Chapter 4.7 - Velocity vector from Displacement vector

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.

---

Now we will see instantaneous velocity
1. Consider fig.4.22(a) below:

Fig.4.22

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):

Fig.4.23

(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

---

So we have seen how to obtain velocity vectors. In the next section, we will see how to obtain acceleration vectors from velocity vectors.

PREVIOUS        CONTENTS          NEXT

Copyright©2018 Higher Secondary Physics. blogspot.in - All Rights Reserved

Chapter 3.4 - Instantaneous Velocity

In the previous section we saw average velocity and average speed. In this section we will see instantaneous velocity. We will write it in steps:
1. Fig.3.34 below shows a position-time graph of an object:

Fig.3.34

• In the above graph a point Q is marked. The following two items are clear:
(i) Q is at a distance of 5.12 m from the origin
(ii) The object passed Q when the stop watch showed 4 s
[Note: It is difficult to obtain the value 5.12 from the graph. So we can use the equation of the curve.
The equation is x = 0.08t³.
Substituting t = 4, we get:
s = 0.08 × (4)³= 0.08 × 64 = 5.12]  
2. Consider the instant at which the object passed Q
• We want to find the exact velocity at that instant
Let us try:
(i) The point of interest is Q. It is the position of the object when it travelled for 4 seconds
(ii) We need two neighbouring points of Q. 
• One before Q and 
• The other after Q
(iii) These new points should satisfy a condition:
• They must be at equal time intervals from Q
(iv)Let us choose a point P before Q such that:
    ♦ The object reached P, 1 s before Q. That is., when the stop watch showed 3 s  
• Let us choose a point R after Q such that:  
    ♦ The object reached R, 1 s after Q. That is., when the stop watch showed 5 s  
■ So the duration of travel from P to R = Δt = (t₂ - t₁) = (5 - 3) = 2 s
This is shown in fig.3.35 below:

Fig.3.35

(v) In the above graph, the distances corresponding to t = 3 and t = 5 are calculated using the same procedure that we used for t = 4:
s(3) = 0.08 × (3)³= 0.08 × 27 = 2.16 m
s(5) = 0.08 × (5)³= 0.08 × 125 = 10 m
(vi) Now we join PR. This is shown in fig.3.36 below:

Fig.3.36

• Slope of PR can be calculated by completing the triangle PRR'
• We get: Slope of PR = ^altitude⁄_base = ^RR'⁄_PR' = ^Δx⁄_Δt = ^(10-2.16)⁄_(5-3) = ^7.84⁄₂ = 3.92 ms^-1.
(vii) But we know that, this slope gives us only the 'average velocity' between P and R
• Though Q falls between P and R, 3.92 ms^-1 is not the instantaneous velocity at Q
• The duration of travel from P to R is 2 s. This '2 s' is not an 'instant'
3. So let us reduce the time interval:
• We need two new neighbouring points of Q. One before Q and the other after Q
• Like before, these new points should also satisfy a condition:
They must be at equal time intervals from Q
(i) Let us choose a point P₁ before Q such that:
    ♦ The object reached P₁, 0.5 s before Q. That is., when the stop watch showed 3.5 s  
Let us choose a point R₁ after Q such that:  
    ♦ The object reached R₁, 0.5 s after Q. That is., when the stop watch showed 4.5 s  
■ So the duration of travel from P₁ to R₁ = Δt = (t₂ - t₁) = (4.5 - 3.5) = 1 s
(ii) After marking P and R, we join them. This is shown in fig.3.37 below:

Fig.3.37

• In the above graph, the distances corresponding to t = 3.5 and t = 4.5 are calculated using the same procedure that we used for t = 4:
s(3) = 0.08 × (3.5)³= 0.08 × 27 = 3.43    
s(5) = 0.08 × (4.5)³= 0.08 × 125 = 7.29
(iii) Slope of P₁R₁ can be calculated by completing the triangle P₁R₁R₁'
• To avoid congestion, we will zoom in on the required area. This is shown in fig.3.38 below:

Fig.3.38

• We get: Slope of DF = ^altitude⁄_base = ^R₁R'₁ ⁄_P1R'1= ^Δx⁄_Δt = ^(7.29-3.43)⁄_(4.5-3.5) = ^3.86⁄₁ = 3.86 ms^-1
(iv) But we know that, this slope gives us only the 'average velocity' between P₁ and R₁
• Though Q falls between P₁ and R₁, 3.86 ms^-1 is not the instantaneous velocity at Q
• The duration of travel from P1 to R1 is 1 s. This '1 s' is not an 'instant'
4. So let us reduce the time interval:
• We need two new neighbouring points of Q. One before Q and the other after Q
• Like before, these new points should also satisfy a condition:
They must be at equal time intervals from Q
(i) Let us choose a point P₂ before Q such that:
    ♦ The object reached P₁, 0.25 s before Q. That is., when the stop watch showed 3.75 s  
Let us choose a point R₂ after Q such that:  
    ♦ The object reached R₂, 0.25 s after Q. That is., when the stop watch showed 4.25 s  
■ So the duration of travel from P₂ to R₂ = Δt = (t₂ - t₁) = (4.25 - 3.75) = 0.5 s
(ii) After marking P₂ and R₂, we join them. This is shown in fig.3.39 below. The triangle is also shown:

