In our present chapter, we are discussing about motions in two dimensions. In the previous section we completed a discussion on relative motion in two dimensions. In this section, we will see problems in river crossing.
Case 1:
1. Consider a man swimming across a river
• The velocity of the man is $\mathbf\small{\vec v_M}$
• The velocity of the river current is $\mathbf\small{\vec v_R}$
2. The $\mathbf\small{\vec v_R}$ is in a direction perpendicular to $\mathbf\small{\vec v_M}$
• So the man will not be able to swim straight ahead. He will be deviated.
• This is shown in fig.4.45(a) below:
3. The new direction will be the direction of $\mathbf\small{\vec v}$, which is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
• So the actual direction of travel is along $\mathbf\small{\vec v}$.
4. Not only the direction, but the 'magnitude of the velocity' with which the travel is made also changes.
• We expect the man to swim with a speed of $\mathbf\small{|\vec v_M|}$
• But the actual speed is $\mathbf\small{|\vec v|}$.
5. Let us apply the above results to an actual case. It is shown in fig.4.45(b)
• The yellow lines are the river banks. The width of the river is w
6. The man starts to swim from A. He wants to reach B, which is directely opposite A
• But due to the current, he will be travelling along $\mathbf\small{\vec v}$.
• Because of this change in direction, he will reach B'
■ The distance BB' is called drift. It is denoted by 'x'
7. If we know the details about $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$, we can easily calculate the details of $\mathbf\small{\vec v}$
• That is., we can find the magnitude $\mathbf\small{|\vec v|}$ and direction θ. We have
(i) $\mathbf\small{|\vec v|=\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}$
(ii) $\mathbf\small{\theta =\tan^{-1}\frac{|\vec{v_M}|}{|\vec{v_R}|}}$
8. Once we find θ, we will be able to find some more useful details. Let us see what they are:
(a) The drift x
(i) Consider fig.b. A perpendicular B'C is dropped from B' to the other bank
(ii) In the right triangle ACB', we have:
$\mathbf\small{\tan \theta =\frac{B'C}{AC}=\frac{w}{x}}$
(iii) So we get Eq.4.34: $\mathbf\small{x=\frac{w}{\tan \theta}}$
(iv) But tan θ is equal to $\mathbf\small{\frac{|\vec{v_M}|}{|\vec{v_R}|}}$ also. So we can write:
$\mathbf\small{\tan \theta =\frac{B'C}{AC}=\frac{w}{x}=\frac{|\vec{v_M}|}{|\vec{v_R}|}}$
From this we get Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
(b) The actual distance travelled:
(i) This is equal to AB'
(ii) Clearly, it is given by: $\mathbf\small{AB'=\sqrt{x^2+w^2}}$
(c) Time of travel T
(i) The travel is with a uniform speed of $\mathbf\small{|\vec v|}$
(ii) The distance covered is AB'
(iii) So time of travel $\mathbf\small{T=\frac{AB'}{|\vec{v}|}}$
(iv) Thus we get Eq.4.36: $\mathbf\small{T=\frac{\sqrt{x^2+w^2}}{\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}}$
(d) Another formula for 'x':
(i) Consider the horizontal travel alone
• The horizontal travel is with a uniform speed of $\mathbf\small{|\vec v_R|}$
• This travel has a duration of T
(ii) So we can write an equation for horizontal distance travelled as:
Eq.4.37: x = $\mathbf\small{|\vec v_R| \times T}$
Now let us see another case:
Case 2:
1. Consider a man swimming across a river
• The velocity of the river current is $\mathbf\small{\vec v_R}$
• The velocity of the man is $\mathbf\small{\vec v_M}$
2. This time, the man aims to a point on the upstream.
• That is., $\mathbf\small{\vec v_R}$ is not directed towards the exact opposite point. It is directed towards a point on the upstream
• Due to the river current $\mathbf\small{\vec v_R}$, he will be deviated from his intended path
■ If the direction of $\mathbf\small{\vec v_M}$ is adjusted carefully, he can achieve an interesting result:
• The resulting $\mathbf\small{\vec v}$ will be directed towards the exact opposite point on the other bank
♦ Thus he will reach the exact opposite point on the other bank
This is shown in fig.4.46(a) below:
3. The new direction will be the direction of $\mathbf\small{\vec v}$, which is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
• So the actual direction of travel is along $\mathbf\small{\vec v}$.
