Chapter 4.10 - Projectile Motion

In the previous section we saw 2-dimensional motion under constant acceleration. In this section, we will see Projectile motion.
1. Consider fig.4.29(a) below:
 

Fig.4.29

• A stone is thrown into the air. 
• It is not thrown straight up. But at an angle (less than 90^o) with the horizontal. 
2. The stop watch is turned on at the instant when the stone is thrown. 
• The position of the stone at that instant is taken as the origin ‘O’. 
• A horizontal line through O is taken as the x axis.
• A vertical line through O is taken as y axis.
• The velocity of the stone at O is called initial velocity of the projectile. It is denoted as $\mathbf\small{\vec v_0}$   
3. Once the stone is thrown, it is on it’s own. That is., once it is thrown, no propelling force acts on it.
(In a rocket, it’s engine produces exhaust gas which propels it forward. Such a motion is not considered as projectile motion)
4. We see that the stone is thrown at an angle.
• So the initial velocity will have a vertical component $\mathbf\small{\vec v_{0y}}$ and a horizontal component $\mathbf\small{\vec v_{0x}}$
5. The vertical component is responsible for taking the stone ‘vertically away’ from O
• But this vertical component will be affected by the acceleration due to gravity ‘g’
    ♦ This is the acceleration vector. We can denote it as (g)$\mathbf\small{\hat j}$
• As a result, magnitude of the vertical velocity component will go on decreasing.   
6. The horizontal component is responsible for taking the stone ‘horizontally away’ from the origin
• This component is not affected by ‘g’
• So the horizontal component of will remain constant during the entire journey.
■ Note that, the air resistance can cause opposition to the projectile motion. But for our present discussion, air resistance is considered to be negligible. So we will not take it into account here.
7. At O, let $\mathbf\small{\theta_0}$ be the angle made by $\mathbf\small{\vec v_0}$ with the horizontal. Then at O:
• The horizontal component of $\mathbf\small{\vec v_0}$ is given by: $\mathbf\small{\vec v_{0x}}$ = $\mathbf\small{(\left | \vec{v_0} \right |\cos \theta _0)}\hat{i}$ 
• The vertical component of $\mathbf\small{\vec v_0}$ is given by: $\mathbf\small{\vec v_{0y}}$ = $\mathbf\small{(\left | \vec{v_0} \right |\sin \theta _0)}\hat{j}$
8. The stone was thrown when the stop watch showed '0' s. What happens to these components when the stop watch shows a reading of 't' seconds?
Ans: The horizontal component will remain the same because, there is no acceleration in the horizontal direction
■ The vertical component will have a smaller value because there is negative acceleration (due to gravity) in the vertical direction
• We can find it's exact value at time = 't' s
• For that, we use the familiar equation:  v = v₀ + at
• Thus we can write: $\mathbf\small{\vec{v_y}=\vec{v_{0y}}+\vec{a_y}\,t}$
$\mathbf\small{\Rightarrow \vec{v_y}=(\left | \vec v_0 \right |\sin\theta _0)\hat{j}-(g)\hat{j}t}$    
$\mathbf\small{\Rightarrow \vec{v_y}=(\left | \vec v_0 \right |\sin\theta _0-gt)\hat{j}}$
• So we can write:
At any time 't', after the beginning of the journey, the magnitude of the vertical component of velocity is given by Eq.4.12: $\mathbf\small{\left | \vec{v_y} \right |=(\left | \vec v_0 \right |\sin\theta _0-gt)}$
9. At time = 't' seconds:
• The magnitude of the horizontal component remains the same
• The vertical component has a lower magnitude as given by Eq.4.12 above.
■ As a result, the resultant velocity $\mathbf\small{\vec v}$ (which is the resultant of the horizontal and vertical components) will have a smaller magnitude than $\mathbf\small{\vec v_0}$. This is shown in fig.4.29(b). We see the following:
• At time = 't' seconds:
    ♦ The stone has reached P
    ♦ $\mathbf\small{\vec v}$ has a smaller length than $\mathbf\small{\vec v_0}$
    ♦ $\mathbf\small{\theta}$ is different from $\mathbf\small{\theta_0}$
10. We saw how the 'velocity of the stone' varies during it's travel. Next we will see how 'it's distance from O' varies
■ First we will see the horizontal travel
(i) We have seen that the horizontal velocity remains the same.
