Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Motion in a Straight Line (1D Kinematics) for CBSE Class 11 Physics, including all special cases and calculus applications.
Key Concepts
1D Kinematics
Every formula and special case you need
1
Distance vs. Displacement
Definition
Distance (Path Length) is the actual total path covered (scalar, always positive). Displacement is the shortest straight-line distance from initial to final position (vector, can be positive, negative, or zero).
$$|\text{Displacement}| \le \text{Distance}$$
2
Average Speed & Average Velocity
Formula
Average speed is total distance over total time. Average velocity is total displacement over total time. They are only equal if the object moves in a straight line without reversing direction.
$$v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_2 – x_1}{t_2 – t_1}$$
3
Special Cases of Average Speed
Case Formulas
Case 1: Object covers equal distances with speeds $v_1$ and $v_2$. (Harmonic Mean)
Case 2: Object travels for equal time intervals with speeds $v_1$ and $v_2$. (Arithmetic Mean)
Case 2: Object travels for equal time intervals with speeds $v_1$ and $v_2$. (Arithmetic Mean)
$$\text{Equal Distance: } v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2} \quad | \quad \text{Equal Time: } v_{\text{avg}} = \frac{v_1 + v_2}{2}$$
4
Instantaneous Velocity & Acceleration
Calculus
Velocity is the rate of change of position. Acceleration is the rate of change of velocity. For motion defined by a function $x(t)$, use differentiation.
$$v = \frac{dx}{dt} \quad | \quad a = \frac{dv}{dt} = \frac{d^2x}{dt^2} = v\frac{dv}{dx}$$
5
Kinematic Equations (Constant Acceleration)
Formula
The “Big Three” equations relate initial velocity ($u$), final velocity ($v$), acceleration ($a$), displacement ($s$), and time ($t$). Valid ONLY when acceleration is constant!
$$v = u + at$$
$$s = ut + \frac{1}{2}at^2$$
$$v^2 = u^2 + 2as$$
6
Displacement in the $n^{\text{th}}$ Second
Formula
The distance covered strictly during a specific 1-second interval (e.g., between $t = 3\text{s}$ and $t = 4\text{s}$ for the 4th second).
$$S_n = u + \frac{a}{2}(2n – 1)$$
7
Motion Under Gravity (Free Fall)
Case Formulas
Taking upward as positive ($+y$) and downward as negative, $a = -g$. For an object thrown upwards with velocity $u$ from the ground:
$$\text{Time to Max Height: } t = \frac{u}{g}$$
$$\text{Total Time of Flight: } T = \frac{2u}{g}$$
$$\text{Maximum Height: } H_{\text{max}} = \frac{u^2}{2g}$$
8
Velocity of an Object Dropped from Height $h$
Formula
If an object is dropped ($u=0$), the velocity with which it hits the ground depends purely on the height it was dropped from (ignoring air resistance).
$$v = \sqrt{2gh} \quad | \quad t = \sqrt{\frac{2h}{g}}$$
9
Stopping Distance of a Vehicle
Formula
The distance a vehicle travels before coming to a complete halt ($v=0$) after the brakes are applied. If reaction time ($t_r$) is given, total stopping distance includes $u \cdot t_r$.
$$d_s = \frac{u^2}{2|a|} \quad | \quad \text{Total } d = (u \cdot t_r) + \frac{u^2}{2|a|}$$
10
Relative Velocity in 1D
Formula
The velocity of object A as observed by someone on object B. If moving in the same direction, subtract. If moving in opposite directions, their relative speeds add up.
$$v_{AB} = v_A – v_B \quad | \quad v_{BA} = v_B – v_A$$
Concept Deep Dive
01
Reading Kinematic Graphs
The visual shortcuts of motionGraphs hide calculus in plain sight.
Position-Time ($x-t$) Graph: The slope gives the velocity ($v = dx/dt$). A straight line means constant velocity. A curved line (parabola) means the object is accelerating.
