Motion in a Straight Line Class 11 Notes (Chapter 3)

Key Terms and Definitions

Path Length (Distance)

The total length of the path traversed by an object between its initial and final positions. It is a scalar quantity.

Displacement

The shortest distance between the initial and final positions of an object. It is a vector quantity, having both magnitude and direction. It can be positive, negative, or zero.

Average Velocity

The ratio of the total displacement to the total time interval. Formula: v_avg = Δx / Δt.

Instantaneous Velocity

The velocity of an object at a particular instant of time. It is the derivative of position with respect to time. Formula: v = dx/dt.

Average Acceleration

The ratio of the change in velocity to the time interval. Formula: a_avg = Δv / Δt.

Instantaneous Acceleration

The acceleration of an object at a particular instant. It is the derivative of velocity with respect to time. Formula: a = dv/dt = d²x/dt².

Frame of Reference

A coordinate system with respect to which the position or motion of an object is described. It consists of an origin and a set of axes.

Uniform Motion

Motion in which an object covers equal distances in equal intervals of time, however small the intervals may be. Velocity is constant and acceleration is zero.

Must-Remember Formulas & Concepts

• Distance ≥ |Displacement|. The equality holds only for motion in a straight line without a change in direction.
• Average Speed = Total Path Length / Total Time. It is always non-negative.
• Average Velocity = Total Displacement / Total Time. It can be positive, negative, or zero.
• For constant acceleration 'a', the three Kinematic Equations are: v = u + at; s = ut + (1/2)at²; v² = u² + 2as.
• Displacement in the nth second: s_n = u + a(n - 1/2).
• For motion under gravity, replace 'a' with '-g' for upward motion and '+g' for downward motion (assuming upward direction is positive).
• The slope of a Position-Time (x-t) graph gives instantaneous velocity (v = dx/dt).
• The slope of a Velocity-Time (v-t) graph gives instantaneous acceleration (a = dv/dt).
• The area under a Velocity-Time (v-t) graph gives displacement (Δx = ∫v dt).
• The area under an Acceleration-Time (a-t) graph gives the change in velocity (Δv = ∫a dt).

Distance vs. Displacement and Speed vs. Velocity

Understanding the distinction between scalar and vector quantities is fundamental in kinematics. Distance and Speed are scalars, while Displacement and Velocity are vectors. Let's clarify this with an example. Imagine you walk 40 meters East from your home to a shop and then walk 30 meters West to a friend's house.

• Distance (Path Length): This is the total ground you covered. You walked 40 m + 30 m = 70 meters. Distance is a scalar quantity; it only has magnitude and is always positive.

• Displacement: This is the net change in your position from the starting point. You started at home and ended at your friend's house, which is 40 m (East) - 30 m (West) = 10 meters East of your home. Displacement is a vector quantity. It has both magnitude (10 m) and direction (East). If you had returned home, your displacement would be zero, even though you walked a significant distance.

This same logic applies to speed and velocity.

• Speed is the rate of change of distance (Speed = Distance / Time). Since distance is a scalar, speed is also a scalar. In our example, if the entire trip took 70 seconds, your average speed would be 70 m / 70 s = 1 m/s.

• Velocity is the rate of change of displacement (Velocity = Displacement / Time). Since displacement is a vector, velocity is also a vector. In our example, your average velocity would be 10 m (East) / 70 s ≈ 0.143 m/s East. The direction is crucial. An object can have a constant speed (like in uniform circular motion) but a changing velocity because its direction of motion is changing.

Understanding Motion Graphs

Deriving Kinematic Equations using Calculus

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Quick Solved Examples

• {"title":"Example 1: Uniform Acceleration","content":"A car starting from rest accelerates uniformly to a speed of 72 km/h in 10 seconds. Find the acceleration.\nSolution:\nInitial velocity, u = 0 m/s\nFinal velocity, v = 72 km/h = 72 (5/18) m/s = 20 m/s\nTime, t = 10 s\nUsing v = u + at:\n20 = 0 + a 10\na = 20 / 10 = 2 m/s²."}
• {"title":"Example 2: Motion Under Gravity","content":"A ball is thrown vertically upwards with a velocity of 20 m/s. What is the maximum height reached by the ball? (Take g = 10 m/s²)\nSolution:\nInitial velocity, u = 20 m/s\nAt maximum height, final velocity, v = 0 m/s\nAcceleration, a = -g = -10 m/s²\nLet height be 'h'. Using v² = u² + 2as:\n0² = 20² + 2(-10)h\n0 = 400 - 20h\n20h = 400\nh = 20 m."}

Common Exam Traps

Sign Convention is King: Always establish a clear sign convention at the start of a problem. Typically, up/right is positive and down/left is negative. For motion under gravity, if you take upward direction as positive, then acceleration a will always be -g, regardless of whether the object is moving up or down. A common error is to use +g for downward motion after setting upward as positive.

Constant Acceleration Only: Remember, the three main kinematic equations (v = u + at, etc.) are valid ONLY for motion with constant acceleration. If acceleration is a function of time or position, you MUST use calculus (integration/differentiation) by starting from a = dv/dt or v = dx/dt.

Quick Revision Check

• When can the displacement of a moving object be zero, but the distance travelled is not zero? When the object returns to its starting point. For example, completing one lap of a circular track.
• What is the acceleration of a car moving with a constant velocity of 60 km/h due East? Zero. Constant velocity means there is no change in speed or direction, hence acceleration is zero.
• A position-time (x-t) graph for a particle is a parabola opening upwards. What can you say about its velocity? The velocity is continuously increasing. The slope of the x-t graph is velocity, and for a parabola opening upwards, the slope is continuously increasing.
• Can a body have zero velocity and still be accelerating? Yes. At the highest point of its trajectory, a vertically thrown object momentarily has zero velocity, but it is still accelerating downwards due to gravity (a = -g).

Frequently Asked Questions