In the previous section we saw how we can obtain velocity vectors from displacement vectors. In this section, we will see how to obtain acceleration vectors from those velocity vectors.
1.Consider fig.4.24(a) below:
■ The object is moving along the path shown in green color
• When the stop watch showed t1 s, it is at P1
♦ So $\mathbf\small{\vec{v_1}}$ is the velocity vector when t = t1 s
♦ This $\mathbf\small{\vec{v_1}}$ is tangential to the path at P1
♦ That is., if we draw a tangent to the path at P1, $\mathbf\small{\vec{v_1}}$ will fall along that tangent
• When the stop watch showed t2 s, it is at P2
♦ So $\mathbf\small{\vec{v_2}}$ is the velocity vector when t = t2 s
♦ This $\mathbf\small{\vec{v_2}}$ is tangential to the path at P2.
2. So, in a duration of [Δt = (t2-t1) s], the velocity changes from $\mathbf\small{\vec{v_1}}$ to $\mathbf\small{\vec{v_2}}$
• Since there is a change in velocity, there must be an acceleration. Our aim is to find this acceleration.
3. We know that, $\mathbf{\text{Average acceleration} = \frac{\text{change in velocity}}{\text{change in time}}}$
• That is., $\bar{\vec{a}}=\frac{\vec{\Delta v}}{\Delta t}$
• So first, we have to find the change in velocity $\vec{\Delta v}$
• It is given by: $\vec{\Delta v}$ = $\mathbf\small{\vec{v_2}}$ - $\mathbf\small{\vec{v_1}}$
• That is., we have to find: [$\mathbf\small{\vec{v_2}}$ + (- $\mathbf\small{\vec{v_1}}$)]
4. We will find it graphically. Consider fig.4.24(b)
• $\mathbf\small{\vec{v_2}}$, $\mathbf\small{\vec{v_1}}$ and -($\mathbf\small{\vec{v_1}}$) are shown
• shift -($\mathbf\small{\vec{v_1}}$) so that, it's tail coincides with the head of $\mathbf\small{\vec{v_2}}$. This is shown in fig.c
• Draw a new vector (shown in cyan colour in fig.c) from the tail of $\mathbf\small{\vec{v_2}}$ to the head of -($\mathbf\small{\vec{v_1}}$)
• This new vector is the required $\vec{\Delta v}$
5. Consider fig.d:
• $\mathbf\small{\vec{v_1}}$ and $\mathbf\small{\vec{v_2}}$ are placed in such a way that, their tails coincide.
• Draw a new vector from the head of $\mathbf\small{\vec{v_1}}$ to head of $\mathbf\small{\vec{v_2}}$
• We see that, this new vector in fig.d, is same as the $\vec{\Delta v}$ that we obtained in fig.c
■ We can write a summary in a Question and Answer form:
• How do we find the $\vec{\Delta v}$ analytically?
Ans: Find (${\vec{v_2}}$ - ${\vec{v_1}}$)
• How do we find the $\vec{\Delta v}$ graphically?