Fig.3.39

• We get: Slope of DF = ^altitude⁄_base = ^R₂R'₂ ⁄_P2R'2
= ^Δx⁄_Δt = ^{(6.1413-4.2188)}⁄_(4.25-3.75) = ^1.9225⁄_0.5 = 3.845 ms^-1.
(iv) But we know that, this slope gives us only the 'average velocity' between P₂ and R₂
• Though Q falls between P₂ and R₂, 3.845 ms^-1 is not the instantaneous velocity at Q
• The duration of travel from P₂ to R₂ is 0.5 s. This '0.5 s' is not an 'instant'

---

Let us write a summary of what we have done so far:
Step 1: We took points P and R which are centred at 4 s
• The duration of travel (Δt) from P to R was 2 s
• The average velocity with which the object travelled from P to R was 3.92 ms^-1
• But the '2 s' is not an instant. So '3.92 ms^-1' was discarded
Step 2: We took points P₁ and R₁ which are centred at 4 s
• The duration of travel (Δt) from P₁ to R₁ was 1 s
• The average velocity with which the object travelled from P₁ to R₁ was 3.86 ms^-1
• But the '1 s' is not an instant. So '3.86 ms^-1' was discarded
Step 3: We took points P₂ and R₂ which are centred at 4 s
• The duration of travel (Δt) from P₂ to R₂ was 0.5 s
• The average velocity with which the object travelled from P₂ to R₂ was 3.845 ms^-1
• But the '0.5 s' is not an instant. So '3.845 ms^-1' was discarded

---

• From the above 3 steps, it is clear that, with each step, the following two items are decreasing:
    ♦ Distance between P and R
    ♦ Time duration required for the travel between P and R
This can be shown as in fig.3.40 below:

Fig.3.40

• We can see that, the lines are becoming more and more aligned with the curve
• We must continue the steps until the time of travel between P and R is very small
• But it is not convenient to use graphs for further steps. because, the points will be very close to each other. We can use a table to do the calculations. It is shown below:

• The three steps that we already did are given in the first 3 rows of the table.
• Two more steps are done in the table. In the fifth step, Δt is very small. It is 0.01
■ But even this is not sufficient. Let us see the reason:
• Consider the ratio ^Δx⁄_Δt . We know that this ratio gives the average velocity
• To get the instantaneous velocity, the denominator Δt must be very small (like 0.00000001 s)
• It must be very small. We call it: 'infinitesimal time' . The meaning of 'infinitesimal' can be seen here.
    ♦ And at the same time, it must not be equal to zero. Because, division by zero is undefined
• Mathematically, such a small denominator is indicated as: '^Δx⁄_Δt when Δt → 0'
    ♦ That is.,we need the ratio when 'Δt tends to zero'
• Such a ratio can be easily calculated using the principles of calculus which we will study in maths classes.
• The reader may try to write the answer after learning calculus. The answer will be obtained as 3.84 ms^-1.
We know that, '^Δx⁄_Δt' is a slope.

---

• When we use calculus, we get a value of ^Δx⁄_Δt
    ♦ (The value of the ratio when the denominator Δt is very small)
• This value of the ratio, 'when Δt is very small', is also a slope
• It is the slope of the tangent to the curve at the point where we seek the instantaneous velocity
• The lines PR, P₁R₁, P₂R₂ etc., touches the curve at two points
• But the tangent will touch the curve only at one point, which is Q

---

Another example:
In the above example, the equation of motion was x = 0.08t³. We used this equation to find the distances at various times. Now consider an object moving according to the equation: 
x = 8.5 + 2.5t².
• What is the instantaneous velocity at t = 0 s ?
• What is the instantaneous velocity at t = 2 s ?
• What is the average velocity with which it travels between t = 2 and t = 4 s
Solution:
A. To find instantaneous velocities:
1. Based on the principles of calculus, the ratio '^Δx⁄_Δt when Δt → 0' is: 5t. 
2. So we get:
• At t = 0, the instantaneous velocity = 5 × 0 = 0 ms^-1 
• At t = 2, the instantaneous velocity = 5 × 2 = 10 ms^-1
B. To find average velocity:
1. Distance travelled when t = 2 s
= x(2) = (8.5 + 2.5×(2)²) = 18.5 m 
• Distance travelled when t = 4 s
= x(4) = (8.5 + 2.5×(4)²) = 48.5 m
2. So Δx = (48.5 - 18.5) = 30
• Δt = (t₂ - t₁) = (4 - 2) = 2 s 
3. Average velocity = ^Δx⁄_Δt = ³⁰⁄₂ = 15 ms^-1.

---

In the next section, we will see acceleration.

PREVIOUS        CONTENTS          NEXT

Copyright©2018 Higher Secondary Physics. blogspot.in - All Rights Reserved