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.
• It is moving along the path shown by green colour.
• It is moving under a constant acceleration →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 →v0.
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 →v
■ We want to know how →v0 became →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 →v0x has become →vx
How did that happen?
Ans: The x component of →a, which is →ax, acted on the object for a duration of 't' seconds.
• As a result, →v0x became →vx
• We can use the familiar equation: v = v0 + at
• Thus we can write: |→vx|=|→v0x|+|→ax|t
• So we get the magnitude of →vx
• It's direction is ofcourse, parallel to the x axis
• Thus we get the complete →vx
■ Consider the y components:
• The y component →v0y has become →vy
How did that happen?
Ans: The y component of →a, which is →ay, acted on the object for a duration of 't' seconds.
• As a result, →v0y became →vy
• We can use the same familiar equation again: v = v0 + at
• Thus we can write: |→vy|=|→v0y|+|→ay|t
• So we get the magnitude of →vy
• It's direction is ofcourse, parallel to the y axis
• Thus we get the complete →vy
5. So we independently derived →vx and →vy
• Now we find the resultant of →vx and →vy
• The resultant thus obtained is →v
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
• Position vector of P0 is →r0
• Position vector of P is →r
2. We can easily draw the displacement vector →Δr (Details here)
• It is shown in fig.b
3. The rectangular components of this →Δr 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 →Δrx
• That is., P0P' = |→Δrx|
(ii) The vertical distance travelled by the object during the 't' seconds is equal to the magnitude of →Δry
• That is., P'P = |→Δry|
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 →v0x
• So, for the travel from to P', the initial velocity is equal to →v0x
(ii) The object travelling with this initial velocity is acted upon by a constant acceleration of →ax
• Also this travel took 't' seconds
(iii) We can use the familiar equation: s=v0t+12at2
• Thus we can write: Distance P0P' = |→v0x|t+12|→ax|t2
(iv) So we get the magnitude of →Δrx
• It's direction is ofcourse, parallel to the x axis
• Thus we get the complete →Δrx
■ Now we will see P'P:
(i) Consider the vertical component of the velocity
• At P', the object is having a velocity of →v0y
• 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 →ay
• Also this travel took 't' seconds
(iii) We can use the familiar equation again: s=v0t+12at2
• Thus we can write: Distance P'P = |→v0y|t+12|→ay|t2
(iv) So we get the magnitude of →Δry
• It's direction is ofcourse, parallel to the y axis
• Thus we get the complete →Δry
5. Thus we get the two component vectors.
• Their resultant will give the required displacement vector →Δr
■ 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 5ˆims−1 and moves in x-y plane under action of a force which produces a constant acceleration of (3ˆi+2ˆj)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 |→v0x| = 5 ms-1
• Constant acceleration |→ax| = 3 ms-2
■ Motion in y direction:
• Initial y coordinate = 0
• Initial velocity |→v0y| = 0 ms-1
• Constant acceleration |→ay| = 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: s=v0t+12at2
• Thus we can write: 84 m = |→v0x|t+12|→ax|t2
• 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: s=v0t+12at2
• Thus we can write: |→Δry| = |→v0y|t+12|→ay|t2
• Substituting the values, we get: |→Δry| = 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: |→vx|=|→v0x|+|→ax|t
• Substituting the values, we get: |→vx| = 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: |→vy|=|→v0y|+|→ay|t
• Substituting the values, we get: |→vy| = 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:
|→v|=√|→vx|2+|→vy|2
• Substituting the values, we get: speed = √[232+122] = 26 ms-1
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 under a constant acceleration →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 →v0.
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 →v
■ We want to know how →v0 became →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 →v0x has become →vx
How did that happen?
Ans: The x component of →a, which is →ax, acted on the object for a duration of 't' seconds.
• As a result, →v0x became →vx
• We can use the familiar equation: v = v0 + at
• Thus we can write: |→vx|=|→v0x|+|→ax|t
• So we get the magnitude of →vx
• It's direction is ofcourse, parallel to the x axis
• Thus we get the complete →vx
■ Consider the y components:
• The y component →v0y has become →vy
How did that happen?
Ans: The y component of →a, which is →ay, acted on the object for a duration of 't' seconds.
• As a result, →v0y became →vy
• We can use the same familiar equation again: v = v0 + at
• Thus we can write: |→vy|=|→v0y|+|→ay|t
• So we get the magnitude of →vy
• It's direction is ofcourse, parallel to the y axis
• Thus we get the complete →vy
5. So we independently derived →vx and →vy
• Now we find the resultant of →vx and →vy
• The resultant thus obtained is →v
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 P is →r
2. We can easily draw the displacement vector →Δr (Details here)
• It is shown in fig.b
3. The rectangular components of this →Δr 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 →Δrx
• That is., P0P' = |→Δrx|
(ii) The vertical distance travelled by the object during the 't' seconds is equal to the magnitude of →Δry
• That is., P'P = |→Δry|
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 →v0x
• So, for the travel from to P', the initial velocity is equal to →v0x
(ii) The object travelling with this initial velocity is acted upon by a constant acceleration of →ax
• Also this travel took 't' seconds
(iii) We can use the familiar equation: s=v0t+12at2
• Thus we can write: Distance P0P' = |→v0x|t+12|→ax|t2
(iv) So we get the magnitude of →Δrx
• It's direction is ofcourse, parallel to the x axis
• Thus we get the complete →Δrx
■ Now we will see P'P:
(i) Consider the vertical component of the velocity
• At P', the object is having a velocity of →v0y
• 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 →ay
• Also this travel took 't' seconds
(iii) We can use the familiar equation again: s=v0t+12at2
• Thus we can write: Distance P'P = |→v0y|t+12|→ay|t2
(iv) So we get the magnitude of →Δry
• It's direction is ofcourse, parallel to the y axis
• Thus we get the complete →Δry
5. Thus we get the two component vectors.
• Their resultant will give the required displacement vector →Δr
■ 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 5ˆims−1 and moves in x-y plane under action of a force which produces a constant acceleration of (3ˆi+2ˆj)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 |→v0x| = 5 ms-1
• Constant acceleration |→ax| = 3 ms-2
■ Motion in y direction:
• Initial y coordinate = 0
• Initial velocity |→v0y| = 0 ms-1
• Constant acceleration |→ay| = 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: s=v0t+12at2
• Thus we can write: 84 m = |→v0x|t+12|→ax|t2
• 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: s=v0t+12at2
• Thus we can write: |→Δry| = |→v0y|t+12|→ay|t2
• Substituting the values, we get: |→Δry| = 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: |→vx|=|→v0x|+|→ax|t
• Substituting the values, we get: |→vx| = 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: |→vy|=|→v0y|+|→ay|t
• Substituting the values, we get: |→vy| = 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:
|→v|=√|→vx|2+|→vy|2
• Substituting the values, we get: speed = √[232+122] = 26 ms-1
Based on the above discussion, we can now learn about Projectile motion. We will see it in the next section.
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