Ans: Shift the two vectors so that, their tails coincide
♦ Draw a new vector such that:
♦ It's tail coincides with the head of ${\vec{v_1}}$
♦ It's head coincides with the head of ${\vec{v_2}}$
This new vector is the required $\vec{\Delta v}$
■Once we obtain $\vec{\Delta v}$, we can easily calculate $\bar{\vec{a}}$ by the equation: $\bar{\vec{a}}=\frac{\vec{\Delta v}}{\Delta t}$
■ Note that:
• The direction of $\vec{\Delta v}$ is very different from the direction of ${\vec{v_1}}$
• The direction of $\vec{\Delta v}$ is very different from the direction of ${\vec{v_2}}$ also
■ So the direction of $\bar{\vec{a}}$ will suffer the same condition:
• The direction of $\bar{\vec{a}}$ will be very different from the direction of ${\vec{v_1}}$
• The direction of $\bar{\vec{a}}$ will be very different from the direction of ${\vec{v_2}}$ also
6. We know that $\bar{\vec{a}}$ is a vector. So it will have two rectangular components. Let us find them:
• To find the $\bar{\vec{a}}$, we are dividing the $\vec{\Delta v}$ by Δt
• This $\vec{\Delta v}$ have it's own rectangular components. We can write:
${\vec{\Delta v}={\left (\Delta v_x \right ) \hat i}+{\left (\Delta v_y \right ) \hat j}}$
• So we get: $\mathbf{\bar{\vec{a}}=\frac{{\left (\Delta v_x \right ) \hat i}+{\left (\Delta v_y \right ) \hat j}}{\Delta t}}$
■ This can be written as: $\mathbf{\bar{\vec{a}}={\left ( \frac{\Delta v_x}{\Delta t} \right )\hat{i}}+{\left ( \frac{\Delta v_y}{\Delta t} \right )\hat{j}}}$
• $\mathbf{\frac{\Delta v_x}{\Delta t}}$ is the average acceleration in the x direction.
♦ We can denote it as: $\mathbf{|\bar{\vec{a}_x}|}$
♦ It is the 'magnitude of the x component' of $\mathbf{\bar{\vec{a}}}$
• $\mathbf{\frac{\Delta v_y}{\Delta t}}$ is the average acceleration in the y direction.
♦ We can denote it as: $\mathbf{|\bar{\vec{a}_y}|}$
♦ It is the 'magnitude of the y component' of $\mathbf{\bar{\vec{a}}}$
Thus we get Eq.4.10:
$\mathbf{\bar{\vec{a}}=|\bar{\vec{a_x}}|\hat{i}+|\bar{\vec{a_y}}|\hat{j}}$
1.Consider fig.4.24(a) below:
 |
| Fig.4.24 |
• When the stop watch showed t1 s, it is at P1
♦ So $\mathbf\small{\vec{v_1}}$ is the velocity vector when t = t1 s
♦ This $\mathbf\small{\vec{v_1}}$ is tangential to the path at P1
♦ That is., if we draw a tangent to the path at P1, $\mathbf\small{\vec{v_1}}$ will fall along that tangent
• When the stop watch showed t2 s, it is at P2
♦ So $\mathbf\small{\vec{v_2}}$ is the velocity vector when t = t2 s
♦ This $\mathbf\small{\vec{v_2}}$ is tangential to the path at P2.
2. So, in a duration of [Δt = (t2-t1) s], the velocity changes from $\mathbf\small{\vec{v_1}}$ to $\mathbf\small{\vec{v_2}}$
• Since there is a change in velocity, there must be an acceleration. Our aim is to find this acceleration.
3. We know that, $\mathbf{\text{Average acceleration} = \frac{\text{change in velocity}}{\text{change in time}}}$
• That is., $\bar{\vec{a}}=\frac{\vec{\Delta v}}{\Delta t}$
• So first, we have to find the change in velocity $\vec{\Delta v}$
• It is given by: $\vec{\Delta v}$ = $\mathbf\small{\vec{v_2}}$ - $\mathbf\small{\vec{v_1}}$
• That is., we have to find: [$\mathbf\small{\vec{v_2}}$ + (- $\mathbf\small{\vec{v_1}}$)]
4. We will find it graphically. Consider fig.4.24(b)
• $\mathbf\small{\vec{v_2}}$, $\mathbf\small{\vec{v_1}}$ and -($\mathbf\small{\vec{v_1}}$) are shown
• shift -($\mathbf\small{\vec{v_1}}$) so that, it's tail coincides with the head of $\mathbf\small{\vec{v_2}}$. This is shown in fig.c
• Draw a new vector (shown in cyan colour in fig.c) from the tail of $\mathbf\small{\vec{v_2}}$ to the head of -($\mathbf\small{\vec{v_1}}$)
• This new vector is the required $\vec{\Delta v}$
5. Consider fig.d:
• $\mathbf\small{\vec{v_1}}$ and $\mathbf\small{\vec{v_2}}$ are placed in such a way that, their tails coincide.