4. Not only the direction, but the 'magnitude of the velocity' with which the travel is made also changes.
• We expect the man to swim with a speed of $\mathbf\small{|\vec v_M|}$
• But the actual speed is $\mathbf\small{|\vec v|}$
5. Let us apply the above results to an actual case. It is shown in fig.4.46(b)
• The yellow lines are the river banks. The width of the river is w.
6. The man starts to swim from A. He wants to reach B, which is directely opposite A
■ If he swim directely towards B, he will be carried away frm B due to the current
• So he aims for a point on the upstream
• We want to know the angle θ which will enable him to reach B
7. For obtaining θ, we can use the following steps:
$\mathbf\small{\vec v}$ is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
Let us write the steps for obtaining $\mathbf\small{\vec v}$:
(i) Considering horizontal components:
$\mathbf\small{\vec v_x=[\vec v_{Mx}+\vec v_{Rx}]=[-(|\vec v_{M}|\cos \theta )\hat{i}+|\vec v_{R}|\hat{i}]=[-(|\vec v_{M}|\cos \theta )+|\vec v_{R}|]\hat{i}}$
• The negative sign is due to the fact that, the horizontal component of $\mathbf\small{\vec v_M}$ is directed towards the left
(ii) Considering vertical components:
$\mathbf\small{\vec v_y=[\vec v_{My}+\vec v_{Ry}]=[(|\vec v_{M}|\sin \theta )\hat{j}+0]=[(|\vec v_{M}|\sin \theta )]\hat{j}}$
• The zero value comes in because, $\mathbf\small{\vec v_R}$ has no vertical component
8. Now, $\mathbf\small{\vec v=0}$ does not have horizontal component. That means: $\mathbf\small{\vec v_x=0}$
So we can equate 7(i) to zero. We get:
$\mathbf\small{[-(|\vec v_{M}|\cos \theta )+|\vec v_{R}|]\hat{i}=0}$
$\mathbf\small{\Rightarrow -(|\vec v_{M}|\cos \theta )+|\vec v_{R}|=0}$
$\mathbf\small{\Rightarrow |\vec v_{R}|=|\vec v_{M}|\cos \theta}$
$\mathbf\small{\Rightarrow \cos \theta=\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
So we get Eq.4.38: $\mathbf\small{\theta=\cos^{-1}\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
■ Thus we successfully calculated θ
9. Let us continue and find the magnitude of $\mathbf\small{\vec v}$ also:
(i) The [vertical component of $\mathbf\small{\vec v}$] is $\mathbf\small{\vec v}$ itself. That means: $\mathbf\small{\vec v_y=\vec v}$
(ii) So we can equate 7(ii) to $\mathbf\small{\vec v}$. We get:
$\mathbf\small{\vec v=[(|\vec v_{M}|\sin \theta )]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec v|=|\vec v_{M}|\sin \theta}$
10. So, if we have 'sin θ', we can multiply it with $\mathbf\small{|\vec v_M|}$ to obtain $\mathbf\small{|\vec v|}$
(i) We have already obtained 'cos θ' in step 8. From that, we can easily calculate 'sin θ'
(ii) From math classes we know that $\mathbf\small{\sin \theta = \sqrt{1-\cos^2\theta }}$
So we get: $\mathbf\small{\sin \theta = \sqrt{1-\left( \frac{|\vec v_{R}|}{|\vec v_{M}|} \right )^2}}$
$\mathbf\small{\Rightarrow \sin \theta = \frac{\sqrt{(|\vec{v_M}|^2-|\vec{v_R}|^2})}{|\vec{v_M}|}}$
(iii) Thus, from the result in (9), we get:
$\mathbf\small{|\vec{v}| = |\vec{v_M}|\times \frac{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}{|\vec{v_M}|}}$
■ So we can write Eq.4.39: $\mathbf\small{|\vec{v}| =\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}$
11. Once we obtain $\mathbf\small{|\vec v|}$, we can calculate the time of travel
(i) The travel is with a uniform speed of $\mathbf\small{|\vec v|}$
(ii) The distance covered is AB = w
(iii) So time of travel $\mathbf\small{T=\frac{AB}{|\vec{v}|}}$
(iv) Thus we get Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
Now we will see some solved examples
Solved example 4.19
A man can row a boat at a speed of 4 km/h in still water. He is crossing a river where the speed of current is 2 km/h
(a) In what direction should he be headed if he wants to reach a point directly opposite to his starting point?