• So we can use the familiar 'equation for uniform motion': s = vt
(ii) Thus we get:
Horizontal displacement in time 't' s = $\mathbf\small{\vec{\Delta r_x}=(\left | \vec{v_0} \right |\cos\theta_0 )\hat{i}\times t}$
$\mathbf\small{\Rightarrow \vec{\Delta r_x}=[(\left | \vec{v_0} \right |\cos\theta_0 )t]\hat{i}}$  
(iii) That means, the magnitude of $\mathbf\small{\vec{\Delta r_x}}$ = $\mathbf\small{\left | \vec{\Delta r_x} \right |=\left ( \left | \vec{v_0} \right | \cos \theta _0 \right )t}$
(iv) This magnitude is the distance OP'. But the distance OP' is the x coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\left ( \left | \vec{v_0} \right | \cos \theta _0 \right )t}$' from the y axis
• In other words, at any time 't', after the beginning of the journey, the x coordinate of the object is given by Eq.4.13: x = $\mathbf\small{\left ( \left | \vec{v_0} \right | \cos \theta _0 \right )t}$ 
■ Now we will see the vertical travel
(i) The vertical travel is affected by an acceleration 'g'. So we will use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• Thus we can write: 
Vertical displacement in time 't' s = $\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$
(iv) This magnitude is the distance P'P. But the distance PP' is the y coordinate of P
• So we can write: At any time 't', after the beginning of the journey, the object will be at a parallel distance of '$\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$' from the x axis
• In other words, at any time 't', after the beginning of the journey, the y coordinate of the object is given by Eq.4.14: y = $\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$
11. So we are now able to specify the position of a projectile at any time 't'.
• We are able to do it by using x and y coordinates.
■ If we can eliminate 't' from Eqs.4.13 and 4.14, we will get a direct relation between x and y.
Let us try:
(i) From Eq.4.13, we get: $\mathbf\small{t=\frac{x}{\left |\vec{v_0}  \right |\cos \theta _0}}$
• We can use this instead of 't' in Eq.4.14. 
• We get Eq.4.15: $\mathbf\small{y=\left [ \tan\theta_0  \right ]x-\left [ \frac{g}{2\left ( \left | \vec{v_0} \right |\cos \theta_0  \right )^2} \right ]x^2}$
(ii) Consider the quantity inside the first pair of square brackets
• $\mathbf\small{\theta_0}$ is the initial angle with which the stone is thrown at the beginning
• Once the stone is thrown at a particular initial angle, it will not be altered
• That is., $\mathbf\small{\theta_0}$ is a constant. 
• So $\mathbf\small{\tan \theta_0}$ is a constant. We will denote it as 'a'   
(iii) Consider the quantities inside the second pair of square brackets
    ♦ 'g' and '2' are constants
    ♦ $\mathbf\small{\theta_0}$ is a constant as we saw above
• Now  $\mathbf\small{\left | \vec{v_0} \right |}$ remains
    ♦ Once the stone is thrown at a particular initial velocity, it will not be altered
    ♦ That is., $\mathbf\small{\left | \vec{v_0} \right |}$ is a constant
• So every thing inside the second pair are constants
• So the final result inside that second pair is a constant. We will denote it as 'b'
(iv) Eq.4.15 becomes: $\mathbf\small{y=ax+bx^2}$
Where $\mathbf\small{a=\left [ \tan\theta_0  \right ]\: \: \text{and}\; \; b=-\left [ \frac{g}{2\left ( \left | \vec{v_0} \right |\cos \theta_0  \right )^2} \right ]}$
(v) But $\mathbf\small{y=ax+bx^2}$ is the equation of a parabola. So we can write:
■ The path of a projectile is a parabola
• This is shown in fig.c
12. Time required to reach the maximum height:
• Consider the path of the projectile shown in fig.c
• We see a peak point M. After M, we see no further upward motion
• That means, at this peak point M, the magnitude of the vertical component is zero
(i) Let $\mathbf\small{t_m}$ be the time required to reach M
(ii) Consider the vertical component of the velocity. We can use Eq.4.12: $\mathbf\small{\left | \vec{v_y} \right |=(\left | \vec v_0 \right |\sin\theta _0-gt_m)}$
(iii) Substituting the known values, we get: $\mathbf\small{0=(\left | \vec v_0 \right |\sin\theta _0-gt_m)}$   
■ From this we get Eq.4.16: $\mathbf\small{t_m=\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$