• Draw a new vector from the head of $\mathbf\small{\vec{v_1}}$ to head of $\mathbf\small{\vec{v_2}}$
• We see that, this new vector in fig.d, is same as the $\vec{\Delta v}$ that we obtained in fig.c
■ We can write a summary in a Question and Answer form:
• How do we find the $\vec{\Delta v}$ analytically?
Ans: Find (${\vec{v_2}}$ - ${\vec{v_1}}$)
• How do we find the $\vec{\Delta v}$ graphically?
Ans: Shift the two vectors so that, their tails coincide
♦ Draw a new vector such that:
♦ It's tail coincides with the head of ${\vec{v_1}}$
♦ It's head coincides with the head of ${\vec{v_2}}$
This new vector is the required $\vec{\Delta v}$
■Once we obtain $\vec{\Delta v}$, we can easily calculate $\bar{\vec{a}}$ by the equation: $\bar{\vec{a}}=\frac{\vec{\Delta v}}{\Delta t}$
■ Note that:
• The direction of $\vec{\Delta v}$ is very different from the direction of ${\vec{v_1}}$
• The direction of $\vec{\Delta v}$ is very different from the direction of ${\vec{v_2}}$ also
■ So the direction of $\bar{\vec{a}}$ will suffer the same condition:
• The direction of $\bar{\vec{a}}$ will be very different from the direction of ${\vec{v_1}}$
• The direction of $\bar{\vec{a}}$ will be very different from the direction of ${\vec{v_2}}$ also
6. We know that $\bar{\vec{a}}$ is a vector. So it will have two rectangular components. Let us find them:
• To find the $\bar{\vec{a}}$, we are dividing the $\vec{\Delta v}$ by Δt
• This $\vec{\Delta v}$ have it's own rectangular components. We can write:
${\vec{\Delta v}={\left (\Delta v_x \right ) \hat i}+{\left (\Delta v_y \right ) \hat j}}$
• So we get: $\mathbf{\bar{\vec{a}}=\frac{{\left (\Delta v_x \right ) \hat i}+{\left (\Delta v_y \right ) \hat j}}{\Delta t}}$
■ This can be written as: $\mathbf{\bar{\vec{a}}={\left ( \frac{\Delta v_x}{\Delta t} \right )\hat{i}}+{\left ( \frac{\Delta v_y}{\Delta t} \right )\hat{j}}}$
• $\mathbf{\frac{\Delta v_x}{\Delta t}}$ is the average acceleration in the x direction.
♦ We can denote it as: $\mathbf{|\bar{\vec{a}_x}|}$
♦ It is the 'magnitude of the x component' of $\mathbf{\bar{\vec{a}}}$
• $\mathbf{\frac{\Delta v_y}{\Delta t}}$ is the average acceleration in the y direction.
♦ We can denote it as: $\mathbf{|\bar{\vec{a}_y}|}$
♦ It is the 'magnitude of the y component' of $\mathbf{\bar{\vec{a}}}$
Thus we get Eq.4.10:
$\mathbf{\bar{\vec{a}}=|\bar{\vec{a_x}}|\hat{i}+|\bar{\vec{a_y}}|\hat{j}}$
Now we will see instantaneous acceleration
1. Consider fig.4.25 below:
An object travels along the green coloured path
(i) The object is at P when the stop watch shows 't' s
♦ At that instant, the velocity of the object is $\mathbf{\vec v}$
(ii) The object is at P1 when the stop watch shows 't1' s
♦ At that instant, the velocity of the object is $\mathbf{\vec v_1}$
• The time duration for travel from P to P1 = $\Delta t_1 = (t_1 - t)$
(iii) We can easily draw the 'change in velocity vector' $\vec{\Delta v_{p-p1}}$
We have learned how to draw it. We just need to draw a vector between $\mathbf{\vec v}$ and $\mathbf{\vec v_1}$. The direction must be from $\mathbf{\vec v}$ to $\mathbf{\vec v_1}$. It is shown below fig.a
(iv) When we divide this $\vec{\Delta v_{p-p1}}$ by $\Delta t_1$, we get the 'average acceleration vector' \(\mathbf{\vec{\bar{a}_{p-p1}}}\)
• Direction of $\vec{\Delta v_{p-p1}}$ is very different from both $\mathbf{\vec v}$ and $\mathbf{\vec v_1}$.
