In the previous section we saw the formula to find the resultant of two vectors. In this section, we will see motion in two dimension.
First we will see Displacement. We will write it in steps:
1.Fig.4.20(a) shows the position 'P' of an object in the xy plane.
• The object is at 'P' when the stop watch shows a reading of 't' s.
2.We can draw the position vector $\small{\vec{r}}$. We have seen the details about position vector in a previous section.
3. We have also learned about the 'rectangular components of any given vector'. So we can write:
$\small{\vec{r}=\text{x}\hat{i}+\text{y}\hat{j}}$. This is also shown in fig.a
• The horizontal component of $\small{\vec{r}}$ is x$\small{\hat{i}}$
♦ Clearly, this horizontal component has a magnitude of x units
♦ So the distance OP' = x units
• The horizontal component of $\small{\vec{r}}$ is y$\small{\hat{j}}$
♦ Clearly, this vertical component has a magnitude of y units
♦ So the distance OP'' = y units
So we can write: The coordinates of the object at time 't' s are (x,y)
4. Now consider fig.b
■ The object is moving along the path shown in green color
• When the stop watch showed t1 s, it is at P1
♦ So $\small{\vec{r_1}}$ is the position vector when t = t1 s
• When the stop watch showed t2 s, it is at P2
♦ So $\small{\vec{r_2}}$ is the position vector when t = t2 s
5. Put a vector between P1 and P2, pointing from P1 towards P2. It is shown in fig.c
■ It is a vector joining the initial and final positions. It is shown in fig.c
• So it is a displacement vector. We will denote it as $\vec{\Delta r}$
• We can write:
$\vec{\Delta r}$ is the displacement vector corresponding to the 'motion in a time duration of (t2-t1) s'
6. The horizontal and vertical dashed lines in fig.c help us to find the rectangular components of $\vec{\Delta r}$. We find that:
♦ The horizontal component of $\small{\vec{\Delta r}}$ has a magnitude of Δx
♦ The vertical component of $\small{\vec{\Delta r}}$ has a magnitude of Δy
■ Thus we can write:
$\vec{\Delta r}={\left (\Delta x \right ) \hat i}+{\left (\Delta y \right ) \hat j}$
7. But from our math classes, we know how to find the actual values of Δx and Δy
• We have: Δx = (x2-x1) and Δy = (y2-y1)
Where,
♦ (x1,y1) are the coordinates of P1
♦ (x2,y2) are the coordinates of P2.
■ Thus we get:
$\vec{\Delta r}={(x_2-x_1)\hat i}+{(y_2-y_1)\hat j}$
8. Now we want the relation between $\small{\vec{r_1}}$ , $\small{\vec{r_2}}$ and $\small{\vec{\Delta r}}$
• Fig.4.21(a) below shows the three vectors
• We see that:
(i) The tail of $\small{\vec{\Delta r}}$ coincides with the head of $\small{\vec{r_1}}$
(ii) $\small{\vec{r_2}}$ starts from the tail of $\small{\vec{r_1}}$ and ends at the head of $\small{\vec{\Delta r}}$
■ Thus, by the triangle method of addition, what we have in fig.4.21(a) is:
$\small{\vec{r_1}}$ + $\small{\vec{\Delta r}}$ = $\small{\vec{r_2}}$
9. Let us find ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$)
• That is., we want: [$\small{\vec{r_2}}$ + (-$\small{\vec{r_1}}$) ]
10. So we must first find (-$\small{\vec{r_1}}$)
For that, reverse ($\small{\vec{r_1}}$) to form (-$\small{\vec{r_1}}$). This is shown in fig.b
11. Then shift (-$\small{\vec{r_1}}$) so that it's tail coincide with the head of $\small{\vec{r_2}}$. This is shown in fig.c
12. Fig.d shows the result ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$)
Compare ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$) in fig.d with $\small{\vec{\Delta r}}$ in fig.a. We see that, both are same
• So it is proved:
($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$) = $\small{\vec{\Delta r}}$
■ We can write a summary in the form of Question and Answer:
• How do we find the displacement vector analytically?
Ans: Find ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$)
• How do we find the displacement vector graphically?
Ans: Draw a new vector such that:
♦ It's tail coincides with the head of $\small{\vec{r_1}}$
♦ It's head coincides with the head of $\small{\vec{r_2}}$
This new vector is the required displacement vector
First we will see Displacement. We will write it in steps:
1.Fig.4.20(a) shows the position 'P' of an object in the xy plane.
 |
| Fig.4.20 |
2.We can draw the position vector $\small{\vec{r}}$. We have seen the details about position vector in a previous section.