13. Maximum height reached by the stone (h_m):
(i) For this, we need the height of M from the x axis
(ii) Let us use Eq.4.14: y = $\mathbf\small{\left | \vec{v_{0y}} \right |t-\frac{1}{2}g t^2}$
• In this equation, if we put t = t_m, we will get the vertical distance traveled during 't_m'
(iii) But the 'vertical distance traveled during t_m' is the height of M. So we get:
$\mathbf\small{y=h_m=\left ( \left |\vec v_0 \right | \sin \theta_0 \right )\left ( \frac{\left |\vec v_0 \right | \sin \theta_0}{g} \right )-\frac{g}{2}\left ( \frac{\left |\vec v_0 \right | \sin \theta_0}{g} \right )^2}$
• From this, we get Eq.4.17: $\mathbf\small{h_m=\frac{\left (\left | \vec v_0 \right| \sin \theta_0  \right )^2}{2g}}$
14. Time required for the whole flight (T_f):
(i) After M, the stone continues the flight for some more time. In the end, it falls back to the ground.
• Let us consider the vertical motion after M. We want the time 't' required for this motion.
• The vertical distance traveled in this motion is h_m.
(ii) We can use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$ 
• In this motion, the initial velocity is zero. It is like the stone just dropped from a height of h_m
• So we can put v₀ = 0
• We get: $\mathbf\small{h_m=0 \times t+\frac{1}{2}g{t}^2}$
$\mathbf\small{\Rightarrow \frac{\left (\left | \vec v_0 \right| \sin \theta_0  \right )^2}{2g}=\frac{1}{2}gt^2}$
$\mathbf\small{\Rightarrow \frac{\left (\left | \vec v_0 \right| \sin \theta_0  \right )^2}{g^2}=t^2}$
$\mathbf\small{\Rightarrow t=\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$
• So total time of flight = T_f = (t_m+t) = $\mathbf\small{\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}+\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$ 
• So we get Eq.4.18: $\mathbf\small{T_f=\frac{2\left | \vec{v_0} \right |\sin \theta_0}{g}}$
■ Note:
• From Eq.4.16, we have: Time required for the upward travel from O to M = $\mathbf\small{\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$   
• In the above step (13), we have: Time required for the downward travel from M to the ground = $\mathbf\small{\frac{\left | \vec{v_0} \right |\sin \theta_0}{g}}$   
• So the times for upward travel and downward travel are the same
15. Horizontal range of a projectile ($\mathbf\small{|\vec R|}$):
(i) For this we consider the horizontal motion
• The horizontal component of the velocity (which is a constant value) will be effective for the entire time (T_f) of the flight 
(ii) So the horizontal distance = Horizontal component of velocity × time
= $\mathbf\small{\vec v_{0x}}$ × $\mathbf\small{T_f}$ = $\mathbf\small{(\left | \vec{v_0} \right |\cos \theta _0)}\hat{i}$ × $\mathbf\small{\frac{2\left | \vec{v_0} \right |\sin \theta_0}{g}}$ =$\mathbf\small{\frac{\left ( \left | \vec{v_0} \right |^2 2 \sin \theta_0 \cos \theta_0 \right )\hat{i}}{g}}$
• But from math classes, we have: $\mathbf\small{2 \sin \theta_0 \cos \theta_0}$ = $\mathbf\small{\sin 2\theta_0}$
(iii) So we can write: $\mathbf\small{\vec{R}=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )\hat{i}}{g}}$
(iv) Magnitude of $\mathbf\small{\vec R}$ is the actual distance
■ Thus we get Eq.4.19: Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
16. Maximum possible range for a given velocity:
• Suppose that a machine can throw an object only at a certain 'fixed speed' $\mathbf\small{\left |\vec v_0 \right |}$
    ♦ But the angle of projection can be changed to any value.
[That is., magnitude of $\mathbf\small{\vec v_0}$ is fixed. But the direction can change]
■ Then what angle would we choose to obtain 'maximum range'?
Solution:
1. We have:
Range of the projectile = $\mathbf\small{\left |\vec{R} \right |=\frac{\left ( \left | \vec{v_0} \right |^2 \sin 2\theta_0 \right )}{g}}$
2. In our present case, $\mathbf\small{\left |\vec v_0 \right |}$ is a constant
• So the only variable is sin 2θ₀.
3. That means, for maximum range, sin 2θ₀ must be maximum
• The maximum value possible for sin 2θ₀ is '1'. 
4. This '1' is obtained when '2θ₀' is 90^o.
• So θ₀ must be 45^o.
■ We can write:
The maximum range is obtained when the angle of projection θ₀ is 45^o.

