Question

The velocity of a boat in still water is 5 m/s. It crossed a river of width 60m, always directed at 1270 to river. If velocity of river is 3.6 m/s. The drift of the boat on reaching the other bank is (tan 37 = 3/4)​

Answer 1

Answer:

32°

Explanation:

i don't know it exactly

Answer 2

Answer:

The drift of the boat on reaching the other bank is = 9 meters.

Explanation:

Given: The velocity if the boat be = $v_B$ = 5m/s

The velocity of the river is = $v_R = 3.6m/s$

As the boat is directed at $127^o$ to river, this means that the boat makes $37^o$ angle with the perpendicular.

Now, $tan(37^o)=\frac{3}{4}$

$\implies sin(37^o)=\frac{3}{5}$

$\implies cos(37^o)=\frac{4}{5}$

We can break the components of the speed of the boat into horizontal and vertical, and solve the problem for horizontal and vertical motion of the boat.

Let's look at the perpendicular motion of the boat:

Speed of boat perpendicular to the river = $v_B*cos(37^o)=5*\frac{4}{5}=4m/s$

So, time taken by boat to cross the river = $\frac{distnace}{speed} =\frac{60}{4} =15m/s$

Let's look at the horizontal motion of the boat:

The horizontal speed of the boat(along the river)=speed of river + speed of the boat = $v_R+(-v_Bsin(37^o))$

$=3.6-5*\frac{3}{5}=3.6-3=0.6m/s$

The drift of the boat on reaching the other bank is = Drift of boat in the time taken by boat to reach the other bank = $speed*time = 0.6*15=9m$

#SPJ2