In our present chapter, we are discussing about motions in two dimensions. In the previous section we completed a discussion on uniform circular motion. In this section, we will see relative velocity in two dimensions.
We saw 'relative velocity in the case of motion in one dimension' in a previous chapter. Our present discussion is based on that.
1. Consider two objects A and B shown in fig.4.41(a) below:
• They are moving in two different directions and with two different magnitudes
♦ A is moving with a velocity of →vA
♦ B is moving with a velocity of →vB
2. We can write two cases:
(i) Case 1: When viewed from A, The other object B will be moving with a velocity of [→vB−→vA]
• That is., Relative velocity of B with respect to A = →vBA=→vB−→vA
(ii) Case 2: When viewed from B, The other object A will be moving with a velocity of [→vA−→vB]
• That is., Relative velocity of A with respect to B = →vAB=→vA−→vB
3. So we need to do vector subtraction to find relative velocities. We already know the methods to do such subtractions. We have both graphical and analytical methods.
■ Here we will use analytical method:
• We use the horizontal and vertical components. Fig.4.41(b) shows those components.
4. Consider the resultant →vBA in case 1
♦ The x component of this resultant will be: →vBx−→vAx
♦ The y component of this resultant will be: →vBy−→vAy
Once the components are obtained, we can easily calculate the resultant
• Consider the resultant →vAB in case 2
♦ The x component of this resultant will be: →vAx−→vBx
♦ The y component of this resultant will be: →vAy−→vBy
Once the components are obtained, we can easily calculate the original
We saw 'relative velocity in the case of motion in one dimension' in a previous chapter. Our present discussion is based on that.
1. Consider two objects A and B shown in fig.4.41(a) below:
![]() |
Fig.4.41 |
♦ A is moving with a velocity of →vA
♦ B is moving with a velocity of →vB
2. We can write two cases:
(i) Case 1: When viewed from A, The other object B will be moving with a velocity of [→vB−→vA]
• That is., Relative velocity of B with respect to A = →vBA=→vB−→vA
(ii) Case 2: When viewed from B, The other object A will be moving with a velocity of [→vA−→vB]
• That is., Relative velocity of A with respect to B = →vAB=→vA−→vB
3. So we need to do vector subtraction to find relative velocities. We already know the methods to do such subtractions. We have both graphical and analytical methods.
■ Here we will use analytical method:
• We use the horizontal and vertical components. Fig.4.41(b) shows those components.
4. Consider the resultant →vBA in case 1
♦ The x component of this resultant will be: →vBx−→vAx
♦ The y component of this resultant will be: →vBy−→vAy
Once the components are obtained, we can easily calculate the resultant
• Consider the resultant →vAB in case 2
♦ The x component of this resultant will be: →vAx−→vBx
♦ The y component of this resultant will be: →vAy−→vBy
Once the components are obtained, we can easily calculate the original
Solved example 4.16
Rain is falling vertically with a speed of 35 ms-1. A woman rides a bicycle with a speed of 12 ms-1 in east to west direction. What is the direction in which she should hold her umbrella ?
Solution:
1. Let the rain fall be denoted by →vR
• Let the motion of bicycle be denoted by →vB
• The two vectors are shown in the fig.4.42(a) below:
2. We want to find 'how the velocity of the rain will appear, when viewed from the bicycle'
![]() |
Fig.4.42 |
♦ That is., we want to find the velocity of the rain with respect to the bicycle.
♦ That is., we want to find →vRB
♦ That is., we want to find →vR−→vB
3. For that, we need to find the following two:
(i) x component of (→vR−→vB)
♦ It is given by: →vRx−→vBx
(ii) y component of (→vR−→vB)
♦ It is given by: →vRy−→vBy
4. Let us write the values:
• →vRx = 0 (∵ the rain is falling vertically and so has no horizontal component)
• →vBx = -12 ˆi (The travel is from left to right. So it is taken as negative)
• →vRy = -35 ˆj (The travel is from top to bottom. So it is taken as negative)
• →vBy = 0 (∵ the bicycle is travelling horizontally and so has no vertical component)
5. Substituting the values in 3(i), we get:
• x component of (→vR−→vB) = →vRBx = [0 - (-12 ˆi)] = 12 ˆi
♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
♦ This is shown in fig.b
♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
♦ This is shown in fig.b
Substituting the values in 3(ii), we get:
• y component of (→vR−→vB) = →vRBy = [(-35 ˆi - 0)] = -35 ˆj
♦ We get a negative value. So this vector is directed towards the negative side of the y axis
♦ This is also shown in fig.b
• y component of (→vR−→vB) = →vRBy = [(-35 ˆi - 0)] = -35 ˆj
♦ We get a negative value. So this vector is directed towards the negative side of the y axis
♦ This is also shown in fig.b
6. Now we can find the magnitude:
Magnitude of →vR−→vB = |→vR−→vB| = |→vRB| = √122+(−35)2 = 37 ms-1.
