In the previous section we saw how to obtain displacement vectors from position vectors. In this section, we will see how to obtain velocity vectors from those displacement vectors.
The steps are given below:
1. If we divide a displacement vector ($\small\mathbf{\vec{\Delta r}}$) by 'the time (Δt) during which that displacement took place', we will get the average velocity.
• That is., \(\mathbf{\vec{\bar{v}}}\) = $\mathbf{\frac{\vec{\Delta r}}{\Delta t}}$
(Note that, an arrow is provided above the bar sign. This is to indicate that, average velocity is indeed a vector quantity.)
2. But we have seen that $\small\mathbf{\vec{\Delta r}={\left (\Delta x \right ) \hat i}+{\left (\Delta y \right ) \hat j}}$
• So we get: $\mathbf{\vec{\bar{v}}=\frac{{\left (\Delta x \right ) \hat i}+{\left (\Delta y \right ) \hat j}}{\Delta t}}$
■ This can be written as: $\mathbf{\vec{\bar{v}}={\left ( \frac{\Delta x}{\Delta t} \right )\hat{i}}+{\left ( \frac{\Delta y}{\Delta t} \right )\hat{j}}}$
• $\mathbf{\frac{\Delta x}{\Delta t}}$ is the average velocity in the x direction. We can denote it as: $\mathbf{|\bar{\vec{v_x}}|}$
♦ It is the 'magnitude of the x component' of $\mathbf{\vec{\bar{v}}}$
• $\mathbf{\frac{\Delta y}{\Delta t}}$ is the average velocity in the y direction. We can denote it as: $\mathbf{|\bar{\vec{v_y}}|}$
♦ It is the 'magnitude of the y component' of $\mathbf{\vec{\bar{v}}}$
■ Thus we get Eq.4.8:
$\mathbf{\bar{\vec{v}}=|\bar{\vec{v_x}}|\hat{i}+|\bar{\vec{v_y}}|\hat{j}}$
3. The direction of $\mathbf{\vec{\bar{v}}}$ is same as the direction of $\small\mathbf{\vec{\Delta r}}$. This is shown in fig.4.22(a) below.
The steps are given below:
1. If we divide a displacement vector ($\small\mathbf{\vec{\Delta r}}$) by 'the time (Δt) during which that displacement took place', we will get the average velocity.
• That is., \(\mathbf{\vec{\bar{v}}}\) = $\mathbf{\frac{\vec{\Delta r}}{\Delta t}}$
(Note that, an arrow is provided above the bar sign. This is to indicate that, average velocity is indeed a vector quantity.)
2. But we have seen that $\small\mathbf{\vec{\Delta r}={\left (\Delta x \right ) \hat i}+{\left (\Delta y \right ) \hat j}}$
• So we get: $\mathbf{\vec{\bar{v}}=\frac{{\left (\Delta x \right ) \hat i}+{\left (\Delta y \right ) \hat j}}{\Delta t}}$
■ This can be written as: $\mathbf{\vec{\bar{v}}={\left ( \frac{\Delta x}{\Delta t} \right )\hat{i}}+{\left ( \frac{\Delta y}{\Delta t} \right )\hat{j}}}$
• $\mathbf{\frac{\Delta x}{\Delta t}}$ is the average velocity in the x direction. We can denote it as: $\mathbf{|\bar{\vec{v_x}}|}$
♦ It is the 'magnitude of the x component' of $\mathbf{\vec{\bar{v}}}$
• $\mathbf{\frac{\Delta y}{\Delta t}}$ is the average velocity in the y direction. We can denote it as: $\mathbf{|\bar{\vec{v_y}}|}$
♦ It is the 'magnitude of the y component' of $\mathbf{\vec{\bar{v}}}$
■ Thus we get Eq.4.8:
$\mathbf{\bar{\vec{v}}=|\bar{\vec{v_x}}|\hat{i}+|\bar{\vec{v_y}}|\hat{j}}$
3. The direction of $\mathbf{\vec{\bar{v}}}$ is same as the direction of $\small\mathbf{\vec{\Delta r}}$. This is shown in fig.4.22(a) below.