(b) If the width of the river is 4 km, how long will it take to reach the opposite bank if he heads in the direction derived in (a)?
(c) In what direction should he be headed if he wants to reach the opposite bank in the shortest possible time? How much is this shortest time?
Solution:
Part (a)
1. This comes under case 2 that we saw above
• We can use Eq.4.38: $\mathbf\small{\theta=\cos^{-1}\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
2. Substituting the values, we get: $\mathbf\small{\theta=\cos^{-1}\frac{2}{4}}$
Thus θ = cos-1 (0.5) = 60°.
■ So the man should row the boat in such a way that, his direction makes an angle of 60° with the bank on his left side
Part (b)
1. We can use Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
2. Substituting the values, we get: $\mathbf\small{T=\frac{4}{\sqrt{4^2-2^2}}=\frac{4}{\sqrt{12}}=\frac{2}{\sqrt{3}}=1.155\: \text{h}}$
Part (c)
1. The shortest possible time is achieved when the direction is headed exactly to the opposite point
• So this is case 1
2. We can use Eq.4.36: $\mathbf\small{T=\frac{\sqrt{x^2+w^2}}{\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}}$
3. But, first we have to calculate the drift 'x'. We can use Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
• Substituting the values, we get: $\mathbf\small{x = \frac{4 \times 2}{4}=2\: \text{km}}$
4. Substituting in (2), we get: $\mathbf\small{T=\frac{\sqrt{2^2+4^2}}{\sqrt{4^2+2^2}}=\frac{\sqrt{20}}{\sqrt{20}}=1\: \text{h}}$
■ Note: To achieve the least possible time, the direction should be headed to the exact opposite point. Any other direction will take up a longer time.
Solved example 4.20
A man crosses a river in a boat. If he choose to cross within the least possible time, he can do so in 10 minutes. But there will be a drift of 120 m. If he choose to cross with the least possible distance, he can do so in 12.5 minutes. Find (a) Width of the river (b) Speed of the boat (c) Speed of the river current
Solution:
1. Consider the 'crossing in least possible time'. This is case 1
• We can use Eq.4.37: drift = x = $\mathbf\small{|\vec v_R| \times T}$
2. Substituting the values, we get: 120 = $\mathbf\small{|\vec v_R| \times 10}$
• So we get speed of the river current = $\mathbf\small{|\vec v_R|}$ = 12 m/min
3. Now we can use Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
Substituting the values, we get: $\mathbf\small{120 = \frac{w \times 12}{|\vec{v_M}|}}$
So we get: $\mathbf\small{\frac{w}{|\vec{v_M}|}=10}$
4. Now we consider 'crossing in least possible distance'. This is case 2
• We can use Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
• Substituting the values, we get: $\mathbf\small{12.5=\frac{10\times|\vec{v_M}|}{\sqrt{|\vec{v_M}|^2-12^2}}}$
• Squaring both sides: $\mathbf\small{12.5^2 \times\left ( |\vec{v_M}|^2-12^2 \right )=10^2 \times |\vec{v_M}|^2}$
$\mathbf\small{\Rightarrow (12.5^2-10^2)\times |\vec{v_M}|^2=(12.5 \times 12)^2}$
• Solving this, we get $\mathbf\small{|\vec{v_M}|}$ = 20 m/min
5. Substituting this value in (3), we get: w = 200 m
Case 1:
1. Consider a man swimming across a river
• The velocity of the man is $\mathbf\small{\vec v_M}$
• The velocity of the river current is $\mathbf\small{\vec v_R}$
2. The $\mathbf\small{\vec v_R}$ is in a direction perpendicular to $\mathbf\small{\vec v_M}$
• So the man will not be able to swim straight ahead. He will be deviated.
