In the previous section we saw the three equations of motion. In this section we will see relative velocity.
1. Consider ourselves being seated in a train moving with uniform velocity.
2. Let another train move parallel to our train. Also let it move in the same direction as our train.
3. If that train overtakes us, it means that, that train is moving faster than ours.
• But we feel that, that other train is moving slowly
4. If that other train is moving with the same uniform velocity as ours, we will feel that, that other train is not moving at all.
• But for a person standing on the ground, both the trains will appear to be moving with the same speed.
■ Such observations can be explained by using the concept of relative velocity.
• Relative velocity can be easily understood with the help of graphs.
• We have seen the details about position-time graph in a previous section.
• Here, we are going to see such a graph. But this time, the distances covered by two objects will be shown in a single graph. Consider fig.3.52 below:
1. Two vehicles are travelling with the same uniform velocity of 7 ms-1
• The lower yellow line is the position time graph of vehicle 1
• The upper yellow line is the position time graph of vehicle 2
2. Draw a vertical dashed line through any convenient point on the time axis. Say t = 4 s
• This vertical line intersect the graphs at P and Q
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
♦ The coordinates of P are (4,68)
♦ The coordinates of Q are (4,118)
3. So we can write:
• At the instant when the stop watch reading is '4 s', vehicle 1 is at a distance of 68 m from the origin O
• At the instant when the stop watch reading is '4 s', vehicle 2 is at a distance of 118 m from the origin O
4. From this, we get:
■ At the instant when the stop watch reading is '4', the distance between the two vehicles is (118-68) = 50 m
5. Draw another vertical dashed line through any other convenient point on the time axis. Say t = 10 s
• This vertical line intersect the graphs at R and S
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
6. So we can write:
• At the instant when the stop watch reading is '10', vehicle 1 is at a distance of 110 m from the origin O
• At the instant when the stop watch reading is '10', vehicle 2 is at a distance of 160 m from the origin O
7. From this, we get:
■ At the instant when the stop watch reading is '10', the distance between the two vehicles is (160-110) = 50 m
(same as before)
8. So we can write:
At the instant when t = 4, the distance between the two vehicles is 50 m
At the instant when t = 10, the distance between the two vehicles is 50 m
9. We can check at any instant. We will find that, the distance between the two vehicles is always 50 m. This is shown in the table below:
10. So, for a person seated in vehicle 1, the other vehicle 2 will always appear to be at a distance of 50 m
■ So, for that person, the vehicle 2 will appear to be stationary
11. The reverse is also true:
For a person seated in vehicle 2, the other vehicle 1 will always appear to be at a distance of 50 m
■ So, for that person, the vehicle 1 will appear to be stationary
Now we will consider another case. See fig.3.53 below:
1. Two vehicles are travelling with uniform velocities
• The uniform velocity of vehicle 1 is 5 ms-1
• The uniform velocity of vehicle 2 is 9 ms-1
• The lower yellow line is the position time graph of vehicle 1
• The upper yellow line is the position time graph of vehicle 2
2. Draw a vertical dashed line through any convenient point on the time axis. Say t = 4 s
• This vertical line intersect the graphs at P and Q
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
♦ The coordinates of P are (4,50)
♦ The coordinates of Q are (4,106)
3. So we can write:
• At the instant when the stop watch reading is '4', vehicle 1 is at a distance of 50 m from the origin O
• At the instant when the stop watch reading is '4', vehicle 2 is at a distance of 106 m from the origin O
4. From this, we get:
■ At the instant when the stop watch reading is '4', the distance between the two vehicles is (106-50) = 56 m
5. Draw another vertical dashed line through any other convenient point on the time axis. Say t = 10 s
• This vertical line intersect the graphs at R and S
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
6. So we can write:
• At the instant when the stop watch reading is '10', vehicle 1 is at a distance of 80 m from the origin O
• At the instant when the stop watch reading is '10', vehicle 2 is at a distance of 160 m from the origin O
7. From this, we get:
■ At the instant when the stop watch reading is '10', the distance between the two vehicles is (160-80) = 80 m
8. So we can write:
• At the instant when t = 4, the distance between the two vehicles is 56 m
• At the instant when t = 10, the distance between the two vehicles is 80 m
9. So as time increases, the distance between the two vehicles go on increasing.
• We can check at any instant. The table below gives more evidence.
