Friday, October 19, 2018

Chapter 4.17 - River crossing Problems

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 vM
• The velocity of the river current is vR
2. The vR is in a direction perpendicular to vM
• So the man will not be able to swim straight ahead. He will be deviated. 
• This is shown in fig.4.45(a) below:
The swimmer or boat will not reach the exact opposite point
Fig.4.45
3. The new direction will be the direction of v, which is the resultant of vM and vR
• So the actual direction of travel is along 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 |vM|
• But the actual speed is |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 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 vM and vR, we can easily calculate the details of v
• That is., we can find the magnitude |v| and direction θWe have
(i) |v|=|vM|2+|vR|2
(ii) θ=tan1|vM||vR|
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:
tanθ=BCAC=wx
(iii) So we get Eq.4.34: x=wtanθ
(iv) But tan θ is equal to |vM||vR| also. So we can write:
tanθ=BCAC=wx=|vM||vR|
From this we get Eq.4.35x=w×|vR||vM|
(b) The actual distance travelled:
(i) This is equal to AB'
(ii) Clearly, it is given by: AB=x2+w2
(c) Time of travel T
(i) The travel is with a uniform speed of |v| 
(ii) The distance covered is AB'
(iii) So time of travel T=AB|v|
(iv) Thus we get Eq.4.36T=x2+w2|vM|2+|vR|2
(d) Another formula for 'x':
(i) Consider the horizontal travel alone
• The horizontal travel is with a uniform speed of |vR| 
• This travel has a duration of T
(ii) So we can write an equation for horizontal distance travelled as: 
Eq.4.37: x = |vR|×T

Now let us see another case:
Case 2:
1. Consider a man swimming across a river
• The velocity of the river current is vR
• The velocity of the man is vM
2. This time, the man aims to a point on the upstream.
• That is., vR is not directed towards the exact opposite point. It is directed towards a point on the upstream
• Due to the river current vR, he will be deviated from his intended path
■ If the direction of vM is adjusted carefully, he can achieve an interesting result:
• The resulting 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:
With the correct angle, the swimmer or boat will reach the exact opposite point
Fig.4.46
3. The new direction will be the direction of v, which is the resultant of vM and vR
• So the actual direction of travel is along 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 |vM|
• But the actual speed is |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:
v is the resultant of vM and vR
Let us write the steps for obtaining v:
(i) Considering horizontal components:
vx=[vMx+vRx]=[(|vM|cosθ)ˆi+|vR|ˆi]=[(|vM|cosθ)+|vR|]ˆi
• The negative sign is due to the fact that, the horizontal component of vM is directed towards the left
(ii) Considering vertical components:
vy=[vMy+vRy]=[(|vM|sinθ)ˆj+0]=[(|vM|sinθ)]ˆj
• The zero value comes in because, vR has no vertical component
8. Now, v=0 does not have horizontal component. That means: vx=0
So we can equate 7(i) to zero. We get:
[(|vM|cosθ)+|vR|]ˆi=0
(|vM|cosθ)+|vR|=0
|vR|=|vM|cosθ
cosθ=|vR||vM|
So we get Eq.4.38: θ=cos1|vR||vM|
Thus we successfully calculated θ
9. Let us continue and find the magnitude of v also:
(i) The [vertical component of v] is v itself. That means: vy=v
(ii) So we can equate 7(ii) to v. We get:
v=[(|vM|sinθ)]ˆj
|v|=|vM|sinθ
10. So, if we have 'sin θ', we can multiply it with |vM| to obtain |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 sinθ=1cos2θ
So we get: sinθ=1(|vR||vM|)2
sinθ=(|vM|2|vR|2)|vM|
(iii) Thus, from the result in (9), we get:
|v|=|vM|×|vM|2|vR|2|vM|
So we can write Eq.4.39: |v|=|vM|2|vR|2
11. Once we obtain |v|, we can calculate the time of travel
(i) The travel is with a uniform speed of |v| 
(ii) The distance covered is AB = w
(iii) So time of travel T=AB|v|
(iv) Thus we get Eq.4.40T=w|vM|2|vR|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: θ=cos1|vR||vM|
2. Substituting the values, we get: θ=cos124 
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: T=w|vM|2|vR|2
2. Substituting the values, we get:  T=44222=412=23=1.155h
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: T=x2+w2|vM|2+|vR|2
3. But, first we have to calculate the drift 'x'. We can use Eq.4.35: x=w×|vR||vM|
• Substituting the values, we get: x=4×24=2km
4. Substituting in (2), we get: T=22+4242+22=2020=1h
■ 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 = |vR|×T
2. Substituting the values, we get: 120 = |vR|×10   
• So we get speed of the river current = |vR| = 12 m/min
3. Now we can use Eq.4.35: x=w×|vR||vM|
Substituting the values, we get: 120=w×12|vM| 
So we get: w|vM|=10
4. Now we consider 'crossing in least possible distance'. This is case 2
• We can use Eq.4.40: T=w|vM|2|vR|2
• Substituting the values, we get: 12.5=10×|vM||vM|2122
• Squaring both sides: 12.52×(|vM|2122)=102×|vM|2
(12.52102)×|vM|2=(12.5×12)2
• Solving this, we get |vM| = 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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