This MCQ module is based on: Acceleration Kinematic Equations
TOPIC 6 OF 33
Acceleration Kinematic Equations
🎓 Class 11
Physics
CBSE
Theory
Ch 2 – Motion in a Straight Line
⏱ ~14 min
🌐 Language: [gtranslate]
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Acceleration Kinematic Equations
2.5 Acceleration
When the velocity of an object changes with time, the object is said to have acceleration.
Average Acceleration
\[\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}\]
Instantaneous Acceleration
\[a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2 x}{dt^2}\]
Uniform acceleration means the velocity changes by equal amounts in equal time intervals — that is, \(a\) is constant.
Important: Acceleration and velocity can have different signs. When they have opposite signs, the object is decelerating (slowing down). For example, a car moving in the positive direction with negative acceleration is braking.
v-t and a-t Graphs
| Graph Type | Slope Gives | Area Under Curve Gives |
|---|---|---|
| x-t graph | Velocity | — |
| v-t graph | Acceleration | Displacement |
| a-t graph | — | Change in velocity |
2.6 Kinematic Equations for Uniformly Accelerated Motion
For an object moving with constant acceleration \(a\), starting with initial velocity \(u\) at position \(x_0\), the following three equations govern the motion:
1First Equation (Velocity-Time)
\[v = u + at\]
Derivation: From the definition of acceleration: \(a = \frac{v - u}{t}\). Rearranging: \(v = u + at\).
2Second Equation (Position-Time)
\[s = ut + \tfrac{1}{2}at^2\]
Derivation from v-t graph: Displacement = area under v-t line = area of rectangle + area of triangle = \(u \cdot t + \frac{1}{2}(v-u) \cdot t\). Since \(v - u = at\): \(s = ut + \frac{1}{2}at^2\).
3Third Equation (Velocity-Displacement)
\[v^2 = u^2 + 2as\]
Derivation: From equations 1 and 2. From (1): \(t = \frac{v-u}{a}\). Substituting into (2): \(s = u\left(\frac{v-u}{a}\right) + \frac{1}{2}a\left(\frac{v-u}{a}\right)^2\). Simplifying: \(2as = 2u(v-u) + (v-u)^2 = v^2 - u^2\). Hence \(v^2 = u^2 + 2as\).
Position at time t: If the object starts at position \(x_0\), then: \(x = x_0 + ut + \frac{1}{2}at^2\)
Displacement in the n-th second: \(s_n = u + \frac{a}{2}(2n - 1)\). This gives the displacement specifically during the n-th second of motion.
Worked Examples
Worked Example 1 (NCERT Example 2.2): Car Acceleration
A car increases its velocity from 18 km/h to 36 km/h in 5 seconds. Find (a) the acceleration, and (b) the distance covered during this time.
Solution:
Convert to SI:
\(u = 18 \text{ km/h} = 18 \times \frac{5}{18} = 5 \text{ m/s}\)
\(v = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10 \text{ m/s}\)
\(t = 5 \text{ s}\)
(a) Acceleration: \[a = \frac{v - u}{t} = \frac{10 - 5}{5} = \frac{5}{5} = \boxed{1 \text{ m/s}^2}\]
(b) Distance: \[s = ut + \frac{1}{2}at^2 = 5(5) + \frac{1}{2}(1)(25) = 25 + 12.5 = \boxed{37.5 \text{ m}}\]
Verification using 3rd equation: \(v^2 = u^2 + 2as \Rightarrow 100 = 25 + 2(1)(s) \Rightarrow s = 37.5\) m. Checks out!
Convert to SI:
\(u = 18 \text{ km/h} = 18 \times \frac{5}{18} = 5 \text{ m/s}\)
\(v = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10 \text{ m/s}\)
\(t = 5 \text{ s}\)
(a) Acceleration: \[a = \frac{v - u}{t} = \frac{10 - 5}{5} = \frac{5}{5} = \boxed{1 \text{ m/s}^2}\]
(b) Distance: \[s = ut + \frac{1}{2}at^2 = 5(5) + \frac{1}{2}(1)(25) = 25 + 12.5 = \boxed{37.5 \text{ m}}\]
Verification using 3rd equation: \(v^2 = u^2 + 2as \Rightarrow 100 = 25 + 2(1)(s) \Rightarrow s = 37.5\) m. Checks out!