• This is shown in fig.4.45(a) below:
 |
| Fig.4.45 |
• So the actual direction of travel is along $\mathbf\small{\vec v}$.
4. Not only the direction, but the 'magnitude of the velocity' with which the travel is made also changes.
• We expect the man to swim with a speed of $\mathbf\small{|\vec v_M|}$
• But the actual speed is $\mathbf\small{|\vec v|}$.
5. Let us apply the above results to an actual case. It is shown in fig.4.45(b)
• The yellow lines are the river banks. The width of the river is w
6. The man starts to swim from A. He wants to reach B, which is directely opposite A
• But due to the current, he will be travelling along $\mathbf\small{\vec v}$.
• Because of this change in direction, he will reach B'
■ The distance BB' is called drift. It is denoted by 'x'
7. If we know the details about $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$, we can easily calculate the details of $\mathbf\small{\vec v}$
• That is., we can find the magnitude $\mathbf\small{|\vec v|}$ and direction θ. We have
(i) $\mathbf\small{|\vec v|=\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}$
(ii) $\mathbf\small{\theta =\tan^{-1}\frac{|\vec{v_M}|}{|\vec{v_R}|}}$
8. Once we find θ, we will be able to find some more useful details. Let us see what they are:
(a) The drift x
(i) Consider fig.b. A perpendicular B'C is dropped from B' to the other bank
(ii) In the right triangle ACB', we have:
$\mathbf\small{\tan \theta =\frac{B'C}{AC}=\frac{w}{x}}$
(iii) So we get Eq.4.34: $\mathbf\small{x=\frac{w}{\tan \theta}}$
(iv) But tan θ is equal to $\mathbf\small{\frac{|\vec{v_M}|}{|\vec{v_R}|}}$ also. So we can write:
$\mathbf\small{\tan \theta =\frac{B'C}{AC}=\frac{w}{x}=\frac{|\vec{v_M}|}{|\vec{v_R}|}}$
From this we get Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
(b) The actual distance travelled:
(i) This is equal to AB'
(ii) Clearly, it is given by: $\mathbf\small{AB'=\sqrt{x^2+w^2}}$
(c) Time of travel T
(i) The travel is with a uniform speed of $\mathbf\small{|\vec v|}$
(ii) The distance covered is AB'
(iii) So time of travel $\mathbf\small{T=\frac{AB'}{|\vec{v}|}}$
(iv) Thus we get Eq.4.36: $\mathbf\small{T=\frac{\sqrt{x^2+w^2}}{\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}}$
(d) Another formula for 'x':
(i) Consider the horizontal travel alone
• The horizontal travel is with a uniform speed of $\mathbf\small{|\vec v_R|}$
• This travel has a duration of T
(ii) So we can write an equation for horizontal distance travelled as:
Eq.4.37: x = $\mathbf\small{|\vec v_R| \times T}$
Now let us see another case:
Case 2:
1. Consider a man swimming across a river
• The velocity of the river current is $\mathbf\small{\vec v_R}$
• The velocity of the man is $\mathbf\small{\vec v_M}$
2. This time, the man aims to a point on the upstream.
• That is., $\mathbf\small{\vec v_R}$ is not directed towards the exact opposite point. It is directed towards a point on the upstream
• Due to the river current $\mathbf\small{\vec v_R}$, he will be deviated from his intended path
■ If the direction of $\mathbf\small{\vec v_M}$ is adjusted carefully, he can achieve an interesting result:
• The resulting $\mathbf\small{\vec v}$ will be directed towards the exact opposite point on the other bank
♦ Thus he will reach the exact opposite point on the other bank
This is shown in fig.4.46(a) below:
 |
| Fig.4.46 |
• So the actual direction of travel is along $\mathbf\small{\vec v}$.