Now we will see instantaneous velocity
1. Consider fig.4.22(a) below:
An object travels along the green coloured path
(i) The object is at P when the stop watch shows 't' s
(ii) The object is at P1 when the stop watch shows 't1' s
• Then the time duration for travel from P to P1 = $\Delta t_1 = (t_1 - t)$
(iii) We can easily draw the displacement vector $\vec{\Delta r_{p-p1}}$
(iv) When we divide this displacement vector by $\Delta t_1$, we get the 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p1}}}\)
• Direction of this average velocity vector is same as that of the displacement vector $\vec{\Delta r_{p-p1}}$. This is shown by the magenta arrow
2. Now consider a point P2 which is closer to P. This is shown in fig.b
We will repeat the steps:
(i) The object is at P when the stop watch shows 't' s
(ii) The object is at P2 when the stop watch shows 't2' s
• Then the time duration for travel from P to P2 = $\Delta t_2 = (t_2 - t)$
• Obviously, $\Delta t_2$ will be less than $\Delta t_1$. Because, compared to P1, P2 is closer to P
(iii) We can easily draw the displacement vector $\vec{\Delta r_{p-p2}}$
(iv) When we divide this displacement vector by $\Delta t_2$, we get the 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p2}}}\)
• Direction of this average velocity vector is same as that of the displacement vector $\vec{\Delta r_{p-p2}}$. This is shown by the magenta arrow in fig.b
■ Direction of $\vec{\Delta r_{p-p2}}$ is different from that of $\vec{\Delta r_{p-p1}}$. This is because, though P is the same, P1 and P2 are different
■So direction of 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p2}}}\) is different from that of 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p1}}}\)
3. Now consider a point P3 which is still closer to P. This is shown in fig.c
We will repeat the steps:
(i) The object is at P when the stop watch shows 't' s
(ii) The object is at P3 when the stop watch shows 't3' s
• Then the time duration for travel from P to P3 = $\Delta t_2 = (t_3 - t)$
• Obviously, $\Delta t_3$ will be less than $\Delta t_2$. Because, compared to P2, P3 is closer to P
(iii) We can easily draw the displacement vector $\vec{\Delta r_{p-p3}}$
(iv) When we divide this displacement vector by $\Delta t_3$, we get the 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p3}}}\)
• Direction of this average velocity vector is same as that of the displacement vector $\vec{\Delta r_{p-p3}}$. This is shown by the magenta arrow in fig.c
■ Direction of $\vec{\Delta r_{p-p3}}$ is different from that of $\vec{\Delta r_{p-p2}}$. This is because, though P is the same, P2 and P3 are different
■So direction of 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p3}}}\) is different from that of 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p2}}}\)
4. In this way, we can choose points P4, P5, P6, . . . , closer and closer to P
• In each case, the time duration $\Delta t$ will be smaller than the previous case
• In each case, we will get an average velocity vector, which has a direction different from that of the previous case.
5. So what will happen if we continue?
Ans: The $\Delta t$ will become so small that, we can no longer call it a 'duration'
• Instead, we will have to call it an 'instant'
• Note that, $\Delta t$ is in the denominator. So we cannot give it zero value.
• However, it can take very small values which are close to zero
6. In each case, to find the 'average velocity', we calculate the ratio $\frac{\vec{\Delta r}}{\Delta t}$
• When $\Delta t$ becomes very close to zero, we call it: 'the limiting value of the ratio'
• Mathematically, it is written as: $\lim_{t\to 0}\frac{\vec{\Delta r}}{\Delta t}$
• But when $\Delta t$ is very close to zero, it is an instant.
• So what we calculate by the ratio, is instantaneous velocity $\mathbf{\vec v}$.
♦ This is shown in fig.d
• It is not the average velocity \(\mathbf{\vec{\bar{v}}}\).
■ So we can write: $\mathbf{\vec v =\lim_{t\to 0}\frac{\vec{\Delta r}}{\Delta t}}$
■ Note that, instantaneous velocity at a point is tangential to the path at that point
7. The $\vec{\Delta r}$ in the numerator is a vector. We know it's rectangular components:
• x component is $\small\mathbf{\left ( \Delta x \right )\hat{i}}$
• y component is $\small\mathbf{\left ( \Delta y \right )\hat{j}}$
8. So the result in (6) becomes:
$\mathbf{\vec{v} = lim_{t\to 0}\left [ \left ( \frac{\Delta x}{\Delta t} \right )\hat{i}+\left ( \frac{\Delta y}{\Delta t} \right )\hat{j} \right ]}$
This can be written as:
$\mathbf{\vec{v} = \hat{i} \left [lim_{t\to 0}\left ( \frac{\Delta x}{\Delta t} \right ) \right ]+\hat{j} \left [lim_{t\to 0}\left ( \frac{\Delta y}{\Delta t} \right ) \right ]}$
9. There are two terms on the right side.
(i) Consider the first term:
• It is the limiting value of the 'displacement to time ratio' in the x direction.
