ANAND CLASSES study material and notes here discuss a detailed and structured explanation of the Second Equation of Motion with an extensive breakdown for JEE, NEET, and CBSE Class 11 students.
📌 $s = ut + \frac{1}{2} at^2$ : Second Equation of Uniformly Accelerated Motion
The second equation of motion helps us determine the distance or displacement of a body under uniform acceleration. $$s = ut + \frac{1}{2} at^2$$
where:
$s$ = Distance traveled by the object in time $t$,
$u$ = Initial velocity of the object,
$a$ = Acceleration of the object,
$t$ = Time taken.
📍 Why is this equation important?
It allows us to calculate displacement when initial velocity, acceleration, and time are known.
It is used extensively in problems involving free fall, motion on inclined planes, and projectile motion.
This equation is derived from the first equation of motion and helps in predicting the future position of an object under constant acceleration.
🔷 Step-by-Step Derivation of the Second Equation of Motion
We derive this equation using the concept of average velocity.
📌 Step 1: Formula for Average Velocity
For a body under uniform acceleration, its average velocity ($v_{\text{avg}}$) is given by:
The total distance traveled (s) is given by: $$s = v_{\text{avg}} \times t$$
$$s = \frac{(u + v)}{2} \times t$$
📌 Step 3: Substituting $\:v = u + at$
From the first equation of motion: $$v = u + at$$
Substituting this value of $v$ in the equation for $s$:
$$s = \frac{(u + (u + at))}{2} \times t$$
$$s = \frac{(2u + at)}{2} \times t$$
$$s = \frac{2ut + at^2}{2}$$
$$s = ut + \frac{1}{2} at^2$$
Thus, we have derived the second equation of motion! 🎯
🔷 Applications of the Second Equation of Motion
Calculating displacement in uniformly accelerated motion.
Projectile motion to determine vertical height.
Free-fall motion under gravity where $a = g = 9.8 \text{ m/s}^2$.
Stopping distance of vehicles when brakes are applied.
🔷 Numerical Problems for JEE, NEET & CBSE Exams
📝 Example 1: A car starts from rest and accelerates at 3 m/s² for 6 seconds. Find the distance covered.
Solution:
Given:
$u = 0$ (starting from rest)
$a = 3 \text{ m/s}^2$
$t = 6 sec$
Using the second equation of motion:
$$s = ut + \frac{1}{2} at^2 $$
$$s = (0)(6) + \frac{1}{2} (3)(6^2)$$
$$s = 0 + \frac{1}{2} (3 \times 36)$$
$$s = \frac{108}{2} = 54 \text{ m}$$
✅ Answer: The car covers 54 meters.
🔷 Multiple Choice Questions (MCQs)
🔹 Q1: A body is moving with an initial velocity of 5 m/s and a uniform acceleration of 2 m/s². What will be the displacement after 4 seconds?
a) 28 m b) 36 m c) 40 m d) 44 m
Solution:
Using the second equation of motion: $$s = ut + \frac{1}{2} at^2$$
$$s = (5)(4) + \frac{1}{2} (2)(4^2)$$
$$s = 20 + \frac{1}{2} (2 \times 16)$$
$$s = 20 + 16 = 36 \text{ m}$$
✅ Correct Answer: (b) 36 m
🔷 Conceptual Questions with Answers
Q1. Why does the second equation of motion contain both $ut$ and $\frac{1}{2}at^2$?
The term $ut$ represents the displacement due to the initial velocity.
The term $\frac{1}{2}at^2$ represents the additional displacement due to acceleration.
🔷 Do You Know?
📌 The second equation of motion is derived from Galileo’s studies of motion. 📌 If initial velocity is zero ($u = 0$), the equation simplifies to: $s = \frac{1}{2} at^2$
📌 It is widely used in free-fall motion, where acceleration due to gravity ($g$) is 9.8 m/s².
🔷 Worksheet
Solve the following questions:
A train starts from rest and accelerates at 2 m/s² for 10 seconds. Find the distance covered.
A stone is dropped from a height of 40 m. How long will it take to reach the ground?
A vehicle moving with 20 m/s stops in 5 seconds under uniform deceleration. Find the stopping distance.
🔷 Test Paper
📌 Sample Test Questions
Derive the second equation of motion using graphical method. (5 marks)
A car accelerates from 4 m/s to 20 m/s in 8 seconds. Find the distance covered. (5 marks)
A ball is thrown vertically upward with 30 m/s speed. Find how high it will go before stopping. (Take $g = 10 \text{ m/s}^2$) (5 marks)
🔷 Quick Revision Points
Equation: $s = ut + \frac{1}{2} at^2$
Used to calculate displacement when time, initial velocity, and acceleration are known.
Derived from first equation of motion: $v = u + at$.
Applies only under constant acceleration.
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