4. Not only the direction, but the 'magnitude of the velocity' with which the travel is made also changes.
• We expect the man to swim with a speed of $\mathbf\small{|\vec v_M|}$
• But the actual speed is $\mathbf\small{|\vec v|}$
5. Let us apply the above results to an actual case. It is shown in fig.4.46(b)
• The yellow lines are the river banks. The width of the river is w.
6. The man starts to swim from A. He wants to reach B, which is directely opposite A
■ If he swim directely towards B, he will be carried away frm B due to the current
• So he aims for a point on the upstream
• We want to know the angle θ which will enable him to reach B
7. For obtaining θ, we can use the following steps:
$\mathbf\small{\vec v}$ is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
Let us write the steps for obtaining $\mathbf\small{\vec v}$:
(i) Considering horizontal components:
$\mathbf\small{\vec v_x=[\vec v_{Mx}+\vec v_{Rx}]=[-(|\vec v_{M}|\cos \theta )\hat{i}+|\vec v_{R}|\hat{i}]=[-(|\vec v_{M}|\cos \theta )+|\vec v_{R}|]\hat{i}}$
• The negative sign is due to the fact that, the horizontal component of $\mathbf\small{\vec v_M}$ is directed towards the left
(ii) Considering vertical components:
$\mathbf\small{\vec v_y=[\vec v_{My}+\vec v_{Ry}]=[(|\vec v_{M}|\sin \theta )\hat{j}+0]=[(|\vec v_{M}|\sin \theta )]\hat{j}}$
• The zero value comes in because, $\mathbf\small{\vec v_R}$ has no vertical component
8. Now, $\mathbf\small{\vec v=0}$ does not have horizontal component. That means: $\mathbf\small{\vec v_x=0}$
So we can equate 7(i) to zero. We get:
$\mathbf\small{[-(|\vec v_{M}|\cos \theta )+|\vec v_{R}|]\hat{i}=0}$
$\mathbf\small{\Rightarrow -(|\vec v_{M}|\cos \theta )+|\vec v_{R}|=0}$
$\mathbf\small{\Rightarrow |\vec v_{R}|=|\vec v_{M}|\cos \theta}$
$\mathbf\small{\Rightarrow \cos \theta=\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
So we get Eq.4.38: $\mathbf\small{\theta=\cos^{-1}\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
■ Thus we successfully calculated θ
9. Let us continue and find the magnitude of $\mathbf\small{\vec v}$ also:
(i) The [vertical component of $\mathbf\small{\vec v}$] is $\mathbf\small{\vec v}$ itself. That means: $\mathbf\small{\vec v_y=\vec v}$
(ii) So we can equate 7(ii) to $\mathbf\small{\vec v}$. We get:
$\mathbf\small{\vec v=[(|\vec v_{M}|\sin \theta )]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec v|=|\vec v_{M}|\sin \theta}$
10. So, if we have 'sin θ', we can multiply it with $\mathbf\small{|\vec v_M|}$ to obtain $\mathbf\small{|\vec v|}$
(i) We have already obtained 'cos θ' in step 8. From that, we can easily calculate 'sin θ'
(ii) From math classes we know that $\mathbf\small{\sin \theta = \sqrt{1-\cos^2\theta }}$
So we get: $\mathbf\small{\sin \theta = \sqrt{1-\left( \frac{|\vec v_{R}|}{|\vec v_{M}|} \right )^2}}$
$\mathbf\small{\Rightarrow \sin \theta = \frac{\sqrt{(|\vec{v_M}|^2-|\vec{v_R}|^2})}{|\vec{v_M}|}}$
(iii) Thus, from the result in (9), we get:
$\mathbf\small{|\vec{v}| = |\vec{v_M}|\times \frac{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}{|\vec{v_M}|}}$
■ So we can write Eq.4.39: $\mathbf\small{|\vec{v}| =\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}$
11. Once we obtain $\mathbf\small{|\vec v|}$, we can calculate the time of travel
(i) The travel is with a uniform speed of $\mathbf\small{|\vec v|}$
(ii) The distance covered is AB = w
(iii) So time of travel $\mathbf\small{T=\frac{AB}{|\vec{v}|}}$
(iv) Thus we get Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
Now we will see some solved examples
Solved example 4.19
A man can row a boat at a speed of 4 km/h in still water. He is crossing a river where the speed of current is 2 km/h
(a) In what direction should he be headed if he wants to reach a point directly opposite to his starting point?