• When t = 0, distance between the two vehicles is 40 m
• When t = 1, distance between the two vehicles is 44 m. This is an increase of 4 ms-1
• When t = 2, distance between the two vehicles is 48 m. This is a further increase of 4 ms-1
• When t = 3, distance between the two vehicles is 52 m. This is a further increase of 4 ms-1
• When t = 4, distance between the two vehicles is 56 m. This is a further increase of 4 ms-1
So on . . .
10. So, for a person seated in vehicle 1, the other vehicle 2 will appear to be moving further and further away
• The distance increasing by 4 m every second
• Thus, for the person seated in vehicle 1, the other vehicle will appear to be moving at a velocity of 4 ms-1.
11. The reverse is also true:
• For a person seated in vehicle 2, the other vehicle 1 will appear to be moving away in the opposite direction.
• That is., vehicle 1 will appear to be moving with a velocity of -4 ms-1
12. As time passes, the distance between the two vehicles become so large that, they will no longer be able to see each other
• It is same as, one vehicle is stationary and the other is moving away at an uniform velocity of 4 ms-
1.
Is there an easy method to obtain this 'velocity of 4 ms-1' ?
Let us try:
• Both the vehicles are travelling in the same direction.
♦ Which is: towards the positive side of the x axis (This is clear from the rising graphs)
• Vehicle 1 is travelling at 5 ms-1
• Vehicle 2 is travelling at 9 ms-1
■ We find that '4' is the difference between the two individual velocities
Let us write a general form:
1. We know the equation for each of the yellow lines. It is: x = x0 + vt
♦ x is the displacement at any time t
♦ x0 is the y intercept, which is the displacement when t = 0
♦ v is the uniform velocity with which the object is travelling
• This equation gives the displacement at any time 't'
2. So, for an object A, moving with an uniform velocity of vA, the displacement from the origin at any time 't' is given by: xA = x0 + vAt
• Similarly, for an object B, moving with an uniform velocity of vB, the displacement from the origin at any time 't' is given by: xB = x0 + vBt
3. Then at any particular instant when the stop watch shows 't', the displacement of object B with respect to object A
= (xB - xA) = [(x0 + vBt) - (x0 + vAt)]
⟹ (xB - xA) = (vB - vA)t
Consider the quantity (vB - vA) in the above equation
• This is in the familiar form: Distance = velocity × time
4. So at any instant, when viewed from object A, the other object B will appear to be moving with a velocity of (vB - vA) ms-1.
■ We say that: Relative velocity of B with respect to A is given by:
Eq. 3.6: vBA = (vB - vA) ms-1
• Note the order in which the subscripts are written.
• In our present case, if we consider the vehicles 1 and 2 as objects A and B respectively, we get:
Relative velocity of vehicle 2 with respect to vehicle 1 = vBA = (9 - 5) = 4 ms-1.
• That is., when viewed from vehicle 1, the other vehicle 2 will appear to be moving with a velocity of 4 ms-1.
5. In a similar way,
• Relative velocity of A with respect to B is given by:
vAB = (vA - vB) ms-1.
• In our present case:
Relative velocity of vehicle 2 with respect to vehicle 1 = vAB = (5 - 9) = -4 ms-1.
• That is., when viewed from vehicle 2, the other vehicle 1 will appear to be moving with a velocity of -4 ms-1. The -ve sign indicates the opposite direction.
Now we will consider one more case. See fig.3.54 below:
• Note that, in the above graph, in the x axis, 1 unit = 2 s
• In the previous graphs, 1 unit was 1 s
1. Two vehicles are travelling with uniform velocities in opposite directions
• The uniform velocity of vehicle 1 is 6 ms-1
• The uniform velocity of vehicle 2 is -3 ms-1
• The lower yellow line is the position time graph of vehicle 1
• The upper yellow line is the position time graph of vehicle 2
2. Draw a vertical dashed line through any convenient point on the time axis. Say t = 4 s
• This vertical line intersect the graphs at P and Q
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
♦ The coordinates of P are (4,44)
♦ The coordinates of Q are (4,148)
3. So we can write:
• At the instant when the stop watch reading is '4', vehicle 1 is at a distance of 44 m from the origin O
• At the instant when the stop watch reading is '4', vehicle 2 is at a distance of 148 m from the origin O
4. From this, we get:
■ At the instant when the stop watch reading is '4', the distance between the two vehicles is (148-44) = 144 m
5. Draw another vertical dashed line through any other convenient point on the time axis. Say t = 10 s
• This vertical line intersect the graphs at R and S
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
6. So we can write:
• At the instant when the stop watch reading is '10', vehicle 1 is at a distance of 80 m from the origin O
• At the instant when the stop watch reading is '10', vehicle 2 is at a distance of 130 m from the origin O
7. From this, we get:
■ At the instant when the stop watch reading is '10', the distance between the two vehicles is (130-80) = 50 m
8. So we can write:
• At the instant when t = 4, the distance between the two vehicles is 144 m
• At the instant when t = 10, the distance between the two vehicles is 50 m
9. So as time increases, the distance between the two vehicles go on decreasing.
• We can check at any instant. The table below gives more evidence.
