NCERT Motion in a Straight Line Class 11 Physics Notes & Practice

Understanding Kinematic Quantities: Distance vs. Displacement

To analyze class 11 physics motion in a straight line, we must first define the frame of reference. A frame of reference is a coordinate system relative to which physical quantities are measured. For rectilinear (straight-line) motion, we align our system with the X-axis.

Path Length (Distance): This is the actual total path length traversed by an object during its motion. It is a scalar quantity, meaning it has only magnitude and is always positive or zero.

Displacement ($_\Delta x$): This is the shortest straight-line distance from the initial position ($x_1$) to the final position ($x_2$), represented mathematically as $_\Delta x = x_2 - x_1$. Displacement is a vector quantity and can be positive, negative, or zero.

Key Distinction: While distance can never decrease with time for a moving object, displacement can. If an object returns to its starting point, its net displacement is zero, but its total distance is non-zero. Understanding this distinction is vital for analyzing graphs and solving kinematics problems accurately.

Deriving Equations of Motion for Uniformly Accelerated Motion

1. Step 1: Establish the Variables and v-t Graph — Consider an object moving along a straight line with a constant acceleration 'a'. At time t = 0, let its initial velocity be u, and at time t, let its final velocity be v. We plot this on a velocity-time (v-t) graph. The slope of this line represents the constant acceleration, while the area under the curve represents the displacement.
2. Step 2: Deriving v = u + at (Velocity-Time Relation) — By definition, acceleration is the rate of change of velocity: a = (v - u) / (t - 0). Multiplying both sides by t gives at = v - u. Rearranging this equation yields our first kinematic equation: v = u + at.
3. Step 3: Deriving s = ut + 0.5at² (Position-Time Relation) — The displacement s equals the area under the v-t graph. This area is a trapezoid, which can be split into a rectangle of area (u t) and a triangle of area 0.5 t (v - u). Since v - u = at, the triangle's area becomes 0.5 a * t². Adding these gives: s = ut + 0.5at².
4. Step 4: Deriving v² = u² + 2as (Velocity-Displacement Relation) — The average velocity for uniform acceleration is v_avg = (v + u) / 2. Displacement is s = v_avg t = ((v + u) / 2) t. From the first equation, substitute t = (v - u) / a into this expression: s = ((v + u) / 2) * ((v - u) / a) = (v² - u²) / 2a. Rearranging this gives: v² = u² + 2as.

Step-by-Step Solved Numerical Examples

• Example 1 (Free Fall Motion): A ball is thrown vertically upwards with a velocity of 20 m/s from the top of a multi-storey building of height 25 m. How high will the ball rise, and how long will it take to hit the ground? (Take g = 10 m/s²) Step 1: Define the sign convention. Let upward direction be positive and downward direction be negative. Here, initial velocity u = +20 m/s, acceleration due to gravity a = -g = -10 m/s². Step 2: Find the maximum height (h) from the top of the building. At the highest point, final velocity v = 0. Using the equation v² = u² + 2as: 0 = (20)² + 2(-10)(h) 20h = 400 => h = 20 m. Therefore, maximum height above the ground is 25 m + 20 m = 45 m. Step 3: Find total time to hit the ground. When it hits the ground, its net displacement from the point of projection is s = -25 m. Using s = ut + 0.5at²: -25 = 20t + 0.5(-10)t² => -25 = 20t - 5t² Divide by -5: t² - 4t - 5 = 0. Solving the quadratic equation: (t - 5)(t + 1) = 0. Since time cannot be negative, t = 5 s. Final Answer: The ball rises to a height of 20 m from the top of the building (total 45 m from ground) and takes 5 s to hit the ground.
• Example 2 (Average Speed vs Average Velocity): A car travels from point A to B at a speed of 40 km/h and returns from B to A at a speed of 60 km/h. Find the average speed and average velocity of the car for the entire journey. Step 1: Let the distance between A and B be d. Total distance traveled = d + d = 2d. Step 2: Calculate time taken. Time for forward journey t1 = d/40 and time for return journey t2 = d/60. Total time T = t1 + t2 = d/40 + d/60 = (3d + 2d)/120 = 5d/120 = d/24. Step 3: Calculate average speed. Average Speed = Total Distance / Total Time = 2d / (d/24) = 48 km/h. Step 4: Calculate average velocity. Since the car returns to its starting point, its net displacement is zero. Therefore, Average Velocity = Net Displacement / Total Time = 0. Final Answer: Average speed is 48 km/h and average velocity is 0.