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In the next section, we will apply the above equations to an actual projectile.

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Chapter 4.9 - Motion under Constant Acceleration

In the previous section we saw the components of instantaneous acceleration. Also, in the section before that, we saw the components of instantaneous velocity. In this section, we will see how we can put all of them to a practical application.

Motion under constant acceleration

1. In fig.4.27(a) below, a body is moving in the xy plane.

Fig.4.27

• It is moving along the path shown by green colour.
• It is moving under a constant acceleration $\mathbf\small{\vec a}$
2. The experiment began at the instant when stop watch was turned on. 
• At that instant t = 0
• At that instant, the object is at P₀.
• At that instant, the velocity of the object is $\mathbf\small{\vec v_0}$.
3. After 't' seconds, the object is at P.
• At that instant, the reading in the stop watch is 't' s
• At that instant, the velocity of the object is $\mathbf\small{\vec v}$
■ We want to know how $\mathbf{\vec v_0}$ became $\mathbf\small{\vec v}$  
4. In fig.b, the velocity vectors are resolved into their rectangular components.
• We will consider the x components together and treat them separately
• We will consider the y components together and treat them separately
■ Consider the x components:
• The x component $\mathbf\small{\vec v_{0x}}$ has become $\mathbf\small{\vec v_x}$
How did that happen?
Ans: The x component of $\mathbf\small{\vec a}$, which is $\mathbf\small{\vec a_x}$, acted on the object for a duration of 't' seconds.
• As a result, $\mathbf\small{\vec v_{0x}}$ became $\mathbf\small{\vec v_x}$
• We can use the familiar equation: v = v₀ + at
• Thus we can write: $\mathbf\small{\left |\vec{v_x} \right|=\left |\vec{v_{0x}}\right |+\left |\vec{a_x}\right |t}$
• So we get the magnitude of $\mathbf\small{\vec{v_x}}$
• It's direction is ofcourse, parallel to the x axis
• Thus we get the complete $\mathbf\small{\vec v_x}$
■ Consider the y components:
• The y component $\mathbf\small{\vec v_{0y}}$ has become $\mathbf\small{\vec v_y}$
How did that happen?
Ans: The y component of $\mathbf\small{\vec a}$, which is $\mathbf\small{\vec a_y}$, acted on the object for a duration of 't' seconds.
• As a result, $\mathbf\small{\vec v_{0y}}$ became $\mathbf\small{\vec v_y}$
• We can use the same familiar equation again: v = v₀ + at
• Thus we can write: $\mathbf\small{\left |\vec{v_y} \right|=\left |\vec{v_{0y}}\right |+\left |\vec{a_y}\right |t}$
• So we get the magnitude of $\mathbf\small{\vec{v_y}}$
• It's direction is ofcourse, parallel to the y axis
• Thus we get the complete $\mathbf\small{\vec v_y}$
5. So we independently derived $\mathbf\small{\vec v_x}$ and $\mathbf\small{\vec v_y}$
• Now we find the resultant of $\mathbf\small{\vec v_x}$ and $\mathbf\small{\vec v_y}$
• The resultant thus obtained is $\mathbf\small{\vec v}$