• Note that, in the time column, the interval between individual entries is 2 s
• In the previous tables, it was 1 s.
This change is made to accommodate more entries
■ We notice some interesting points:
• When t = 15.55 s, the distance between the two vehicles is zero
• That means, as time increases, the distance between the two vehicles goes on decreasing to such a level that, at a particular point, the distance between them becomes zero
• It is clear that, at the instant when the distance is zero, both the vehicles are at the same point. It is the meeting point between the two vehicles. It is indicated by the point M in the graph
• Also it is highlighted in yellow colour in the table.
• But remember that, the vehicles are travelling in opposite directions. So, as time increases further, the vehicles will move away from each other.
Let us consider the travel upto the point M
• When t = 0, distance between the two vehicles is 140 m
• When t = 2, distance between the two vehicles is 122 m. This is a decrease of 18 ms-1
• When t = 4, distance between the two vehicles is 104 m. This is a further decrease of 18 ms-1
• When t = 6, distance between the two vehicles is 86 m. This is a further decrease of 18 ms-1
• When t = 8, distance between the two vehicles is 68 m. This is a further decrease of 18 ms-1
So on . . .
10. So, for a person seated in vehicle 1, the other vehicle 2 will appear to be coming closer and closer
• The distance decreasing by 18 m every 2 seconds
• '18 m every 2 seconds' is '9 m every one second'
• Thus, for the person seated in vehicle 1, the other vehicle will appear to be coming closer and closer with a velocity of 9 ms-1.
11. The reverse is also true:
• For a person seated in vehicle 2, the other vehicle 1 will appear to be moving in the opposite direction.
• That is., vehicle 1 will appear to be coming closer and closer with a velocity of -9 ms-1
12. As time passes, the distance between the two vehicles falls to zero. That is the meeting point
13. Let us consider the travel after the point M
• When t = 16, distance between the two vehicles is 4 m
• When t = 18, distance between the two vehicles is 22 m. This is an increase of 18 ms-1
• When t = 20, distance between the two vehicles is 40 m. This is a further increase of 18 ms-1
• When t = 22, distance between the two vehicles is 58 m. This is a further increase of 18 ms-1
• When t = 24 distance between the two vehicles is 76 m. This is a further increase of 18 ms-1
So on . . .
14. So, for a person seated in vehicle 1, the other vehicle 2 will appear to be moving away
• The distance increasing by 18 m every 2 seconds
• '18 m every 2 seconds' is '9 m every one second'
• Thus, after point M, for the person seated in vehicle 1, the other vehicle will appear to be moving away with a velocity of 9 ms-1.
15. The reverse is also true:
• For a person seated in vehicle 2, the other vehicle 1 will appear to be moving in the opposite direction.
• That is., vehicle 1 will appear to be moving away with a velocity of -9 ms-1
16. As time passes, the distance between the two vehicles become so large that, they will not be able to see each other.
Is there an easy method to obtain this 'velocity of 9 ms-1' ?
Let us try:
• Vehicle 1 is travelling towards the positive side of the x axis
(This is clear from the rising graph)
• Vehicle 2 is travelling towards the negative side of the x axis
(This is clear from the falling graph)
That is why we wrote earlier:
• The uniform velocity of vehicle 1 is 6 ms-1
• The uniform velocity of vehicle 2 is -3 ms-1
■ We find that '9' is the difference between the two individual velocities (∵ 6 - (-3) = 9)
Let us write a general form. The steps are similar to those in the previous case.
• At any particular instant when the stop watch shows 't', the displacement of object B with respect to object A
= (xB - xA) = [(x0 + vBt) - (x0 + vAt)]
⟹ (xB - xA) = (vB - vA)t
So Eq.3.6 is valid here also
1. Consider ourselves being seated in a train moving with uniform velocity.
2. Let another train move parallel to our train. Also let it move in the same direction as our train.
3. If that train overtakes us, it means that, that train is moving faster than ours.
• But we feel that, that other train is moving slowly
4. If that other train is moving with the same uniform velocity as ours, we will feel that, that other train is not moving at all.