(b) If the width of the river is 4 km, how long will it take to reach the opposite bank if he heads in the direction derived in (a)?
(c) In what direction should he be headed if he wants to reach the opposite bank in the shortest possible time? How much is this shortest time?
Solution:
Part (a)
1. This comes under case 2 that we saw above
• We can use Eq.4.38: $\mathbf\small{\theta=\cos^{-1}\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
2. Substituting the values, we get: $\mathbf\small{\theta=\cos^{-1}\frac{2}{4}}$
Thus θ = cos-1 (0.5) = 60°.
■ So the man should row the boat in such a way that, his direction makes an angle of 60° with the bank on his left side
Part (b)
1. We can use Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
2. Substituting the values, we get: $\mathbf\small{T=\frac{4}{\sqrt{4^2-2^2}}=\frac{4}{\sqrt{12}}=\frac{2}{\sqrt{3}}=1.155\: \text{h}}$
Part (c)
1. The shortest possible time is achieved when the direction is headed exactly to the opposite point
• So this is case 1
2. We can use Eq.4.36: $\mathbf\small{T=\frac{\sqrt{x^2+w^2}}{\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}}$
3. But, first we have to calculate the drift 'x'. We can use Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
• Substituting the values, we get: $\mathbf\small{x = \frac{4 \times 2}{4}=2\: \text{km}}$
4. Substituting in (2), we get: $\mathbf\small{T=\frac{\sqrt{2^2+4^2}}{\sqrt{4^2+2^2}}=\frac{\sqrt{20}}{\sqrt{20}}=1\: \text{h}}$
■ Note: To achieve the least possible time, the direction should be headed to the exact opposite point. Any other direction will take up a longer time.
Solved example 4.20
A man crosses a river in a boat. If he choose to cross within the least possible time, he can do so in 10 minutes. But there will be a drift of 120 m. If he choose to cross with the least possible distance, he can do so in 12.5 minutes. Find (a) Width of the river (b) Speed of the boat (c) Speed of the river current
Solution:
1. Consider the 'crossing in least possible time'. This is case 1
• We can use Eq.4.37: drift = x = $\mathbf\small{|\vec v_R| \times T}$
2. Substituting the values, we get: 120 = $\mathbf\small{|\vec v_R| \times 10}$
• So we get speed of the river current = $\mathbf\small{|\vec v_R|}$ = 12 m/min
3. Now we can use Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
Substituting the values, we get: $\mathbf\small{120 = \frac{w \times 12}{|\vec{v_M}|}}$
So we get: $\mathbf\small{\frac{w}{|\vec{v_M}|}=10}$
4. Now we consider 'crossing in least possible distance'. This is case 2
• We can use Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
• Substituting the values, we get: $\mathbf\small{12.5=\frac{10\times|\vec{v_M}|}{\sqrt{|\vec{v_M}|^2-12^2}}}$
• Squaring both sides: $\mathbf\small{12.5^2 \times\left ( |\vec{v_M}|^2-12^2 \right )=10^2 \times |\vec{v_M}|^2}$
$\mathbf\small{\Rightarrow (12.5^2-10^2)\times |\vec{v_M}|^2=(12.5 \times 12)^2}$
• Solving this, we get $\mathbf\small{|\vec{v_M}|}$ = 20 m/min
5. Substituting this value in (3), we get: w = 200 m
With this we complete our present discussion on two dimensional motion. In the next chapter, we will see laws of motion.
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