1. Consider fig.4.25 below:
 |
| Fig.4.25 |
An object travels along the green coloured path
(i) The object is at P when the stop watch shows 't' s
♦ At that instant, the velocity of the object is $\mathbf{\vec v}$
(ii) The object is at P1 when the stop watch shows 't1' s
♦ At that instant, the velocity of the object is $\mathbf{\vec v_1}$
• The time duration for travel from P to P1 = $\Delta t_1 = (t_1 - t)$
(iii) We can easily draw the 'change in velocity vector' $\vec{\Delta v_{p-p1}}$
We have learned how to draw it. We just need to draw a vector between $\mathbf{\vec v}$ and $\mathbf{\vec v_1}$. The direction must be from $\mathbf{\vec v}$ to $\mathbf{\vec v_1}$. It is shown below fig.a
(iv) When we divide this $\vec{\Delta v_{p-p1}}$ by $\Delta t_1$, we get the 'average acceleration vector' \(\mathbf{\vec{\bar{a}_{p-p1}}}\)
• Direction of $\vec{\Delta v_{p-p1}}$ is very different from both $\mathbf{\vec v}$ and $\mathbf{\vec v_1}$.
• So, direction of \(\mathbf{\vec{\bar{a}_{p-p1}}}\) will also be very different from both $\mathbf{\vec v}$ and $\mathbf{\vec v_1}$.
♦ In fact, it will be same as the direction of $\vec{\Delta v_{p-p1}}$. This is shown by the cyan 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
♦ At that instant, the velocity of the object is $\mathbf{\vec v}$
(ii) The object is at P2 when the stop watch shows 't2' s
♦ At that instant, the velocity of the object is $\mathbf{\vec v_2}$
• 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 'change in velocity vector' $\vec{\Delta v_{p-p2}}$
We just need to draw a vector between $\mathbf{\vec v}$ and $\mathbf{\vec v_2}$. The direction must be from $\mathbf{\vec v}$ to $\mathbf{\vec v_2}$. It is shown below fig.b
(iv) When we divide this $\vec{\Delta v_{p-p2}}$ by $\Delta t_2$, we get the 'average acceleration vector' \(\mathbf{\vec{\bar{a}_{p-p2}}}\)
• Direction of $\vec{\Delta v_{p-p2}}$ is very different from both $\mathbf{\vec v}$ and $\mathbf{\vec v_2}$.
We will repeat the steps:
(i) The object is at P when the stop watch shows 't' s
♦ At that instant, the velocity of the object is $\mathbf{\vec v}$
(ii) The object is at P2 when the stop watch shows 't2' s
♦ At that instant, the velocity of the object is $\mathbf{\vec v_2}$
• 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 'change in velocity vector' $\vec{\Delta v_{p-p2}}$
We just need to draw a vector between $\mathbf{\vec v}$ and $\mathbf{\vec v_2}$. The direction must be from $\mathbf{\vec v}$ to $\mathbf{\vec v_2}$. It is shown below fig.b
(iv) When we divide this $\vec{\Delta v_{p-p2}}$ by $\Delta t_2$, we get the 'average acceleration vector' \(\mathbf{\vec{\bar{a}_{p-p2}}}\)
• Direction of $\vec{\Delta v_{p-p2}}$ is very different from both $\mathbf{\vec v}$ and $\mathbf{\vec v_2}$.