• But for a person standing on the ground, both the trains will appear to be moving with the same speed.
■ Such observations can be explained by using the concept of relative velocity.
• Relative velocity can be easily understood with the help of graphs.
• We have seen the details about position-time graph in a previous section.
• Here, we are going to see such a graph. But this time, the distances covered by two objects will be shown in a single graph. Consider fig.3.52 below:
 |
| Fig.3.52 |
• The lower yellow line is the position time graph of vehicle 1
• The upper yellow line is the position time graph of vehicle 2
2. Draw a vertical dashed line through any convenient point on the time axis. Say t = 4 s
• This vertical line intersect the graphs at P and Q
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
♦ The coordinates of P are (4,68)
♦ The coordinates of Q are (4,118)
3. So we can write:
• At the instant when the stop watch reading is '4 s', vehicle 1 is at a distance of 68 m from the origin O
• At the instant when the stop watch reading is '4 s', vehicle 2 is at a distance of 118 m from the origin O
4. From this, we get:
■ At the instant when the stop watch reading is '4', the distance between the two vehicles is (118-68) = 50 m
5. Draw another vertical dashed line through any other convenient point on the time axis. Say t = 10 s
• This vertical line intersect the graphs at R and S
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
♦ The coordinates of R are (10,110)
♦ The coordinates of S are (10,160)6. So we can write:
• At the instant when the stop watch reading is '10', vehicle 1 is at a distance of 110 m from the origin O
• At the instant when the stop watch reading is '10', vehicle 2 is at a distance of 160 m from the origin O
7. From this, we get:
■ At the instant when the stop watch reading is '10', the distance between the two vehicles is (160-110) = 50 m
(same as before)
8. So we can write:
At the instant when t = 4, the distance between the two vehicles is 50 m
At the instant when t = 10, the distance between the two vehicles is 50 m
9. We can check at any instant. We will find that, the distance between the two vehicles is always 50 m. This is shown in the table below:
10. So, for a person seated in vehicle 1, the other vehicle 2 will always appear to be at a distance of 50 m
■ So, for that person, the vehicle 2 will appear to be stationary
11. The reverse is also true:
For a person seated in vehicle 2, the other vehicle 1 will always appear to be at a distance of 50 m
■ So, for that person, the vehicle 1 will appear to be stationary
Now we will consider another case. See fig.3.53 below:
 |
| Fig.3.53 |
• The uniform velocity of vehicle 1 is 5 ms-1
• The uniform velocity of vehicle 2 is 9 ms-1
• The lower yellow line is the position time graph of vehicle 1
• The upper yellow line is the position time graph of vehicle 2
2. Draw a vertical dashed line through any convenient point on the time axis. Say t = 4 s
• This vertical line intersect the graphs at P and Q
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
♦ The coordinates of P are (4,50)
♦ The coordinates of Q are (4,106)
3. So we can write:
• At the instant when the stop watch reading is '4', vehicle 1 is at a distance of 50 m from the origin O
• At the instant when the stop watch reading is '4', vehicle 2 is at a distance of 106 m from the origin O
4. From this, we get:
■ At the instant when the stop watch reading is '4', the distance between the two vehicles is (106-50) = 56 m
5. Draw another vertical dashed line through any other convenient point on the time axis. Say t = 10 s
• This vertical line intersect the graphs at R and S
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
♦ The coordinates of R are (10,80)
♦ The coordinates of S are (10,160)6. So we can write:
• At the instant when the stop watch reading is '10', vehicle 1 is at a distance of 80 m from the origin O
• At the instant when the stop watch reading is '10', vehicle 2 is at a distance of 160 m from the origin O
7. From this, we get:
■ At the instant when the stop watch reading is '10', the distance between the two vehicles is (160-80) = 80 m
8. So we can write:
• At the instant when t = 4, the distance between the two vehicles is 56 m
• At the instant when t = 10, the distance between the two vehicles is 80 m
9. So as time increases, the distance between the two vehicles go on increasing.
• We can check at any instant. The table below gives more evidence.
• When t = 0, distance between the two vehicles is 40 m
• When t = 1, distance between the two vehicles is 44 m. This is an increase of 4 ms-1
• When t = 2, distance between the two vehicles is 48 m. This is a further increase of 4 ms-1
• When t = 3, distance between the two vehicles is 52 m. This is a further increase of 4 ms-1
• When t = 4, distance between the two vehicles is 56 m. This is a further increase of 4 ms-1
So on . . .