• So it is the instantaneous velocity in the x direction. It is a vector quantity
• We can denote it as $\mathbf{{v_x}\hat i}$
(ii) Consider the second term:
• It is the limiting value of the 'displacement to time ratio' in the y direction.
• So it is the instantaneous velocity in the y direction. It is a vector quantity
• We can denote it as $\mathbf{{v_y}\hat j}$
■ Thus we get Eq.4.9:
$\mathbf{\vec{v}}$ = $\mathbf{v_x \hat{i}}$ + $\mathbf{v_y \hat{j}}$
■ So we can write:
• The instantaneous velocity $\mathbf{\vec v}$ can be resolved into two rectangular components: $\mathbf{{v_x}\hat i}$ and $\mathbf{{v_y}\hat j}$
■ We can write the converse also:
• If we know the two rectangular components $\mathbf{{v_x}\hat i}$, and $\mathbf{{v_y}\hat j}$ of a velocity $\mathbf{\vec v}$, then (see.fig.4.23 below):
(i) Magnitude of $\mathbf{\vec v}$ is given by Eq.4.9(a):
$\mathbf{\left | \vec{v} \right |=\sqrt{{{|\vec{v_x}}|^2}+{|\vec{v_y}|}^2}}$
(ii) Direction of $\mathbf{\vec v}$ is given by Eq.4.9(b):
$\mathbf{\tan\theta =\frac{|\vec{v_y}|}{|\vec{v_x}|}}$
• Where:
♦ $\mathbf{|\vec{v_x}|}$ is the magnitude of $\mathbf{\vec{v_x}}$
♦ $\mathbf{|\vec{v_y}|}$ is the magnitude of $\mathbf{\vec{v_y}}$
♦ $\mathbf{\theta}$ is the angle made by $\mathbf{\vec v}$ with the horizontal
1. Consider fig.4.22(a) below:
 |
| Fig.4.22 |
(i) The object is at P when the stop watch shows 't' s
(ii) The object is at P1 when the stop watch shows 't1' s
• Then the time duration for travel from P to P1 = $\Delta t_1 = (t_1 - t)$
(iii) We can easily draw the displacement vector $\vec{\Delta r_{p-p1}}$
(iv) When we divide this displacement vector by $\Delta t_1$, we get the 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p1}}}\)
• Direction of this average velocity vector is same as that of the displacement vector $\vec{\Delta r_{p-p1}}$. This is shown by the magenta arrow
2. Now consider a point P2 which is closer to P. This is shown in fig.b
We will repeat the steps:
(i) The object is at P when the stop watch shows 't' s
(ii) The object is at P2 when the stop watch shows 't2' s
• Then the time duration for travel from P to P2 = $\Delta t_2 = (t_2 - t)$
• Obviously, $\Delta t_2$ will be less than $\Delta t_1$. Because, compared to P1, P2 is closer to P
(iii) We can easily draw the displacement vector $\vec{\Delta r_{p-p2}}$
(iv) When we divide this displacement vector by $\Delta t_2$, we get the 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p2}}}\)
• Direction of this average velocity vector is same as that of the displacement vector $\vec{\Delta r_{p-p2}}$. This is shown by the magenta arrow in fig.b
■ Direction of $\vec{\Delta r_{p-p2}}$ is different from that of $\vec{\Delta r_{p-p1}}$. This is because, though P is the same, P1 and P2 are different
■So direction of 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p2}}}\) is different from that of 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p1}}}\)
3. Now consider a point P3 which is still closer to P. This is shown in fig.c
We will repeat the steps:
(i) The object is at P when the stop watch shows 't' s
(ii) The object is at P3 when the stop watch shows 't3' s
• Then the time duration for travel from P to P3 = $\Delta t_2 = (t_3 - t)$
• Obviously, $\Delta t_3$ will be less than $\Delta t_2$. Because, compared to P2, P3 is closer to P
(iii) We can easily draw the displacement vector $\vec{\Delta r_{p-p3}}$
(iv) When we divide this displacement vector by $\Delta t_3$, we get the 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p3}}}\)
• Direction of this average velocity vector is same as that of the displacement vector $\vec{\Delta r_{p-p3}}$. This is shown by the magenta arrow in fig.c
■ Direction of $\vec{\Delta r_{p-p3}}$ is different from that of $\vec{\Delta r_{p-p2}}$. This is because, though P is the same, P2 and P3 are different
■So direction of 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p3}}}\) is different from that of 'average velocity vector' \(\mathbf{\vec{\bar{v}_{p-p2}}}\)
4. In this way, we can choose points P4, P5, P6, . . . , closer and closer to P
• In each case, the time duration $\Delta t$ will be smaller than the previous case
• In each case, we will get an average velocity vector, which has a direction different from that of the previous case.