Worked Example 2 (NCERT Example 2.3): Ball Thrown Upward
A ball is thrown vertically upward with an initial velocity of 15 m/s. Find (a) the maximum height reached, and (b) the time taken to reach the maximum height. (Take \(g = 10\) m/s²)
Solution:
Taking upward as positive direction:
\(u = +15\) m/s, \(a = -g = -10\) m/s², \(v = 0\) (at maximum height)
(a) Maximum height: Using \(v^2 = u^2 + 2as\): \[0 = (15)^2 + 2(-10)(s)\] \[0 = 225 - 20s\] \[s = \frac{225}{20} = \boxed{11.25 \text{ m}}\]
(b) Time to reach maximum height: Using \(v = u + at\): \[0 = 15 + (-10)t\] \[t = \frac{15}{10} = \boxed{1.5 \text{ s}}\]
Taking upward as positive direction:
\(u = +15\) m/s, \(a = -g = -10\) m/s², \(v = 0\) (at maximum height)
(a) Maximum height: Using \(v^2 = u^2 + 2as\): \[0 = (15)^2 + 2(-10)(s)\] \[0 = 225 - 20s\] \[s = \frac{225}{20} = \boxed{11.25 \text{ m}}\]
(b) Time to reach maximum height: Using \(v = u + at\): \[0 = 15 + (-10)t\] \[t = \frac{15}{10} = \boxed{1.5 \text{ s}}\]
Worked Example 3: Train Starting from Rest
A train starts from rest and accelerates uniformly at 0.5 m/s² for 2 minutes. Find (a) the final speed, and (b) the distance covered.
Solution:
\(u = 0\) (starts from rest), \(a = 0.5\) m/s², \(t = 2\) min = 120 s
(a) Final speed: \[v = u + at = 0 + 0.5 \times 120 = \boxed{60 \text{ m/s}}\] (That is \(60 \times \frac{18}{5} = 216\) km/h)
(b) Distance covered: \[s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.5)(120)^2 = 0.25 \times 14400 = \boxed{3600 \text{ m} = 3.6 \text{ km}}\]
\(u = 0\) (starts from rest), \(a = 0.5\) m/s², \(t = 2\) min = 120 s
(a) Final speed: \[v = u + at = 0 + 0.5 \times 120 = \boxed{60 \text{ m/s}}\] (That is \(60 \times \frac{18}{5} = 216\) km/h)
(b) Distance covered: \[s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.5)(120)^2 = 0.25 \times 14400 = \boxed{3600 \text{ m} = 3.6 \text{ km}}\]
Worked Example 4: Braking Distance
A car travelling at 20 m/s applies brakes and decelerates at 4 m/s². Find (a) when the car stops, (b) the distance covered before stopping, and (c) the displacement in the 6th second.
Solution:
\(u = 20\) m/s, \(a = -4\) m/s² (deceleration), \(v = 0\) (when stopped)
(a) Time to stop: \[v = u + at \Rightarrow 0 = 20 + (-4)t \Rightarrow t = \frac{20}{4} = \boxed{5 \text{ s}}\]
(b) Braking distance: \[v^2 = u^2 + 2as \Rightarrow 0 = 400 + 2(-4)s \Rightarrow s = \frac{400}{8} = \boxed{50 \text{ m}}\]
(c) Displacement in the 6th second:
The car stops at \(t = 5\) s. After stopping, the car remains stationary (brakes applied). Therefore displacement in the 6th second = \(\boxed{0}\).
Note: The kinematic equation \(s_n = u + \frac{a}{2}(2n-1)\) gives \(s_6 = 20 + \frac{-4}{2}(11) = 20 - 22 = -2\) m. This negative value is unphysical — it would mean the car reverses, which doesn't happen with brakes. The equations are valid only while the car is in motion, i.e., for \(t \le 5\) s.
