Rest and Motion
An object is said to be at rest if its position does not change with time relative to a reference point. Motion is a change in position with time.
Important: Rest and motion are relative. A passenger sitting in a moving train is at rest relative to the train but in motion relative to the ground.
Scalars and Vectors
Scalar quantities: Have only magnitude. Examples: mass, length, time, speed, temperature.
Vector quantities: Have both magnitude and direction. Examples: displacement, velocity, acceleration, force.
Distance and Displacement
| Property | Distance | Displacement |
|---|---|---|
| Definition | Total path length | Shortest path from initial to final position |
| Nature | Scalar | Vector |
| Value | Always positive | Positive, negative, or zero |
| Dependence | Depends on path | Path independent |
Distance is always greater than or equal to the magnitude of displacement.
Speed and Velocity
Speed: Distance travelled per unit time. Scalar quantity.
- Average speed = Total distance / Total time
Velocity: Displacement per unit time. Vector quantity.
- Average velocity = Displacement / Time interval
- Instantaneous velocity =
lim_(Delta t->0) Delta x/Delta t = dx/dt
Acceleration
Rate of change of velocity. Vector quantity.
a = (v - u)/t (constant acceleration)
a = dv/dt = d^2x/dt^2
Uniform acceleration: Velocity changes by equal amounts in equal time intervals.
Non-uniform acceleration: Velocity changes by unequal amounts in equal time intervals.
Kinematic Equations (Uniform Acceleration)
For an object moving with constant acceleration a:
v = u + at(First equation)s = ut + (1/2)at^2(Second equation)v^2 = u^2 + 2as(Third equation)
Where:
u= initial velocityv= final velocitya= constant accelerationt= times= displacement
Derivation of First Equation
a = (v - u)/t => at = v - u => v = u + at
Derivation of Second Equation
s = text(average velocity) x t = ((u+v)/2) x t = ((u+u+at)/2) x t = ut + (1/2)at^2
Derivation of Third Equation
From first: t = (v-u)/a. Substituting in second:
s = u(v-u)/a + (1/2)a((v-u)/a)^2
Simplifying: v^2 = u^2 + 2as
Motion Under Gravity
For objects moving under gravity (with g = 9.8 m/s^2 downward):
v = u - gt(upward positive)s = ut - (1/2)gt^2v^2 = u^2 - 2gs
Key results for free fall: Max height h = u^2/(2g), Time of ascent = Time of descent = u/g.
Graphical Analysis
Displacement-Time Graph
- Slope = velocity
- Straight line = uniform velocity
- Curved line = acceleration/deceleration
Velocity-Time Graph
- Slope = acceleration
- Area under curve = displacement
Acceleration-Time Graph
- Area under curve = change in velocity
Worked Examples
Example 1: A car accelerates from rest at 2 m/s^2 for 5 seconds. Find its final velocity and distance covered.
Solution: u = 0, a = 2 m/s^2, t = 5 s.
v = u + at = 0 + 2*5 = 10 m/s.
s = ut + (1/2)at^2 = 0 + (1/2)*2*25 = 25 m.
Example 2: A ball is thrown upward with 20 m/s. Find max height and time to reach top.
Solution: u = 20 m/s, v = 0, a = -g = -10 m/s^2.
v^2 = u^2 + 2as => 0 = 400 - 2*10*h => h = 20 m.
v = u + at => 0 = 20 - 10*t => t = 2 s.
Example 3: A body starts from rest and moves with constant acceleration a. The ratio of distances covered in 1st, 2nd, and 3rd seconds is?
Solution: s_n = u + a/2(2n-1). With u = 0: s_1 = a/2, s_2 = 3a/2, s_3 = 5a/2. Ratio = 1:3:5.
Common Mistakes
- Velocity vs speed: Velocity considers direction; speed does not.
- Negative acceleration: Does not always mean deceleration. Sign depends on chosen direction.
- Equations apply only for uniform acceleration: Do not use these equations for non-uniform acceleration.
- g is constant near Earth's surface: Ignores air resistance in 'motion under gravity' problems.
ISC Exam Focus
- Theory (70%): Definitions, derivations of equations of motion, graphical analysis.
- Application (30%): Numerical problems using kinematic equations, motion under gravity.
- ISC frequently asks numericals: "A ball thrown upward ... find max height/time of flight."
- Graphical interpretation questions (slope and area) are common in exams.
Self-Test Questions
Q1: A train moving at 72 km/h is brought to rest in 10 seconds. Find retardation.
Answer: u = 72 km/h = 20 m/s, v = 0, t = 10 s. a = (0-20)/10 = -2 m/s^2. Retardation = 2 m/s^2.
Q2: A car accelerates from 10 m/s to 30 m/s over 100 m. Find acceleration.
Answer: u = 10, v = 30, s = 100. v^2 = u^2 + 2as => 900 = 100 + 200a => a = 4 m/s^2.
Q3: A stone dropped from a height of 45 m. Find time taken.
Answer: u = 0, s = 45, g = 10 m/s^2. s = ut + (1/2)gt^2 => 45 = 0 + 5t^2 => t = 3 s.
Q4: Derive v^2 = u^2 + 2as using graphical method.
Answer: From v-t graph, area under curve = (u+v)/2 * t = s. Also t = (v-u)/a. Substituting: s = (u+v)(v-u)/(2a) => v^2 - u^2 = 2as.
Q5: A ball thrown up with 40 m/s. Find velocity after 3 seconds and 5 seconds.
Answer: v = u - gt. At t=3: v = 40 - 30 = 10 m/s (upward). At t=5: v = 40 - 50 = -10 m/s (downward).
Q6: Draw v-t graph for a body thrown upward and returning to ground.
Answer: Straight line with negative slope intersecting t-axis at t=u/g, with total time = 2u/g.