5. So what will happen if we continue?
Ans: The $\Delta t$ will become so small that, we can no longer call it a 'duration'
• Instead, we will have to call it an 'instant'
• Note that, $\Delta t$ is in the denominator. So we cannot give it zero value.
• However, it can take very small values which are close to zero
6. In each case, to find the 'average velocity', we calculate the ratio $\frac{\vec{\Delta r}}{\Delta t}$
• When $\Delta t$ becomes very close to zero, we call it: 'the limiting value of the ratio'
• Mathematically, it is written as: $\lim_{t\to 0}\frac{\vec{\Delta r}}{\Delta t}$
• But when $\Delta t$ is very close to zero, it is an instant.
• So what we calculate by the ratio, is instantaneous velocity $\mathbf{\vec v}$.
♦ This is shown in fig.d
• It is not the average velocity \(\mathbf{\vec{\bar{v}}}\).
■ So we can write: $\mathbf{\vec v =\lim_{t\to 0}\frac{\vec{\Delta r}}{\Delta t}}$
■ Note that, instantaneous velocity at a point is tangential to the path at that point
7. The $\vec{\Delta r}$ in the numerator is a vector. We know it's rectangular components:
• x component is $\small\mathbf{\left ( \Delta x \right )\hat{i}}$
• y component is $\small\mathbf{\left ( \Delta y \right )\hat{j}}$
8. So the result in (6) becomes:
$\mathbf{\vec{v} = lim_{t\to 0}\left [ \left ( \frac{\Delta x}{\Delta t} \right )\hat{i}+\left ( \frac{\Delta y}{\Delta t} \right )\hat{j} \right ]}$
This can be written as:
$\mathbf{\vec{v} = \hat{i} \left [lim_{t\to 0}\left ( \frac{\Delta x}{\Delta t} \right ) \right ]+\hat{j} \left [lim_{t\to 0}\left ( \frac{\Delta y}{\Delta t} \right ) \right ]}$
9. There are two terms on the right side.
(i) Consider the first term:
• It is the limiting value of the 'displacement to time ratio' in the x direction.
• So it is the instantaneous velocity in the x direction. It is a vector quantity
• We can denote it as $\mathbf{{v_x}\hat i}$
(ii) Consider the second term:
• It is the limiting value of the 'displacement to time ratio' in the y direction.
• So it is the instantaneous velocity in the y direction. It is a vector quantity
• We can denote it as $\mathbf{{v_y}\hat j}$
■ Thus we get Eq.4.9:
$\mathbf{\vec{v}}$ = $\mathbf{v_x \hat{i}}$ + $\mathbf{v_y \hat{j}}$
■ So we can write:
• The instantaneous velocity $\mathbf{\vec v}$ can be resolved into two rectangular components: $\mathbf{{v_x}\hat i}$ and $\mathbf{{v_y}\hat j}$
■ We can write the converse also:
• If we know the two rectangular components $\mathbf{{v_x}\hat i}$, and $\mathbf{{v_y}\hat j}$ of a velocity $\mathbf{\vec v}$, then (see.fig.4.23 below):
 |
| Fig.4.23 |
$\mathbf{\left | \vec{v} \right |=\sqrt{{{|\vec{v_x}}|^2}+{|\vec{v_y}|}^2}}$
$\mathbf{\tan\theta =\frac{|\vec{v_y}|}{|\vec{v_x}|}}$
♦ $\mathbf{|\vec{v_x}|}$ is the magnitude of $\mathbf{\vec{v_x}}$
♦ $\mathbf{|\vec{v_y}|}$ is the magnitude of $\mathbf{\vec{v_y}}$
♦ $\mathbf{\theta}$ is the angle made by $\mathbf{\vec v}$ with the horizontal
So we have seen how to obtain velocity vectors. In the next section, we will see how to obtain acceleration vectors from velocity vectors.
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