\(u = 20\) m/s, \(a = -4\) m/s² (deceleration), \(v = 0\) (when stopped)
(a) Time to stop: \[v = u + at \Rightarrow 0 = 20 + (-4)t \Rightarrow t = \frac{20}{4} = \boxed{5 \text{ s}}\]
(b) Braking distance: \[v^2 = u^2 + 2as \Rightarrow 0 = 400 + 2(-4)s \Rightarrow s = \frac{400}{8} = \boxed{50 \text{ m}}\]
(c) Displacement in the 6th second:
The car stops at \(t = 5\) s. After stopping, the car remains stationary (brakes applied). Therefore displacement in the 6th second = \(\boxed{0}\).
Note: The kinematic equation \(s_n = u + \frac{a}{2}(2n-1)\) gives \(s_6 = 20 + \frac{-4}{2}(11) = 20 - 22 = -2\) m. This negative value is unphysical — it would mean the car reverses, which doesn't happen with brakes. The equations are valid only while the car is in motion, i.e., for \(t \le 5\) s.
Worked Example 5: Displacement in the n-th Second
An object starts from rest and accelerates at 2 m/s². Find the displacement in the 1st, 2nd, and 3rd seconds. What pattern do you notice?
Solution:
Using \(s_n = u + \frac{a}{2}(2n-1)\) with \(u = 0\), \(a = 2\) m/s²:
\(s_1 = 0 + \frac{2}{2}(2 \times 1 - 1) = 1 \times 1 = \boxed{1 \text{ m}}\)
\(s_2 = 0 + \frac{2}{2}(2 \times 2 - 1) = 1 \times 3 = \boxed{3 \text{ m}}\)
\(s_3 = 0 + \frac{2}{2}(2 \times 3 - 1) = 1 \times 5 = \boxed{5 \text{ m}}\)
Pattern: The displacements in successive seconds are in the ratio \(1 : 3 : 5 : 7 : \ldots\) (odd numbers). This is a well-known property of uniformly accelerated motion starting from rest.
Total distances: after 1s = 1m, after 2s = 4m, after 3s = 9m — following \(s = \frac{1}{2}at^2 = t^2\). The ratios of cumulative distances are \(1 : 4 : 9 : 16\) (perfect squares).
Using \(s_n = u + \frac{a}{2}(2n-1)\) with \(u = 0\), \(a = 2\) m/s²:
\(s_1 = 0 + \frac{2}{2}(2 \times 1 - 1) = 1 \times 1 = \boxed{1 \text{ m}}\)
\(s_2 = 0 + \frac{2}{2}(2 \times 2 - 1) = 1 \times 3 = \boxed{3 \text{ m}}\)
\(s_3 = 0 + \frac{2}{2}(2 \times 3 - 1) = 1 \times 5 = \boxed{5 \text{ m}}\)
Pattern: The displacements in successive seconds are in the ratio \(1 : 3 : 5 : 7 : \ldots\) (odd numbers). This is a well-known property of uniformly accelerated motion starting from rest.
Total distances: after 1s = 1m, after 2s = 4m, after 3s = 9m — following \(s = \frac{1}{2}at^2 = t^2\). The ratios of cumulative distances are \(1 : 4 : 9 : 16\) (perfect squares).
Worked Example 6: Two-Stage Motion
A motorcycle accelerates from rest at 2 m/s² for 10 s, then continues at constant velocity for 20 s. Find the total distance covered.
Solution:
Stage 1: \(u = 0\), \(a = 2\) m/s², \(t_1 = 10\) s
\[s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(100) = 100 \text{ m}\] Velocity at end of Stage 1: \(v = 0 + 2 \times 10 = 20\) m/s
Stage 2: \(v = 20\) m/s (constant), \(t_2 = 20\) s
\[s_2 = v \times t_2 = 20 \times 20 = 400 \text{ m}\]
Total distance: \(s = s_1 + s_2 = 100 + 400 = \boxed{500 \text{ m}}\)
Stage 1: \(u = 0\), \(a = 2\) m/s², \(t_1 = 10\) s
\[s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(100) = 100 \text{ m}\] Velocity at end of Stage 1: \(v = 0 + 2 \times 10 = 20\) m/s
Stage 2: \(v = 20\) m/s (constant), \(t_2 = 20\) s
\[s_2 = v \times t_2 = 20 \times 20 = 400 \text{ m}\]
Total distance: \(s = s_1 + s_2 = 100 + 400 = \boxed{500 \text{ m}}\)
Activity: Understanding v-t Graphs
L4 Analyse
Predict: If you plot velocity vs time for a ball thrown straight up and caught on its way down, what shape will the graph be?