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In a similar way, the 'distance travelled' from P₀ to P can also be analysed
1. In fig.4.28(a) below, the position vectors of the points P₀ and P are shown

Fig.4.28

• Position vector of P₀ is $\mathbf\small{\vec r_0}$
• Position vector of P is $\mathbf\small{\vec r}$
2. We can easily draw the displacement vector $\mathbf\small{\vec{\Delta r}}$ (Details here)
• It is shown in fig.b
3. The rectangular components of this $\mathbf\small{\vec{\Delta r}}$ are shown in fig.c
The following two points are obvious:
(i) The horizontal distance travelled by the object during the 't' seconds is equal to the magnitude of $\mathbf\small{\vec{{\Delta r}_x}}$
• That is., P₀P' = $\mathbf\small{\left | \vec{{\Delta r}_x} \right |}$
(ii) The vertical distance travelled by the object during the 't' seconds is equal to the magnitude of $\mathbf\small{\vec{{\Delta r}_y}}$
• That is., P'P = $\mathbf\small{\left | \vec{{\Delta r}_y} \right |}$
4. So our next aim is to find P₀P' and P'P
■ First we will see P₀P':
(i) Consider the horizontal component of the velocity
• At P₀, the object is having a velocity of $\mathbf\small{\vec v_{0x}}$
• So, for the travel from to P', the initial velocity is equal to $\mathbf\small{\vec v_{0x}}$
(ii) The object travelling with this initial velocity is acted upon by a constant acceleration of $\mathbf\small{\vec a_x}$
• Also this travel took 't' seconds
(iii) We can use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• Thus we can write: Distance P₀P' = $\mathbf\small{\left | \vec{v_{0x}} \right |t+\frac{1}{2}\left | \vec{a_x}\right |t^2}$
(iv) So we get the magnitude of $\mathbf\small{\vec{{\Delta r}_x}}$
• It's direction is ofcourse, parallel to the x axis
• Thus we get the complete $\mathbf\small{\vec{{\Delta r}_x}}$
■ Now we will see P'P:
(i) Consider the vertical component of the velocity
• At P', the object is having a velocity of $\mathbf\small{\vec v_{0y}}$
• So, for the vertical travel from P, the initial velocity is equal to $\mathbf\small{\vec v_{0y}}
(ii) The object travelling with this initial velocity is acted upon by a constant acceleration of $\mathbf\small{\vec a_y}$
• Also this travel took 't' seconds
(iii) We can use the familiar equation again: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• Thus we can write: Distance P'P = $\mathbf\small{\left | \vec{v_{0y}} \right |t+\frac{1}{2}\left | \vec{a_y}\right |t^2}$
(iv) So we get the magnitude of $\mathbf\small{\vec{{\Delta r}_y}}$
• It's direction is ofcourse, parallel to the y axis
• Thus we get the complete $\mathbf\small{\vec{{\Delta r}_y}}$
5. Thus we get the two component vectors.
• Their resultant will give the required displacement vector $\mathbf\small{\vec{\Delta r}}$
■ So we can write a summary:
• The 2-dimensional motion problems (with constant acceleration) can be effectively solved by considering two independent motions:
    ♦ One parallel to the x axis
    ♦ The other parallel to the y axis