• So, direction of \(\mathbf{\vec{\bar{a}_{p-p2}}}\) will also be very different from both $\mathbf{\vec v}$ and $\mathbf{\vec v_2}$.
♦ In fact, it will be same as the direction of $\vec{\Delta v_{p-p2}}$. This is shown by the cyan arrow in fig.b.
■ Direction of the cyan arrow in fig.b is different from that in fig.a. This is because, though P is the same, P1 and P2 are different
3. Now consider a point P3 which is closer to P. This is shown in fig.c
We will repeat the steps:
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 acceleration 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 a zero value.
• However, it can take very small values which are close to zero
6. In each case, to find the 'average acceleration', we calculate the ratio $\frac{\vec{\Delta v}}{\Delta t}$
• When $\Delta t$ becomes very close to zero, we call it: 'the limiting value of the ratio'
• Mathematically, it is written as: $\lim_{t\to 0}\frac{\vec{\Delta v}}{\Delta t}$
• But when $\Delta t$ is very close to zero, it is an instant.
• So what we calculate by the ratio, is instantaneous acceleration $\mathbf{\vec a}$.
♦ This is shown in fig.d
• It is not the average acceleration \(\mathbf{\vec{\bar{a}}}\).
■ So we can write: $\mathbf{\vec a =\lim_{t\to 0}\frac{\vec{\Delta v}}{\Delta t}}$
7. In linear motion, the two quantities below will have the same direction:
(i) $\mathbf{\vec v}$ of the object
(ii) $\mathbf{\vec a}$ of the object
■ In two dimensional motion, those two quantities may have different directions.
• That is., there can be an angle between those two vectors
• This angle may have any value between 0o to 180o both inclusive
♦ If it is 0o, then it means that $\mathbf{\vec a}$ has the same direction as $\mathbf{\vec v}$
♦ If it is 180o, then it means that $\mathbf{\vec a}$ has the exact opposite direction as $\mathbf{\vec v}$
8. The $\vec{\Delta v}$ in the numerator is a vector. We know it's rectangular components:
• x component is $\small\mathbf{\left ( \Delta v_x \right )\hat{i}}$
• y component is $\small\mathbf{\left ( \Delta v_y \right )\hat{j}}$
9. So the result in (6) becomes:
$\mathbf{\vec{a} = lim_{t\to 0}\left [ \left ( \frac{\Delta v_x}{\Delta t} \right )\hat{i}+\left ( \frac{\Delta v_y}{\Delta t} \right )\hat{j} \right ]}$
This can be written as:
$\mathbf{\vec{a} = \hat{i} \left [lim_{t\to 0}\left ( \frac{\Delta v_x}{\Delta t} \right ) \right ]+\hat{j} \left [lim_{t\to 0}\left ( \frac{\Delta v_y}{\Delta t} \right ) \right ]}$
10. There are two terms on the right side.
(i) Consider the first term:
• It is the limiting value of the 'velocity to time ratio' in the x direction.
• So it is the instantaneous acceleration in the x direction. It is a vector quantity
• We can denote it as $\mathbf{|\vec{a_x}|\hat i}$
(ii) Consider the second term:
• It is the limiting value of the 'velocity to time ratio' in the y direction.