- Throw a ball straight up and catch it. Think about how the velocity changes at each phase: going up (positive, decreasing), at the top (zero), coming down (negative, increasing magnitude).
- Sketch a v-t graph on paper with time on the x-axis and velocity on the y-axis.
- From your graph, identify: (a) the point of zero velocity, (b) the slope (what does it represent?), and (c) what the area under the curve means.
Observation: The v-t graph is a straight line with negative slope passing through zero at the highest point. Going up: positive velocity decreasing linearly. Coming down: negative velocity increasing in magnitude.
Key insights:
- Slope = acceleration = \(-g \approx -9.8\) m/s² (constant throughout)
- Area above the t-axis = upward displacement
- Area below the t-axis = downward displacement
- Total area (with signs) = net displacement (zero if caught at the same height)
Key insights:
- Slope = acceleration = \(-g \approx -9.8\) m/s² (constant throughout)
- Area above the t-axis = upward displacement
- Area below the t-axis = downward displacement
- Total area (with signs) = net displacement (zero if caught at the same height)
Interactive: Kinematic Equation Calculator L3 Apply
Enter any 3 of the 5 variables (u, v, a, t, s) and calculate the remaining two. Leave exactly 2 fields empty.
Competency-Based Questions
A metro train starts from station A and needs to reach station B, 1.2 km away. The train accelerates uniformly at 1 m/s² from rest, then travels at maximum speed, and finally decelerates uniformly at 2 m/s² to stop at station B. The maximum speed reached is 20 m/s.
Q1. L1 Remember Write the three kinematic equations for uniformly accelerated motion.
Answer:
(i) \(v = u + at\)
(ii) \(s = ut + \frac{1}{2}at^2\)
(iii) \(v^2 = u^2 + 2as\)
(i) \(v = u + at\)
(ii) \(s = ut + \frac{1}{2}at^2\)
(iii) \(v^2 = u^2 + 2as\)
Q2. L3 Apply Calculate the distance covered during the acceleration phase and the time taken. (3 marks)
Answer:
Acceleration phase: \(u = 0\), \(v = 20\) m/s, \(a = 1\) m/s²
\(v^2 = u^2 + 2as \Rightarrow 400 = 0 + 2(1)s_1 \Rightarrow s_1 = 200\) m
\(v = u + at \Rightarrow 20 = 0 + 1 \times t_1 \Rightarrow t_1 = 20\) s
Distance = 200 m, Time = 20 s
Acceleration phase: \(u = 0\), \(v = 20\) m/s, \(a = 1\) m/s²
\(v^2 = u^2 + 2as \Rightarrow 400 = 0 + 2(1)s_1 \Rightarrow s_1 = 200\) m
\(v = u + at \Rightarrow 20 = 0 + 1 \times t_1 \Rightarrow t_1 = 20\) s
Distance = 200 m, Time = 20 s
Q3. L3 Apply Find the distance and time for the deceleration phase. (3 marks)
Answer:
Deceleration phase: \(u = 20\) m/s, \(v = 0\), \(a = -2\) m/s²
\(v^2 = u^2 + 2as \Rightarrow 0 = 400 + 2(-2)s_3 \Rightarrow s_3 = 100\) m
\(v = u + at \Rightarrow 0 = 20 + (-2)t_3 \Rightarrow t_3 = 10\) s
Distance = 100 m, Time = 10 s
Deceleration phase: \(u = 20\) m/s, \(v = 0\), \(a = -2\) m/s²
\(v^2 = u^2 + 2as \Rightarrow 0 = 400 + 2(-2)s_3 \Rightarrow s_3 = 100\) m
\(v = u + at \Rightarrow 0 = 20 + (-2)t_3 \Rightarrow t_3 = 10\) s
Distance = 100 m, Time = 10 s
Q4. L4 Analyse Determine the total journey time from A to B. (3 marks)
Answer:
Total distance = 1200 m. Constant speed phase distance: \(s_2 = 1200 - 200 - 100 = 900\) m
Time for constant speed phase: \(t_2 = \frac{900}{20} = 45\) s
Total time = \(t_1 + t_2 + t_3 = 20 + 45 + 10 = \boxed{75 \text{ s}}\)
Total distance = 1200 m. Constant speed phase distance: \(s_2 = 1200 - 200 - 100 = 900\) m
Time for constant speed phase: \(t_2 = \frac{900}{20} = 45\) s
Total time = \(t_1 + t_2 + t_3 = 20 + 45 + 10 = \boxed{75 \text{ s}}\)
