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:
$$v_{\text{avg}} = \frac{\text{Initial velocity} + \text{Final velocity}}{2} $$
$$v_{\text{avg}} = \frac{u + v}{2}$$
📌 Step 2: Distance Traveled Formula
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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