7. Next we find the direction:
• We have: tanθ=|→vRBy||→vRBx|=3512
• Thus we get: θ = 71.076o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
8. But what we want is 'φ' which the resultant makes with the vertical
• It can be easily obtained because, in fig.c, we can see that φ = (90-θ)
• Thus we get: φ = (90-71.076) = 18.924o.
9. So we can write the result:
• When viewed from the bicycle, rain falls towards the cyclist at an angle of 18.924o with the vertical
• So the cyclist must hold the umbrella at an angle of 18.924o with the vertical,
10. We must note the difference between this example and example 4.1.
• In that example, the boy was stationary. The rain was falling at an angle because of the wind
• But here, the rain is falling vertically. But to the cyclist, it appears to be falling at an angle
Solved example 4.17
In a harbour, wind is blowing at a speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
Solution:
1. The speed and direction of the wind is given:
• Speed is 72 km/h and direction is N-E
2. Let us set up the coordinate axes as follows:
♦ E-W direction is the x axis
♦ N-S direction is the y axis
• Then the N-E direction (in which the wind is blowing) will be inclined at 45o with the x axis.
• This is shown in fig.4.43(a) below:
3. Now we can write the details about the velocity vector of wind
• Let us denote it as →vW
• We can write:
♦ |→vW| = 72 km/h
♦ →vW makes an angle 45o with the x axis
• This is also shown in fig.a
4. This velocity of the wind will not change
• But when viewed from a moving boat, the wind will appear to have a different magnitude and different direction
Magnitude of →vR−→vB = |→vR−→vB| = |→vRB| = √122+(−35)2 = 37 ms-1.
7. Next we find the direction:
• We have: tanθ=|→vRBy||→vRBx|=3512
• Thus we get: θ = 71.076o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
8. But what we want is 'φ' which the resultant makes with the vertical
• It can be easily obtained because, in fig.c, we can see that φ = (90-θ)
• Thus we get: φ = (90-71.076) = 18.924o.
9. So we can write the result:
• When viewed from the bicycle, rain falls towards the cyclist at an angle of 18.924o with the vertical
• So the cyclist must hold the umbrella at an angle of 18.924o with the vertical,
10. We must note the difference between this example and example 4.1.
• In that example, the boy was stationary. The rain was falling at an angle because of the wind
• But here, the rain is falling vertically. But to the cyclist, it appears to be falling at an angle
Solved example 4.17
In a harbour, wind is blowing at a speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
Solution:
1. The speed and direction of the wind is given:
• Speed is 72 km/h and direction is N-E
2. Let us set up the coordinate axes as follows:
♦ E-W direction is the x axis
♦ N-S direction is the y axis
• Then the N-E direction (in which the wind is blowing) will be inclined at 45o with the x axis.
• This is shown in fig.4.43(a) below:
![]() |
Fig.4.43 |
• Let us denote it as →vW
• We can write:
♦ |→vW| = 72 km/h
♦ →vW makes an angle 45o with the x axis
• This is also shown in fig.a
4. This velocity of the wind will not change
• But when viewed from a moving boat, the wind will appear to have a different magnitude and different direction
• We want to find 'how the velocity of the wind will appear, when viewed from the moving boat'
♦ That is., we want to find the velocity of the wind with respect to the moving boat.
♦ That is., we want to find →vWB
♦ That is., we want to find →vW−→vB
Where →vB is the velocity of the boat
Where →vB is the velocity of the boat
5. For that, we need to find the following two:
(i) x component of (→vW−→vB)
♦ It is given by: →vWx−→vBx
(ii) y component of (→vW−→vB)
♦ It is given by: →vWy−→vBy
6. Let us write the values:
• →vWx = 72 cos 45 ˆi = 50.912 ˆi km/h
• →vBx = 0 (∵ the boat is travelling north and so has no component towards east)
• →vWy = 72 sin 45 ˆj = 50.912 ˆi km/h
• →vBy = 51 ˆj
• →vWy = 72 sin 45 ˆj = 50.912 ˆi km/h
• →vBy = 51 ˆj
7. Substituting the values in 5(i), we get:
• x component of (→vW−→vB) = →vWBx = [50.912 ˆi - 0] = 50.912 ˆi
♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
♦ This is shown in fig.b
♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
♦ This is shown in fig.b
Substituting the values in 5(ii), we get:
• y component of (→vW−→vB) = →vWBy = [(72 sin 45 ˆj - 51 ˆj)] = -0.0883
♦ We get a negative value. So this vector is directed towards the negative side of the y axis
♦ This is also shown in fig.b
8. Now we can find the magnitude. But it is not asked in the question
9. But we do have to find the direction
• We have: tanθ=|→vWBy||→vWBx|=0.088350.912
• Thus we get: θ = 0.0997o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
10. We see that, the relative velocity vector falls between the east and south directions
• But the deviation from east is only 0.0997o, which is a very small quantity
• So we can write:
■ The flag flutters in the east direction approximately
Solved example 4.18
When a man walks at a rate of 3 km/h, the rain appears to fall vertically. When he walks at the rate of 6 km/h, it appears to fall at an angle of 45° with the vertical. What is the original magnitude and direction of the rain?