CBSE Exam Traps & Sign Convention Guide

Many students lose marks in motion in a straight line ncert notes questions due to incorrect sign conventions or graph misinterpretations. Here are the ultimate rules to avoid losing marks:

1. Consistent Sign Convention: Always choose a positive direction (usually upwards or rightwards) and stick to it. If upward is positive, then acceleration due to gravity $g$ must be entered as negative, and any downward displacement is negative.
2. Graphs Interpretation: The slope of a position-time ($x-t$) graph gives velocity. The slope of a velocity-time ($v-t$) graph gives acceleration. The area under a $v-t$ graph gives displacement, while the total area (treating negative areas as positive) gives distance. Do not mix these up!
3. Relative Velocity: Remember that $v_{AB} = v_A - v_B$. Pay close attention to directions; if they are moving towards each other, their relative speed adds up.

Practice Questions with Solutions

• Q: A particle moves along a straight line such that its position x at any time t is given by x = t³ - 6t² + 9t + 5 meters. Find the velocity of the particle when its acceleration is zero. A: Step 1: Differentiate position x with respect to time t to find velocity (v): v = dx/dt = 3t² - 12t + 9. Step 2: Differentiate velocity with respect to time to find acceleration (a): a = dv/dt = 6t - 12. Step 3: Set acceleration to zero to find the time: 6t - 12 = 0 => t = 2 s. Step 4: Substitute t = 2 back into the velocity equation: v(2) = 3(2)² - 12(2) + 9 = 12 - 24 + 9 = -3 m/s. Final answer: The velocity of the particle when acceleration is zero is -3 m/s.
• Q: A jet airplane traveling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground? A: Step 1: Assign velocities with respect to the ground frame. Let the direction of the jet be positive (+X). Velocity of Jet, v_J = +500 km/h. Step 2: Note the direction of the combustion products. They are ejected in the opposite direction. The velocity of combustion products relative to the jet is v_CJ = -1500 km/h. Step 3: Use the relative velocity formula: v_CJ = v_C - v_J. -1500 = v_C - 500. Step 4: Solve for v_C: v_C = -1500 + 500 = -1000 km/h. Final answer: The speed of the combustion products with respect to an observer on the ground is 1000 km/h in the direction opposite to the motion of the jet.
• Q: A police van moving on a highway with 30 km/h fires a bullet at a thief's car speeding away in the same direction with 192 km/h. If the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief's car? A: Step 1: Convert all speeds into m/s. Speed of police van, v_P = 30 (5/18) = 25/3 m/s. Speed of thief's car, v_T = 192 (5/18) = 160/3 m/s. Muzzle speed of bullet relative to police van, v_BP = 150 m/s. Step 2: Calculate the speed of the bullet with respect to the ground (v_B): v_B = v_BP + v_P = 150 + 25/3 = 475/3 m/s. Step 3: Find the speed of the bullet relative to the thief's car (v_BT): v_BT = v_B - v_T = 475/3 - 160/3 = 315/3 = 105 m/s. Final answer: The bullet hits the thief's car with a speed of 105 m/s.
• Q: A body covers 12 m in the 2nd second and 20 m in the 4th second of its motion. Find the initial velocity and acceleration of the body. A: Step 1: Use the formula for displacement in the nth second: s_n = u + (a/2)(2n - 1). Step 2: Set up equations for the given conditions: For n = 2: 12 = u + (a/2)(2(2) - 1) => 12 = u + 1.5a --- (Equation 1) For n = 4: 20 = u + (a/2)(2(4) - 1) => 20 = u + 3.5a --- (Equation 2) Step 3: Subtract Equation 1 from Equation 2: 20 - 12 = (u + 3.5a) - (u + 1.5a) 8 = 2a => a = 4 m/s². Step 4: Substitute a = 4 back into Equation 1: 12 = u + 1.5(4) => 12 = u + 6 => u = 6 m/s. Final answer: The initial velocity is 6 m/s and the acceleration is 4 m/s².

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