3. We have also learned about the 'rectangular components of any given vector'. So we can write:
$\small{\vec{r}=\text{x}\hat{i}+\text{y}\hat{j}}$. This is also shown in fig.a
• The horizontal component of $\small{\vec{r}}$ is x$\small{\hat{i}}$
♦ Clearly, this horizontal component has a magnitude of x units
♦ So the distance OP' = x units
• The horizontal component of $\small{\vec{r}}$ is y$\small{\hat{j}}$
♦ Clearly, this vertical component has a magnitude of y units
♦ So the distance OP'' = y units
So we can write: The coordinates of the object at time 't' s are (x,y)
4. Now consider fig.b
■ The object is moving along the path shown in green color
• When the stop watch showed t1 s, it is at P1
♦ So $\small{\vec{r_1}}$ is the position vector when t = t1 s
• When the stop watch showed t2 s, it is at P2
♦ So $\small{\vec{r_2}}$ is the position vector when t = t2 s
5. Put a vector between P1 and P2, pointing from P1 towards P2. It is shown in fig.c
■ It is a vector joining the initial and final positions. It is shown in fig.c
• So it is a displacement vector. We will denote it as $\vec{\Delta r}$
• We can write:
$\vec{\Delta r}$ is the displacement vector corresponding to the 'motion in a time duration of (t2-t1) s'
6. The horizontal and vertical dashed lines in fig.c help us to find the rectangular components of $\vec{\Delta r}$. We find that:
♦ The horizontal component of $\small{\vec{\Delta r}}$ has a magnitude of Δx
♦ The vertical component of $\small{\vec{\Delta r}}$ has a magnitude of Δy
■ Thus we can write:
$\vec{\Delta r}={\left (\Delta x \right ) \hat i}+{\left (\Delta y \right ) \hat j}$
7. But from our math classes, we know how to find the actual values of Δx and Δy
• We have: Δx = (x2-x1) and Δy = (y2-y1)
Where,
♦ (x1,y1) are the coordinates of P1
♦ (x2,y2) are the coordinates of P2.
■ Thus we get:
$\vec{\Delta r}={(x_2-x_1)\hat i}+{(y_2-y_1)\hat j}$
8. Now we want the relation between $\small{\vec{r_1}}$ , $\small{\vec{r_2}}$ and $\small{\vec{\Delta r}}$
• Fig.4.21(a) below shows the three vectors
 |
| Fig.4.21 |
(i) The tail of $\small{\vec{\Delta r}}$ coincides with the head of $\small{\vec{r_1}}$
(ii) $\small{\vec{r_2}}$ starts from the tail of $\small{\vec{r_1}}$ and ends at the head of $\small{\vec{\Delta r}}$
■ Thus, by the triangle method of addition, what we have in fig.4.21(a) is:
$\small{\vec{r_1}}$ + $\small{\vec{\Delta r}}$ = $\small{\vec{r_2}}$
9. Let us find ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$)
• That is., we want: [$\small{\vec{r_2}}$ + (-$\small{\vec{r_1}}$) ]
10. So we must first find (-$\small{\vec{r_1}}$)
For that, reverse ($\small{\vec{r_1}}$) to form (-$\small{\vec{r_1}}$). This is shown in fig.b
11. Then shift (-$\small{\vec{r_1}}$) so that it's tail coincide with the head of $\small{\vec{r_2}}$. This is shown in fig.c
12. Fig.d shows the result ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$)
Compare ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$) in fig.d with $\small{\vec{\Delta r}}$ in fig.a. We see that, both are same
• So it is proved:
($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$) = $\small{\vec{\Delta r}}$
■ We can write a summary in the form of Question and Answer:
• How do we find the displacement vector analytically?
Ans: Find ($\small{\vec{r_2}}$ - $\small{\vec{r_1}}$)
• How do we find the displacement vector graphically?
Ans: Draw a new vector such that:
♦ It's tail coincides with the head of $\small{\vec{r_1}}$
♦ It's head coincides with the head of $\small{\vec{r_2}}$
This new vector is the required displacement vector
So we have seen how to obtain displacement vectors from position vectors. In the next section, we will see how to obtain velocity vectors from displacement vectors.
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