• So it is the instantaneous acceleration in the y direction. It is a vector quantity
• We can denote it as $\mathbf{|\vec{a_y}|\hat j}$
■ Thus we get Eq.4.11:
$\mathbf\small{\vec a=|\vec{a_x}|\hat{i}+|\vec{a_y}|\hat{j}}$
■ So we can write:
• The instantaneous acceleration $\mathbf{\vec a}$ can be resolved into two rectangular components: $\mathbf{|\vec{a_x}|\hat i}$ and $\mathbf{|\vec{a_y}|\hat j}$
■ We can write the converse also:
• If we know the two rectangular components $\mathbf{|\vec{a_x}|\hat i}$, and $\mathbf{|\vec{a_y}|\hat j}$ of an acceleration $\mathbf{\vec a}$, then (see.fig.4.26 below):
(i) Magnitude of $\mathbf{\vec a}$ is given by Eq.4.11(a):
$\mathbf{\left | \vec{a} \right |=\sqrt{{|\vec{a_x}|^2}+{|\vec{a_y}|^2}}}$
(ii) Direction of $\mathbf{\vec a}$ is given by Eq.4.11(b):
$\mathbf{\tan\theta =\frac{|\vec{a_y}|}{|\vec{a_x}|}}$
• Where:
♦ $\mathbf{|\vec{a_x}|}$ is the magnitude of $\mathbf{\vec{a_x}}$
♦ $\mathbf{|\vec{a_y}|}$ is the magnitude of $\mathbf{\vec{a_y}}$
♦ $\mathbf{\theta}$ is the angle made by $\mathbf{\vec a}$ with the horizontal
3. Now consider a point P3 which is 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
♦ At that instant, the velocity of the object is $\mathbf{\vec v}$
(ii) The object is at P3 when the stop watch shows 't3' s
♦ At that instant, the velocity of the object is $\mathbf{\vec v_3}$
• The time duration for travel from P to P3 = $\Delta t_3 = (t_3 - t)$
• 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 'change in velocity vector' $\vec{\Delta v_{p-p3}}$
We just need to draw a vector between $\mathbf{\vec v}$ and $\mathbf{\vec v_3}$. The direction must be from $\mathbf{\vec v}$ to $\mathbf{\vec v_3}$. It is shown below fig.c
(iv) When we divide this $\vec{\Delta v_{p-p3}}$ by $\Delta t_3$, we get the 'average acceleration vector' \(\mathbf{\vec{\bar{a}_{p-p3}}}\)
• Direction of $\vec{\Delta v_{p-p3}}$ is very different from both $\mathbf{\vec v}$ and $\mathbf{\vec v_3}$.
■ Direction of the cyan arrow in fig.c is different from that in fig.a. This is because, though P is the same, P2 and P3 are different♦ At that instant, the velocity of the object is $\mathbf{\vec v}$
(ii) The object is at P3 when the stop watch shows 't3' s
♦ At that instant, the velocity of the object is $\mathbf{\vec v_3}$
• The time duration for travel from P to P3 = $\Delta t_3 = (t_3 - t)$
• 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 'change in velocity vector' $\vec{\Delta v_{p-p3}}$
We just need to draw a vector between $\mathbf{\vec v}$ and $\mathbf{\vec v_3}$. The direction must be from $\mathbf{\vec v}$ to $\mathbf{\vec v_3}$. It is shown below fig.c
(iv) When we divide this $\vec{\Delta v_{p-p3}}$ by $\Delta t_3$, we get the 'average acceleration vector' \(\mathbf{\vec{\bar{a}_{p-p3}}}\)
• Direction of $\vec{\Delta v_{p-p3}}$ is very different from both $\mathbf{\vec v}$ and $\mathbf{\vec v_3}$.
• So, direction of \(\mathbf{\vec{\bar{a}_{p-p2}}}\) will also be very different from both $\mathbf{\vec v}$ and $\mathbf{\vec v_3}$.
♦ In fact, it will be same as the direction of $\vec{\Delta v_{p-p3}}$. This is shown by the cyan arrow.
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 acceleration 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 a zero value.
• However, it can take very small values which are close to zero
6. In each case, to find the 'average acceleration', we calculate the ratio $\frac{\vec{\Delta v}}{\Delta t}$
• When $\Delta t$ becomes very close to zero, we call it: 'the limiting value of the ratio'
• Mathematically, it is written as: $\lim_{t\to 0}\frac{\vec{\Delta v}}{\Delta t}$
• But when $\Delta t$ is very close to zero, it is an instant.
