Question

River flows from west to east with speed 4 ms 1. A motorboat going downstream with 6 ms
relative to water overcame a float at a point at t = 0 second. At t = 60 second, the boat turned
back and meet the float again at t = to second. Then the value of to and the distance traversed
by float from t = 0 to t = to, are respectively​

Answer 1

Answer:

river flows from west to east with speed

Answer 2

Given :

The speed of river flowing from west to east is =  4 m per s

Speed of motor boat going downstream relative to water is = 6 m per s

The motorboat overcame a float at a point at time :

t = sec

After 60 sec the boat turned back and meet the flaot again at time :

$t = t_0$ sec

To Find :

Value of $t_0$ and the distance traversed by float from t = o to $t = t_0$ sec are =  ?

Solution ;

Since the boat is going downstream so its velocity will be in same direction as that of the flow .

So, the relative velocity of boat to water is :

$V_{BR} = V_B - V_R$

∴$V_B - V_R = 6 \frac{m}{s}$

Or , $V_B- 4 \frac{m}{s} =6 \frac{m}{s}$

So,$V_B$ ( velocity of boat w.r.t. ground ) = $10 \frac{m}{s}$

And , velocity of float w.r.t. ground = $V_R= 4 \frac{m}{s}$

So, in 60 s , the boat travels a distance :

$S_B = 60 \times V_B$

    $= 60 \times 10$

    $= 600 m$

In same 60 sec , the float travels = $60 \times 4$

                                                       =240 m

So, distance between boat and float = 600 - 240 = 360 m

Now when the boat turns :

$V_{BF} = |V_B| + |V_F|$

       = 6 m per sec

where $V_F$ is velocity of float .

Now , since boat velocity w.r.t. ground becomes :

( 6 -4 ) = 2 $\frac{m}{s}$

∴$360 = 2 \times t _0 - 60$

Or, $420 = 2 t_o$

so , $t_0 = 210$ sec

And meanwhile distance traversed by float is :

= $V_F\times 210$

 =$4 \times 210$

= 840 m

So,  the value of $t_0$ is 210 sec and the distance covered by float is 840 m .