10. So, for a person seated in vehicle 1, the other vehicle 2 will appear to be moving further and further away
• The distance increasing by 4 m every second
• Thus, for the person seated in vehicle 1, the other vehicle will appear to be moving at a velocity of 4 ms-1.
11. The reverse is also true:
• For a person seated in vehicle 2, the other vehicle 1 will appear to be moving away in the opposite direction.
• That is., vehicle 1 will appear to be moving with a velocity of -4 ms-1
12. As time passes, the distance between the two vehicles become so large that, they will no longer be able to see each other
• It is same as, one vehicle is stationary and the other is moving away at an uniform velocity of 4 ms-
1.
Is there an easy method to obtain this 'velocity of 4 ms-1' ?
Let us try:
• Both the vehicles are travelling in the same direction.
♦ Which is: towards the positive side of the x axis (This is clear from the rising graphs)
• Vehicle 1 is travelling at 5 ms-1
• Vehicle 2 is travelling at 9 ms-1
■ We find that '4' is the difference between the two individual velocities
Let us write a general form:
1. We know the equation for each of the yellow lines. It is: x = x0 + vt
♦ x is the displacement at any time t
♦ x0 is the y intercept, which is the displacement when t = 0
♦ v is the uniform velocity with which the object is travelling
• This equation gives the displacement at any time 't'
2. So, for an object A, moving with an uniform velocity of vA, the displacement from the origin at any time 't' is given by: xA = x0 + vAt
• Similarly, for an object B, moving with an uniform velocity of vB, the displacement from the origin at any time 't' is given by: xB = x0 + vBt
3. Then at any particular instant when the stop watch shows 't', the displacement of object B with respect to object A
= (xB - xA) = [(x0 + vBt) - (x0 + vAt)]
⟹ (xB - xA) = (vB - vA)t
Consider the quantity (vB - vA) in the above equation
• This is in the familiar form: Distance = velocity × time
4. So at any instant, when viewed from object A, the other object B will appear to be moving with a velocity of (vB - vA) ms-1.
■ We say that: Relative velocity of B with respect to A is given by:
Eq. 3.6: vBA = (vB - vA) ms-1
• Note the order in which the subscripts are written.
• In our present case, if we consider the vehicles 1 and 2 as objects A and B respectively, we get:
Relative velocity of vehicle 2 with respect to vehicle 1 = vBA = (9 - 5) = 4 ms-1.
• That is., when viewed from vehicle 1, the other vehicle 2 will appear to be moving with a velocity of 4 ms-1.
5. In a similar way,
• Relative velocity of A with respect to B is given by:
vAB = (vA - vB) ms-1.
• In our present case:
Relative velocity of vehicle 2 with respect to vehicle 1 = vAB = (5 - 9) = -4 ms-1.
• That is., when viewed from vehicle 2, the other vehicle 1 will appear to be moving with a velocity of -4 ms-1. The -ve sign indicates the opposite direction.
Now we will consider one more case. See fig.3.54 below:
 |
| Fig.3.54 |
• In the previous graphs, 1 unit was 1 s
1. Two vehicles are travelling with uniform velocities in opposite directions
• The uniform velocity of vehicle 1 is 6 ms-1
• The uniform velocity of vehicle 2 is -3 ms-1
• The lower yellow line is the position time graph of vehicle 1
• The upper yellow line is the position time graph of vehicle 2
2. Draw a vertical dashed line through any convenient point on the time axis. Say t = 4 s
• This vertical line intersect the graphs at P and Q
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
♦ The coordinates of P are (4,44)
♦ The coordinates of Q are (4,148)
3. So we can write:
• At the instant when the stop watch reading is '4', vehicle 1 is at a distance of 44 m from the origin O
• At the instant when the stop watch reading is '4', vehicle 2 is at a distance of 148 m from the origin O
4. From this, we get:
■ At the instant when the stop watch reading is '4', the distance between the two vehicles is (148-44) = 144 m
5. Draw another vertical dashed line through any other convenient point on the time axis. Say t = 10 s
• This vertical line intersect the graphs at R and S
• If we know the 'uniform velocity' with which an object travels, we can write the coordinates of any point on it's position time graph
♦ The coordinates of R are (10,80)
♦ The coordinates of S are (10,130)6. So we can write:
• At the instant when the stop watch reading is '10', vehicle 1 is at a distance of 80 m from the origin O
• At the instant when the stop watch reading is '10', vehicle 2 is at a distance of 130 m from the origin O
7. From this, we get:
■ At the instant when the stop watch reading is '10', the distance between the two vehicles is (130-80) = 50 m
8. So we can write:
• At the instant when t = 4, the distance between the two vehicles is 144 m
• At the instant when t = 10, the distance between the two vehicles is 50 m
9. So as time increases, the distance between the two vehicles go on decreasing.
• We can check at any instant. The table below gives more evidence.