Solution:
1. Consider the simple case when the rain falls in the exact vertical direction
■ For a man moving from right to left, the rain will obviously appear to be falling at a slope
• This slope will be towards the right.
• That is., →vRM will slope towards the right
♦ Where →vRM is the relative velocity of the rain with respect to the man
♦ This is shown in fig.4.44(a) below:
2. But in our present case, even when there is motion at 3 km/h, the rain appears to be exact vertical
■ So it is clear that, the original rain is sloping towards the left. This is shown in fig.b
• We are asked to find this original magnitude and direction
• That is., we need to find the details of →vR in fig.b
3. We know this:
• The two components of →vRM in fig.b are:
(i) →vRMx
(ii) →vRMy
4. So we will first find those components:
(i) We have: →vRMx=→vRx−→vMx
• Let us write the values:
(a) →vRx=(|→vR|cosθ)ˆi
♦ Where θ is the angle made by →vR with the horizontal
♦ This is shown in fig.c
(b) →vMx=3ˆi
• Substituting in 4(i), we get: →vRMx=(|→vR|cosθ)ˆi−3ˆi
(ii) We have: →vRMy=→vRy−→vMy
• Let us write the values:
(a) →vRy=(|→vR|sinθ)ˆj
♦ Where θ is the angle made by →vR with the horizontal
♦ This is shown in fig.c
(b) →vMy=0
• Substituting in 4(ii), we get: →vRMy=(|→vR|sinθ)ˆj
5. Now consider the →vRM in fig.c
• It is perfect vertical. That means, it has no horizontal component
• That means, →vRMx = 0
• From 4(i)b, we can write: →vRMx=(|→vR|cosθ)ˆi−3ˆi = 0
■ From this we get: |→vR|cosθ=3
This is one of the two equations which we will need to find |→vR| and θ.
6. To get the other equation, we consider the motion at 6 km/h
• This time, the →vRM is indeed at a slope. This is shown in fig.d
♦ Given that, it makes an angle of 45° with the vertical
♦ So it will make the same angle of 45° with the horizontal also
Note the two points:
(i) →vR in fig.d is same as that in fig.c. This is because, the original speed and direction of rain does not change
(ii) →vM in fig.d has the same direction as that in fig.c But the magnitude changed from 3 to 6 km/h
7. We know this:
• The two components of →vRM in fig.d are:
(i) →vRMx
(ii) →vRMy
8. So we will first find those components:
(i) We have: →vRMx=→vRx−→vMx
• Let us write the values:
(a) →vRx=(|→vR|cosθ)ˆi
♦ Where θ is the angle made by →vR with the horizontal
♦ This is shown in fig.d
(b) →vMx=6ˆi
• Substituting in 8(i), we get: →vRMx=(|→vR|cosθ)ˆi−6ˆi
(ii) We have: →vRMy=→vRy−→vMy
• Let us write the values:
(a) →vRy=(|→vR|sinθ)ˆj
♦ Where θ is the angle made by →vR with the horizontal
♦ This is shown in fig.d
(b) →vMy=0
• Substituting in 8(ii), we get: →vRMy=(|→vR|sinθ)ˆj
9. Using the 'formula for direction of the resultant', we can write:
tan45=|→vRMy||→vRMx|=||→vR|sinθ||(|→vR|cosθ)−6|
10. But in (5), we obtained: |→vR|cosθ=3
Substituting this in (9), we get: tan45=||→vR|sinθ||(3)−6|=||→vR|sinθ||(−3)|=|→vR|sinθ3
11. But tan 45 = 1√2
Thus we get: |→vR|sinθ=3√2
12.So we have two results:
(i) |→vR|cosθ=3
(ii) |→vR|sinθ=3√2
Dividing (ii) by (i), we get: tanθ=1√2
Thus we get θ = 45°.
