Chapter 3, Problem 66GP

(a)

To determine

To Formulate:The time needed to make a roundtripof total distance when the boat is movingupstream and back downstream.

Answer to Problem 66GP

The total time taken to cover D distance of roundtrip is $t=\left(\frac{Dv}{v^2-u^2}\right)$

Explanation of Solution

Given data:

Speed of boat in still water is v .

The speed of river current is u .

Total distance of roundtrip is D.

Formula used:

The speed-distance relation is given by:

  $s=\frac{d}{t}$

Where, s is the speed of an object travelling d distance in t time.

Calculation:

Assume speed of boat in still water, v is more than the speed of river current, u , otherwise the boat will not move upstream.

The distance covered when the boat moves upstream is $d_1=\frac{D}{2}$ .

While moving upstream, the boat moves against the river current.

Therefore, speed of the boat, $s_1=v-u$

Hence, the time taken to move upstream,

  $t_1=\frac{d_1}{s_1}=\frac{D/2}{v-u}=\frac{D}{2(v-u)}$

While moving downstream, the boat moves with the river current.

Therefore, speed of the boat, $s_2=v+u$

The distance covered when the boat moves downstream is $d_2=\frac{D}{2}$ .

Hence, the time taken to move downstream,

  $t_2=\frac{d_2}{s_2}=\frac{D/2}{v+u}=\frac{D}{2(v+u)}$

Therefore, total time taken to cover D distance of roundtrip is

  $\begin{array}{c}t=t_1+t_2=\frac{D}{2(v-u)}+\frac{D}{2(v+u)}\\ =\frac{D}{2}\left(\frac{1}{(v-u)}+\frac{1}{(v+u)}\right)\\ =\frac{D}{2}\left(\frac{v+u+v-u}{v^2-u^2}\right)\\ =\frac{D}{2}\left(\frac{2v}{v^2-u^2}\right)\\ =\left(\frac{Dv}{v^2-u^2}\right)\end{array}$

Conclusion:

The total time taken to cover D distance of roundtrip is $t=\left(\frac{Dv}{v^2-u^2}\right)$

(b)

To determine

To Formulate:The time needed to make a roundtripof total distance when the boat is movingdirectly across the river and back.

Answer to Problem 66GP

The totaltime needed to make a roundtripwhen the boat is movingdirectly across the river and back is $t=\frac{D}{\sqrt{v^2-u^2}}$ .

Explanation of Solution

Given data:

Speed of boat in still water is v .

The speed of river current is u .

Total distance of roundtrip is D.

Formula used:

The speed-distance relation is given by:

  $s=\frac{d}{t}$

Where s is the speed of an object travelling d distance in t time.

Calculation:

For the boat to move directly across the water, it must at a certain angle with river current such that net velocity is straight across the river.

Assume speed of boat in still water, v is more than the speed of river current, u .

That is, ${\overrightarrow{v}}_{\text{boat relative to shore}}=\overrightarrow{v}+\overrightarrow{u}$

  

Now, the velocity of boat relative to shore, ${\overrightarrow{v}}_{\text{boat relative to shore}}=\sqrt{v^2-u^2}$ (as per the diagram)

The distance covered when the boat moves across the river is $d_1=\frac{D}{2}$ .

Hence, the time taken to move directly across the river

  $t_1=\frac{D/2}{\sqrt{v^2-u^2}}=\frac{D}{2\sqrt{v^2-u^2}}$

Similarly the time taken in coming back is

  $t=t_1+t_2=\frac{D}{2\sqrt{v^2-u^2}}+\frac{D}{2\sqrt{v^2-u^2}}=\frac{D}{\sqrt{v^2-u^2}}$

Therefore, total time taken to cover D distance of roundtrip is

  $t=t_1+t_2=\frac{D}{2\sqrt{v^2-u^2}}+\frac{D}{2\sqrt{v^2-u^2}}=\frac{D}{\sqrt{v^2-u^2}}$

Conclusion:

The total time needed to make a roundtripwhen the boat is movingdirectly across the river and back is $t=\frac{D}{\sqrt{v^2-u^2}}$