Velocity-Time ($v-t$) Graph: The slope gives the acceleration ($a = dv/dt$). The area under the curve gives the total displacement ($\Delta x = \int v \, dt$). If the area dips below the x-axis, that displacement is negative.
Acceleration-Time ($a-t$) Graph: The area under the curve gives the change in velocity ($\Delta v = \int a \, dt$).
Position-Time ($x-t$) Graph: The slope gives the velocity ($v = dx/dt$). A straight line means constant velocity. A curved line (parabola) means the object is accelerating.
Velocity-Time ($v-t$) Graph: The slope gives the acceleration ($a = dv/dt$). The area under the curve gives the total displacement ($\Delta x = \int v \, dt$). If the area dips below the x-axis, that displacement is negative.
Acceleration-Time ($a-t$) Graph: The area under the curve gives the change in velocity ($\Delta v = \int a \, dt$).
02
Integration vs Differentiation in Kinematics
Which way do you step on the ladder?Think of Position ($x$), Velocity ($v$), and Acceleration ($a$) as rungs on a ladder.
To step DOWN the ladder ($x \rightarrow v \rightarrow a$), you Differentiate.
To step UP the ladder ($a \rightarrow v \rightarrow x$), you Integrate.
Important Rule: When stepping up (integrating), you always generate a constant of integration ($+C$). This constant represents the initial condition (initial velocity $u$ or initial position $x_0$). Do not forget to use boundary conditions at $t=0$ to find it!
To step DOWN the ladder ($x \rightarrow v \rightarrow a$), you Differentiate.
To step UP the ladder ($a \rightarrow v \rightarrow x$), you Integrate.
Important Rule: When stepping up (integrating), you always generate a constant of integration ($+C$). This constant represents the initial condition (initial velocity $u$ or initial position $x_0$). Do not forget to use boundary conditions at $t=0$ to find it!
$$v = \int a \, dt \quad \text{and} \quad x = \int v \, dt$$
Compare & Contrast
✗ Retardation (Deceleration)
- Means the speed of the object is decreasing.
- Happens ONLY when Velocity and Acceleration have opposite signs.
- Example: Moving right ($+v$) but accelerating left ($-a$), like hitting brakes.
- Example: Moving left ($-v$) but accelerating right ($+a$).
✓ Negative Acceleration
- Simply means the acceleration vector points in the negative axis direction.
- It can cause speeding up OR slowing down!
- If moving left ($-v$) and accelerating left ($-a$), the object is speeding up (NOT decelerating).
- Sign convention dictates direction, not behavior.
Remember
At the highest point of an upward throw, the velocity is exactly $v = 0$. However, the acceleration is NOT zero. It is still $g$ ($9.8 \text{ m/s}^2$ downwards). If acceleration were zero at the top, the ball would freeze mid-air!
Common Mistakes to Avoid
Mistake 1
Using Equations of Motion for Variable Acceleration: Formulas like $v = u + at$ and $s = ut + \frac{1}{2}at^2$ are strictly derived assuming $a$ is CONSTANT. If a problem states $a = 2t$ or $a = 3x^2$, you cannot use these algebraic equations. You MUST use calculus integration to solve the problem.
Mistake 2
Mixing up $n^{\text{th}}$ second distance vs. $n$ seconds distance: $s_n = ut + \frac{1}{2}at^2$ gives the total distance covered from $t=0$ to $t=n$. But $S_{n^{\text{th}}} = u + \frac{a}{2}(2n-1)$ gives the distance covered only in that specific 1-second window (e.g., from $t=4$ to $t=5$).
Mistake 3
Inconsistent Free Fall Signs: If you drop a stone from a building, and take the ground as $y=0$ (meaning initial position is $+H$), then displacement $s = -H$, and $a = -g$. Do not randomly switch $g$ to positive halfway through the calculation. Pick a convention (up is positive) and stick to it strictly!
Exam Tips
Tip 1
Galileo’s Law of Odd Numbers: If an object falls from rest ($u=0$) under uniform acceleration ($g$), the distances covered in successive equal time intervals (e.g., 1st sec, 2nd sec, 3rd sec) are in the ratio of odd numbers: 1 : 3 : 5 : 7 : 9… This is a massive time-saver for MCQs.