13. Substituting this value of θ in 12(i), we get:
|→vR|cos45=3
⇒|→vR|=31√2=3√2=4.243km/h
14. So we can write:
• The rain is actually falling at an angle of 45° towards the left
• It has a magnitude of 4.243 km/h
• But when the man walks ar the rate of 6 km/h, the rain appears to be falling at an angle of 45° towards the right
• y component of (→vW−→vB) = →vWBy = [(72 sin 45 ˆj - 51 ˆj)] = -0.0883
♦ We get a negative value. So this vector is directed towards the negative side of the y axis
♦ This is also shown in fig.b
8. Now we can find the magnitude. But it is not asked in the question
9. But we do have to find the direction
• We have: tanθ=|→vWBy||→vWBx|=0.088350.912
• Thus we get: θ = 0.0997o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
10. We see that, the relative velocity vector falls between the east and south directions
• But the deviation from east is only 0.0997o, which is a very small quantity
• So we can write:
■ The flag flutters in the east direction approximately
Solved example 4.18
When a man walks at a rate of 3 km/h, the rain appears to fall vertically. When he walks at the rate of 6 km/h, it appears to fall at an angle of 45° with the vertical. What is the original magnitude and direction of the rain?
Solution:
1. Consider the simple case when the rain falls in the exact vertical direction
■ For a man moving from right to left, the rain will obviously appear to be falling at a slope
• This slope will be towards the right.
• That is., →vRM will slope towards the right
♦ Where →vRM is the relative velocity of the rain with respect to the man
♦ This is shown in fig.4.44(a) below:
![]() |
Fig.4.44 |
■ So it is clear that, the original rain is sloping towards the left. This is shown in fig.b
• We are asked to find this original magnitude and direction
• That is., we need to find the details of →vR in fig.b
3. We know this:
• The two components of →vRM in fig.b are:
(i) →vRMx
(ii) →vRMy
4. So we will first find those components:
(i) We have: →vRMx=→vRx−→vMx
• Let us write the values:
(a) →vRx=(|→vR|cosθ)ˆi
♦ Where θ is the angle made by →vR with the horizontal
♦ This is shown in fig.c
(b) →vMx=3ˆi
• Substituting in 4(i), we get: →vRMx=(|→vR|cosθ)ˆi−3ˆi
(ii) We have: →vRMy=→vRy−→vMy
• Let us write the values:
(a) →vRy=(|→vR|sinθ)ˆj
♦ Where θ is the angle made by →vR with the horizontal
♦ This is shown in fig.c
(b) →vMy=0
• Substituting in 4(ii), we get: →vRMy=(|→vR|sinθ)ˆj
5. Now consider the →vRM in fig.c
• It is perfect vertical. That means, it has no horizontal component
• That means, →vRMx = 0
• From 4(i)b, we can write: →vRMx=(|→vR|cosθ)ˆi−3ˆi = 0
■ From this we get: |→vR|cosθ=3
This is one of the two equations which we will need to find |→vR| and θ.
6. To get the other equation, we consider the motion at 6 km/h
• This time, the →vRM is indeed at a slope. This is shown in fig.d
♦ Given that, it makes an angle of 45° with the vertical
♦ So it will make the same angle of 45° with the horizontal also
Note the two points:
(i) →vR in fig.d is same as that in fig.c. This is because, the original speed and direction of rain does not change
(ii) →vM in fig.d has the same direction as that in fig.c But the magnitude changed from 3 to 6 km/h
7. We know this:
• The two components of →vRM in fig.d are:
(i) →vRMx
(ii) →vRMy
8. So we will first find those components:
(i) We have: →vRMx=→vRx−→vMx
• Let us write the values:
(a) →vRx=(|→vR|cosθ)ˆi
♦ Where θ is the angle made by →vR with the horizontal
♦ This is shown in fig.d
(b) →vMx=6ˆi
• Substituting in 8(i), we get: →vRMx=(|→vR|cosθ)ˆi−6ˆi
(ii) We have: →vRMy=→vRy−→vMy
• Let us write the values:
(a) →vRy=(|→vR|sinθ)ˆj
♦ Where θ is the angle made by →vR with the horizontal
♦ This is shown in fig.d
(b) →vMy=0
• Substituting in 8(ii), we get: →vRMy=(|→vR|sinθ)ˆj
9. Using the 'formula for direction of the resultant', we can write:
tan45=|→vRMy||→vRMx|=||→vR|sinθ||(|→vR|cosθ)−6|
10. But in (5), we obtained: |→vR|cosθ=3
Substituting this in (9), we get: tan45=||→vR|sinθ||(3)−6|=||→vR|sinθ||(−3)|=|→vR|sinθ3
11. But tan 45 = 1√2
Thus we get: |→vR|sinθ=3√2
12.So we have two results:
(i) |→vR|cosθ=3
(ii) |→vR|sinθ=3√2
Dividing (ii) by (i), we get: tanθ=1√2
Thus we get θ = 45°.
13. Substituting this value of θ in 12(i), we get:
|→vR|cos45=3
⇒|→vR|=31√2=3√2=4.243km/h
14. So we can write:
• The rain is actually falling at an angle of 45° towards the left
• It has a magnitude of 4.243 km/h
• But when the man walks ar the rate of 6 km/h, the rain appears to be falling at an angle of 45° towards the right
In the next section, we will see a few more solved examples.
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