Third Equation of a Uniformly Accelerated Motion

1. Introduction

The third equation of motion relates the square of the final velocity ([v]) of a uniformly accelerated object to its initial velocity ([u]), acceleration ([a]), and displacement ([s]).

It is given by:

[v^2 = u^2 + 2as]

This equation is extremely useful because it links velocity and displacement directly, eliminating the need for time ([t]).

2. Derivation of the Third Equation of Motion

(a) Using the First Two Equations of Motion

We know:

1. First equation: ([v = u + at])
2. Second equation: ([s = ut + \dfrac{1}{2}at^2])

From (1),

[
t = \dfrac{v – u}{a}
]

Substitute (t) into (2):

[s = u\left(\dfrac{v – u}{a}\right) + \dfrac{1}{2}a\left(\dfrac{v – u}{a}\right)^2]

Simplify step by step:

[s = \dfrac{u(v – u)}{a} + \dfrac{1}{2a}(v – u)^2]

[s = \dfrac{uv – u^2}{a} + \dfrac{(v^2 – 2uv + u^2)}{2a}]

[s = \dfrac{2uv – 2u^2 + v^2 – 2uv + u^2}{2a}]

[s = \dfrac{v^2 – u^2}{2a}]

Finally, multiplying both sides by (2a):

[v^2 = u^2 + 2as]

(b) Using the Graphical Method

In a velocity–time graph for uniformly accelerated motion:

• The area under the graph gives displacement (s).
• Using geometry of the trapezium, we arrive at the same result:
  [
  v^2 = u^2 + 2as
  ]

3. Conditions of Applicability

• The motion must be along a straight line.
• The acceleration must remain constant.
• The quantities (v), (u), (s), and (a) must be along the same direction.

4. SI Units and Dimensional Formula

5. Practice Questions (With Solutions)

1. A car starts from rest and accelerates at [3\ \text{m/s}^2]. Find the velocity after it has travelled [54\ \text{m}].

Solution:
Given: [u = 0,\ a = 3\ \text{m/s}^2,\ s = 54\ \text{m}]
Using ([v^2 = u^2 + 2as]):
[
v^2 = 0 + 2 \times 3 \times 54 = 324
]
[
v = 18\ \text{m/s}
]

2. A train moving at [20\ \text{m/s}] is stopped by applying brakes after [200\ \text{m})]. Find the acceleration.

Solution:
Given: [v = 0,\ u = 20\ \text{m/s},\ s = 200\ \text{m}]
Using ([v^2 = u^2 + 2as]):
[
0 = 400 + 2a \times 200
]
[
a = \dfrac{-400}{400} = -1\ \text{m/s}^2
]

3. A ball is thrown upward with a velocity of [25\ \text{m/s}]. Find the maximum height reached. (Take [g = 9.8\ \text{m/s}^2])

Solution:
At maximum height, [v = 0,\ u = 25\ \text{m/s},\ a = -9.8\ \text{m/s}^2)]
Using ([v^2 = u^2 + 2as]):
[
0 = 625 + 2(-9.8)s
]
[
s = \dfrac{625}{19.6} \approx 31.9\ \text{m}
]

6. Conceptual Questions

7. FAQs / Common Misconceptions