• Note that, in the time column, the interval between individual entries is 2 s
• In the previous tables, it was 1 s.
This change is made to accommodate more entries
• When t = 15.55 s, the distance between the two vehicles is zero
• That means, as time increases, the distance between the two vehicles goes on decreasing to such a level that, at a particular point, the distance between them becomes zero
• It is clear that, at the instant when the distance is zero, both the vehicles are at the same point. It is the meeting point between the two vehicles. It is indicated by the point M in the graph
• Also it is highlighted in yellow colour in the table.
• But remember that, the vehicles are travelling in opposite directions. So, as time increases further, the vehicles will move away from each other.
Let us consider the travel upto the point M
• When t = 0, distance between the two vehicles is 140 m
• When t = 2, distance between the two vehicles is 122 m. This is a decrease of 18 ms-1
• When t = 4, distance between the two vehicles is 104 m. This is a further decrease of 18 ms-1
• When t = 6, distance between the two vehicles is 86 m. This is a further decrease of 18 ms-1
• When t = 8, distance between the two vehicles is 68 m. This is a further decrease of 18 ms-1
So on . . .
10. So, for a person seated in vehicle 1, the other vehicle 2 will appear to be coming closer and closer
• The distance decreasing by 18 m every 2 seconds
• '18 m every 2 seconds' is '9 m every one second'
• Thus, for the person seated in vehicle 1, the other vehicle will appear to be coming closer and closer with a velocity of 9 ms-1.
11. The reverse is also true:
• For a person seated in vehicle 2, the other vehicle 1 will appear to be moving in the opposite direction.
• That is., vehicle 1 will appear to be coming closer and closer with a velocity of -9 ms-1
12. As time passes, the distance between the two vehicles falls to zero. That is the meeting point
13. Let us consider the travel after the point M
• When t = 16, distance between the two vehicles is 4 m
• When t = 18, distance between the two vehicles is 22 m. This is an increase of 18 ms-1
• When t = 20, distance between the two vehicles is 40 m. This is a further increase of 18 ms-1
• When t = 22, distance between the two vehicles is 58 m. This is a further increase of 18 ms-1
• When t = 24 distance between the two vehicles is 76 m. This is a further increase of 18 ms-1
So on . . .
14. So, for a person seated in vehicle 1, the other vehicle 2 will appear to be moving away
• The distance increasing by 18 m every 2 seconds
• '18 m every 2 seconds' is '9 m every one second'
• Thus, after point M, for the person seated in vehicle 1, the other vehicle will appear to be moving away with a velocity of 9 ms-1.
15. The reverse is also true:
• For a person seated in vehicle 2, the other vehicle 1 will appear to be moving in the opposite direction.
• That is., vehicle 1 will appear to be moving away with a velocity of -9 ms-1
16. As time passes, the distance between the two vehicles become so large that, they will not be able to see each other.
Is there an easy method to obtain this 'velocity of 9 ms-1' ?
Let us try:
• Vehicle 1 is travelling towards the positive side of the x axis
(This is clear from the rising graph)
• Vehicle 2 is travelling towards the negative side of the x axis
(This is clear from the falling graph)
That is why we wrote earlier:
• The uniform velocity of vehicle 1 is 6 ms-1
• The uniform velocity of vehicle 2 is -3 ms-1
■ We find that '9' is the difference between the two individual velocities (∵ 6 - (-3) = 9)
Let us write a general form. The steps are similar to those in the previous case.
• At any particular instant when the stop watch shows 't', the displacement of object B with respect to object A
= (xB - xA) = [(x0 + vBt) - (x0 + vAt)]
⟹ (xB - xA) = (vB - vA)t
So Eq.3.6 is valid here also
In the next section, we will see a special case of relative velocity.
Copyright©2018 Higher Secondary Physics. blogspot.in - All Rights Reserved
No comments:
Post a Comment