• So what we calculate by the ratio, is instantaneous acceleration $\mathbf{\vec a}$.
♦ This is shown in fig.d
• It is not the average acceleration \(\mathbf{\vec{\bar{a}}}\).
■ So we can write: $\mathbf{\vec a =\lim_{t\to 0}\frac{\vec{\Delta v}}{\Delta t}}$
7. In linear motion, the two quantities below will have the same direction:
(i) $\mathbf{\vec v}$ of the object
(ii) $\mathbf{\vec a}$ of the object
■ In two dimensional motion, those two quantities may have different directions.
• That is., there can be an angle between those two vectors
• This angle may have any value between 0o to 180o both inclusive
♦ If it is 0o, then it means that $\mathbf{\vec a}$ has the same direction as $\mathbf{\vec v}$
♦ If it is 180o, then it means that $\mathbf{\vec a}$ has the exact opposite direction as $\mathbf{\vec v}$
8. The $\vec{\Delta v}$ in the numerator is a vector. We know it's rectangular components:
• x component is $\small\mathbf{\left ( \Delta v_x \right )\hat{i}}$
• y component is $\small\mathbf{\left ( \Delta v_y \right )\hat{j}}$
9. So the result in (6) becomes:
$\mathbf{\vec{a} = lim_{t\to 0}\left [ \left ( \frac{\Delta v_x}{\Delta t} \right )\hat{i}+\left ( \frac{\Delta v_y}{\Delta t} \right )\hat{j} \right ]}$
This can be written as:
$\mathbf{\vec{a} = \hat{i} \left [lim_{t\to 0}\left ( \frac{\Delta v_x}{\Delta t} \right ) \right ]+\hat{j} \left [lim_{t\to 0}\left ( \frac{\Delta v_y}{\Delta t} \right ) \right ]}$
10. There are two terms on the right side.
(i) Consider the first term:
• It is the limiting value of the 'velocity to time ratio' in the x direction.
• So it is the instantaneous acceleration in the x direction. It is a vector quantity
• We can denote it as $\mathbf{|\vec{a_x}|\hat i}$
(ii) Consider the second term:
• It is the limiting value of the 'velocity to time ratio' in the y direction.
• So it is the instantaneous acceleration in the y direction. It is a vector quantity
• We can denote it as $\mathbf{|\vec{a_y}|\hat j}$
■ Thus we get Eq.4.11:
$\mathbf\small{\vec a=|\vec{a_x}|\hat{i}+|\vec{a_y}|\hat{j}}$
■ So we can write:
• The instantaneous acceleration $\mathbf{\vec a}$ can be resolved into two rectangular components: $\mathbf{|\vec{a_x}|\hat i}$ and $\mathbf{|\vec{a_y}|\hat j}$
■ We can write the converse also:
• If we know the two rectangular components $\mathbf{|\vec{a_x}|\hat i}$, and $\mathbf{|\vec{a_y}|\hat j}$ of an acceleration $\mathbf{\vec a}$, then (see.fig.4.26 below):
$\mathbf{\left | \vec{a} \right |=\sqrt{{|\vec{a_x}|^2}+{|\vec{a_y}|^2}}}$
$\mathbf{\tan\theta =\frac{|\vec{a_y}|}{|\vec{a_x}|}}$
♦ $\mathbf{|\vec{a_x}|}$ is the magnitude of $\mathbf{\vec{a_x}}$
♦ $\mathbf{|\vec{a_y}|}$ is the magnitude of $\mathbf{\vec{a_y}}$
♦ $\mathbf{\theta}$ is the angle made by $\mathbf{\vec a}$ with the horizontal
So we have seen the acceleration of an object in 2-dimensional motion. In the next section, we will see a practical application of this acceleration.
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