Solved example 4.5
A particle starts from origin at t = 0 with a velocity $\mathbf\small{5 \hat{i}\,ms^{-1}}$ and moves in x-y plane under action of a force which produces a constant acceleration of $\mathbf\small{(3 \hat{i}+2 \hat{j})\,ms^{-2}}$. (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m ? (b) What is the speed of the particle at this time ?
Solution:
We will consider the motions in x and y directions separately:
■ Motion in x direction:
• Initial x coordinate = 0
• Initial velocity $\mathbf\small{ | \vec{v_{0x}} | }$ = 5 ms^-1
• Constant acceleration $\mathbf\small{ | \vec{a_{x}} | }$ = 3 ms^-2
■ Motion in y direction:
• Initial y coordinate = 0
• Initial velocity $\mathbf\small{ | \vec{v_{0y}} | }$ = 0 ms^-1
• Constant acceleration $\mathbf\small{ | \vec{a_{y}} | }$ = 2 ms^-2.
Now we can write the steps:
Part (a):
1. The particle traveled 84 m horizontally. Let 't' be the time required for this travel.  
• We can use the familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• Thus we can write: 84 m = $\mathbf\small{\left | \vec{v_{0x}} \right |t+\frac{1}{2}\left | \vec{a_x}\right |t^2}$
• Substituting the values, we get: 84 = 5t + 1.5t²
• Solving this quadratic equation, we get, t = 6 s
• So the particle traveled for 6 s horizontally to reach a point whose x coordinate is 84 m
2. During those 6 s, there was vertical motion also
• We can use the same familiar equation: $\mathbf\small{s=v_0 t+\frac{1}{2}at^2}$
• Thus we can write: $\mathbf\small{\left | \vec{{\Delta r}_y} \right |}$ = $\mathbf\small{\left | \vec{v_{0y}} \right |t+\frac{1}{2}\left | \vec{a_y}\right |t^2}$
• Substituting the values, we get: $\mathbf\small{\left | \vec{{\Delta r}_y} \right |}$ = 0 + 0.5×2×6² = 36 m
■ Thus the required y coordinate is 36 m
Part (b):
1. Consider the motion in x direction:
• We can use the familiar equation: v = v₀ + at
• Thus we can write: $\mathbf\small{\left |\vec{v_x} \right|=\left |\vec{v_{0x}}\right |+\left |\vec{a_x}\right |t}$ 
• Substituting the values, we get: $\mathbf\small{\left | \vec{{v}_x} \right |}$ = 5 + 3 × 6 = 23 ms^-1 
2. Consider the motion in y direction:
• We can use the same familiar equation: v = v₀ + at
• Thus we can write: $\mathbf\small{\left |\vec{v_y} \right|=\left |\vec{v_{0y}}\right |+\left |\vec{a_y}\right |t}$ 
• Substituting the values, we get: $\mathbf\small{\left | \vec{{v}_y} \right |}$ = 0 + 2 × 6 = 12 ms^-1. 
3. The required speed is the magnitude of the resultant of the two velocities. It is given by the equation:
$\mathbf\small{ | \vec{v} |=\sqrt{| \vec{v_x} |^2+| \vec{v_y} |^2} }$
• Substituting the values, we get: speed = √[23²+12²] = 26 ms^-1

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Based on the above discussion, we can now learn about Projectile motion. We will see it in the next section.

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Chapter 4.8 - Acceleration vector from Velocity vectors

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:

Fig.4.24

■ 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:
$\mathbf{\bar{\vec{a}}=|\bar{\vec{a_x}}|\hat{i}+|\bar{\vec{a_y}}|\hat{j}}$

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Now we will see instantaneous acceleration
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 $\mathbf{\vec v_2}$. 

• 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:

(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 $\mathbf{\vec v_3}$. 

• 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.

■ 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
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 0^o to 180^o both inclusive 
    ♦ If it is 0^o, then it means that $\mathbf{\vec a}$ has the same direction as $\mathbf{\vec v}$ 
    ♦ If it is 180^o, 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    

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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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