Tip 2
When a train of length $L_1$ crosses a bridge of length $L_2$, the total distance the train must travel to completely clear the bridge is $D = L_1 + L_2$. If crossing another moving train, apply relative velocity to the speed, but still add their lengths for the distance!
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · 1D Kinematics · CBSE Exam1
A ball is thrown vertically upwards with a velocity of $20 \text{ m/s}$. From the same point, another ball is thrown upwards with the same velocity $2$ seconds later. When and where will they meet? ($g = 10 \text{ m/s}^2$) [3 marks]
Answer
$t = 3 \text{ s}$ for 1st ball; Height $= 15 \text{ m}$
📝
Explanation
Let the 1st ball travel for time $t$. Its displacement $y_1 = ut – \frac{1}{2}gt^2 = 20t – 5t^2$.
The 2nd ball is thrown 2s later, so it travels for time $(t – 2)$. Its displacement $y_2 = 20(t – 2) – 5(t – 2)^2$.
When they meet, their displacements from ground are equal: $y_1 = y_2$.
$20t – 5t^2 = 20(t – 2) – 5(t^2 – 4t + 4)$
$20t – 5t^2 = 20t – 40 – 5t^2 + 20t – 20$
$0 = 20t – 60 \implies 20t = 60 \implies t = 3 \text{ s}$.
Substitute $t=3$ into $y_1$: $y = 20(3) – 5(3^2) = 60 – 45 = 15 \text{ m}$.
2
The position of an object moving along the x-axis is given by $x = 8.5 + 2.5t^2$, where x is in meters and t is in seconds. Calculate its velocity at $t = 2.0 \text{ s}$ and the average velocity between $t = 2.0 \text{ s}$ and $t = 4.0 \text{ s}$. [3 marks]
Answer
Inst. Vel $= 10 \text{ m/s}$; Avg Vel $= 15 \text{ m/s}$
📝
Explanation
Instantaneous Velocity: $v = \frac{dx}{dt} = \frac{d}{dt}(8.5 + 2.5t^2) = 0 + 5.0t$.
At $t = 2.0 \text{ s}$, $v = 5.0(2.0) = 10 \text{ m/s}$.
Average Velocity: $v_{\text{avg}} = \frac{x(4) – x(2)}{4 – 2}$.
$x(4) = 8.5 + 2.5(4^2) = 8.5 + 40 = 48.5 \text{ m}$.
$x(2) = 8.5 + 2.5(2^2) = 8.5 + 10 = 18.5 \text{ m}$.
$v_{\text{avg}} = \frac{48.5 – 18.5}{2} = \frac{30.0}{2} = 15 \text{ m/s}$.
3
Two trains A and B of length $400 \text{ m}$ each are moving on two parallel tracks with a uniform speed of $72 \text{ km/h}$ in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by $1 \text{ m/s}^2$. If after $50 \text{ s}$, the guard of B just brushes past the driver of A, what was the original distance between them? [3 marks]
Answer
$1250 \text{ m}$
📝
Explanation
We use the concept of relative motion. Let’s observe the motion from Train A’s frame.
Initial relative velocity of B with respect to A: $u_{BA} = u_B – u_A = 72 \text{ km/h} – 72 \text{ km/h} = 0 \text{ m/s}$.
Relative acceleration of B w.r.t A: $a_{BA} = a_B – a_A = 1 \text{ m/s}^2 – 0 = 1 \text{ m/s}^2$.
Time taken $t = 50 \text{ s}$.
Using $S_{rel} = u_{rel}t + \frac{1}{2}a_{rel}t^2$
$S_{rel} = 0 + \frac{1}{2}(1)(50)^2 = 1250 \text{ m}$.
This total relative distance covers the initial gap between them.
Concept Map
1D Kinematics connects to →
Mechanics
Motion in a Plane (2D Vectors)
Newton’s Laws of Motion
Work, Energy & Power
Differential Calculus
Integral Calculus
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