Q5. L5 Evaluate A student uses \(s = ut + \frac{1}{2}at^2\) with \(u = 20\) m/s, \(a = -4\) m/s², and \(t = 8\) s to calculate the stopping distance of a car. They get \(s = 20(8) + \frac{1}{2}(-4)(64) = 160 - 128 = 32\) m. Is this correct? If not, explain the error. (3 marks)
Answer:
The answer is incorrect. The car stops when \(v = 0\):
\(v = u + at \Rightarrow 0 = 20 - 4t \Rightarrow t = 5\) s
The car stops at \(t = 5\) s, not 8 s. After stopping, the car does not reverse (brakes hold it). Using \(t = 8\) s in the equation assumes the car goes backward after stopping, which is unphysical.
Correct stopping distance: \(s = 20(5) + \frac{1}{2}(-4)(25) = 100 - 50 = \boxed{50 \text{ m}}\).
The kinematic equations are valid only while the given acceleration applies — that is, until the car stops.
The answer is incorrect. The car stops when \(v = 0\):
\(v = u + at \Rightarrow 0 = 20 - 4t \Rightarrow t = 5\) s
The car stops at \(t = 5\) s, not 8 s. After stopping, the car does not reverse (brakes hold it). Using \(t = 8\) s in the equation assumes the car goes backward after stopping, which is unphysical.
Correct stopping distance: \(s = 20(5) + \frac{1}{2}(-4)(25) = 100 - 50 = \boxed{50 \text{ m}}\).
The kinematic equations are valid only while the given acceleration applies — that is, until the car stops.
Assertion-Reason Questions
Assertion (A): For uniform acceleration, the v-t graph is a straight line.
Reason (R): Acceleration is the slope of the v-t graph, and a constant slope produces a straight line.
Answer: A. If acceleration (slope of v-t graph) is constant, then v varies linearly with t, producing a straight line. R correctly explains A.
Assertion (A): An object can have negative acceleration and still be speeding up.
Reason (R): Negative acceleration means the object is decelerating.
Answer: C. A is true: if an object moves in the negative direction with negative acceleration, it speeds up (both velocity and acceleration point in the same direction). R is false: negative acceleration does NOT always mean deceleration. The object decelerates only when acceleration and velocity have opposite signs.
Assertion (A): The area under a v-t graph between two time instants gives the displacement of the object in that interval.
Reason (R): Displacement equals the integral of velocity with respect to time: \(s = \int v \, dt\).
Answer: A. Both statements are true. The area under the v-t curve is the graphical representation of the integral \(\int v\,dt\), which by definition gives the displacement. R correctly explains A.
Frequently Asked Questions - Acceleration Kinematic Equations
What is the main concept covered in Acceleration Kinematic Equations?
In NCERT Class 11 Physics Chapter 2 (Motion in a Straight Line), "Acceleration Kinematic Equations" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Acceleration Kinematic Equations useful in real-life applications?
Real-life applications of Acceleration Kinematic Equations from NCERT Class 11 Physics Chapter 2 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Acceleration Kinematic Equations?
Key formulas in Acceleration Kinematic Equations (NCERT Class 11 Physics Chapter 2 Motion in a Straight Line) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 2?
NCERT Class 11 Physics Chapter 2 (Motion in a Straight Line) is structured so each part builds on the previous one. Acceleration Kinematic Equations connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Acceleration Kinematic Equations?
CBSE board questions from Acceleration Kinematic Equations typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Acceleration Kinematic Equations lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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