $\sf{e}$ kilometres away from A.

If $\sf{v_w}$ is the speed of flow of water and $\sf{v_b}$ is speed of boat (both in km/h), then resultant speed of boat is $\sf{v_b+v_w}$ as it's going downstream, and speed of raft is $\sf{v_w}$ since it has no speed wrt water flow.

Time taken by boat to move from A to B is 30 minutes or 0.5 hours. So,

$\sf{\longrightarrow d=0.5(v_b+v_w)\quad\quad\dots(1)}$

When the boat turned back at B it starts going upstream, so it's new speed will be $\sf{v_b-v_w.}$

After a time $\sf{t}$ hours of turning back at B, it reaches point C, $\sf{(d-e)}$ kilometres away from B, and meets raft again. So,

$\sf{\longrightarrow d-e=(v_b-v_w)t\quad\quad\dots(2)}$

During this time $\sf{(0.5+t)}$ hours, the raft moves from A to C with speed $\sf{v_w.}$ So,

$\sf{\longrightarrow e=v_w(0.5+t)\quad\quad\dots(3)}$

Adding (2) and (3),

$\sf{\longrightarrow d-e+e=(v_b-v_w)t+v_w(0.5+t)}$

$\sf{\longrightarrow d=v_b\cdot t-v_w\cdot t+0.5v_w+v_w\cdot t}$

From (1),

$\sf{\longrightarrow 0.5(v_b+v_w)=v_b\cdot t+0.5v_w}$

$\sf{\longrightarrow t=0.5\ hr}$

Then (3) becomes, taking $\sf{v_w=4\ km\,h^{-1}}$ as per the question,

$\sf{\longrightarrow e=4(0.5+0.5)}$

$\sf{\longrightarrow\underline{\underline{e=4\ km}}}$