Motion in a Straight Line: NEET notes.

01 · NCERT § "Introduction"

Background: position, displacement and distance (prerequisite recap)

• Background recap: the ideas in this block (displacement, path length, average velocity and average speed) are prerequisites the rest of the chapter builds on, summarised here so the later sections read clearly.
• Motion along a straight line is described by one coordinate x measured from a chosen origin, with one direction taken as positive; the choice is free, but it must be kept for the whole problem.
• A body can be handled as a single point (a point object) whenever its own size is negligible next to how far it travels during the motion being studied.
• Displacement is the change in position, Δx = x₂ − x₁. It carries a sign and depends only on the start and end points.
• Path length (distance) is the total length actually covered. It is never negative and is never less than the magnitude of the displacement.
• For a round trip ending at the start point the displacement is zero even though the distance covered is not.
• Average velocity is displacement divided by the time taken, while average speed is path length divided by the time taken; the two are equal in size only if the body never reverses direction.

02 · NCERT § "Instantaneous Velocity and Speed"

Instantaneous velocity and speed

• Instantaneous velocity, v = dx/dt, is the value the average velocity approaches as the time interval shrinks to zero.
• On a position-time (x-t) graph, the slope of the tangent drawn to the curve at any instant gives the velocity at that instant.
• A straight x-t line means constant velocity; a curved x-t graph means the velocity is changing.
• Instantaneous speed is the magnitude of instantaneous velocity. This equality holds at an instant, not for averages over an interval.
• Velocity can be found numerically (by shrinking Δt in a table of values) or by calculus when x is given as a function of t.
• A negative velocity only says the body is moving towards the negative x-direction; it says nothing about whether it is slowing down.

03 · NCERT § "Acceleration"

Acceleration

• Dividing the change in velocity by the time interval gives the average acceleration; the instantaneous acceleration is a = dv/dt = d²x/dt².
• Acceleration is the slope of the velocity-time (v-t) graph at a given instant; SI unit m s⁻².
• The sign of acceleration is fixed by the chosen positive direction. A body speeds up when v and a have the same sign and slows down when they have opposite signs.
• A body can have zero velocity and non-zero acceleration at the same instant, for example at the highest point of a vertical throw.
• With uniform (constant) acceleration, plotting v against t gives a straight line; for uniform motion (a = 0) that line runs parallel to the time axis.
• The x-t graph curves upward when a > 0 and downward when a < 0; for a = 0 it is a straight line.
• The area under the v-t graph over an interval equals the displacement in that interval, with area below the time axis counted as negative.

04 · NCERT § "Acceleration"

Reading motion graphs

• x-t graph: slope gives velocity. v-t graph: slope gives acceleration and area gives displacement. a-t graph: area gives change in velocity.
• To find distance from a v-t graph, add the magnitudes of the areas above and below the time axis; to find displacement, add them with their signs.
• A point where the v-t graph crosses the time axis is where the body momentarily stops and reverses direction.
• An x-t graph can never show two positions at the same instant, and a real v-t graph cannot jump suddenly, since that would need infinite acceleration.
• In NCERT's idealised graphs the sharp corners (instant changes in velocity) are a simplification; real motion rounds them off.
• When x is a polynomial in t, differentiate once for v and twice for a; if a depends on t, the kinematic equations of constant acceleration do not apply.

05 · NCERT § "Kinematic Equations for Uniformly Accelerated Motion"

Kinematic equations for uniformly accelerated motion

• For constant acceleration a, with initial velocity v₀ (at t = 0) and velocity v at time t, three relations connect x, v, a and t: v = v₀ + at, x = v₀t + ½at², and v² = v₀² + 2ax.
• Here x is the displacement from the starting position, not the distance travelled; if the body turns around, find the turning point first.
• The average velocity over the interval is (v₀ + v)/2 for constant acceleration only, which gives x = (v₀ + v)t/2.
• The equations can be obtained graphically (area under the v-t graph) or by integrating a = dv/dt and v = dx/dt with constant a.
• Choose the positive direction first, then give every vector quantity (v₀, v, a, x) its sign from that choice.
• Convert all speeds to m s⁻¹ before substituting: multiply km/h by 5/18.
• For a body that accelerates and then decelerates from rest to rest, the same peak speed links both phases; the v-t graph is a triangle whose area is the total displacement.

06 · NCERT § "Kinematic Equations for Uniformly Accelerated Motion"

Free fall and vertical motion under gravity

• Near the earth's surface, with air resistance neglected, every body has the same downward acceleration g ≈ 9.8 m s⁻², whatever its mass.
• If upward is taken positive, a = −g throughout the flight, on the way up, at the top and on the way down.
• A body released from rest falls a distance ½gt² in time t, and its speed after falling a height h is √(2gh).
• A body thrown up with speed u rises to a height u²/2g in time u/g; with no air resistance it takes the same time to come down to the launch level and returns with the same speed.
• When a body is thrown up from a height above the ground, use one equation from launch to landing with the landing displacement taken as negative; there is no need to split the flight.
• Galileo's law of odd numbers: for a body starting from rest with constant acceleration, distances covered in successive equal time intervals are in the ratio 1 : 3 : 5 : 7 and so on.
• In free fall, v plotted against t is a straight line with slope −g (upward positive), while x plotted against t is a parabola.

07 · NCERT § "Kinematic Equations for Uniformly Accelerated Motion"

Stopping distance and reaction time

• A vehicle moving at speed v₀ that brakes with constant retardation a stops in a distance d = v₀²/2a.
• Because d varies as v₀², doubling the initial speed makes the stopping distance four times larger for the same braking.
• Reaction time is the delay between seeing a signal and acting on it; during this time the vehicle keeps moving at its original speed, adding v₀ × (reaction time) to the stopping distance.
• A simple way to estimate reaction time is to catch a falling ruler: if it falls a distance d before being caught, the reaction time is √(2d/g).
• These are applications of the same constant-acceleration equations, not new laws.

08 · NCERT § "Kinematic Equations for Uniformly Accelerated Motion"

Using calculus in kinematics

• Velocity is the time derivative of position and acceleration is the time derivative of velocity.
• Going the other way, displacement is the time integral of velocity and change in velocity is the time integral of acceleration.
• For motion given as x(t), the body turns around where v = dx/dt = 0; path length over an interval must be added piecewise between turning points.
• When acceleration depends on time, integrate directly instead of using the constant-acceleration equations.
• The kinematic equations stay valid for any sign of constant acceleration, including deceleration, as long as the sign is put in consistently.

Revise

Must-know facts

1. 01Displacement can be positive, negative or zero; distance (path length) is always positive or zero.
2. 02Magnitude of displacement ≤ distance, with equality only when the body moves in one direction without turning back.
3. 03Average speed equals the magnitude of average velocity only if the direction of motion never reverses.
4. 04Slope of the x-t graph = velocity; slope of the v-t graph = acceleration; area under the v-t graph = displacement.
5. 05At the top of a vertical throw, v = 0 but a = g downward.
6. 06v = v₀ + at, x = v₀t + ½at², v² = v₀² + 2ax hold only for constant acceleration.
7. 07km/h × 5/18 = m/s; for example 72 km/h = 20 m/s and 54 km/h = 15 m/s.
8. 08Maximum height of a body thrown up with speed u is u²/2g, reached in time u/g.
9. 09Time of ascent equals time of descent to the same level when air resistance is neglected.
10. 10Distances in successive equal time intervals from rest: 1 : 3 : 5 : 7 (Galileo's law of odd numbers).
11. 11Distances covered from rest in times t, 2t, 3t are in the ratio 1 : 4 : 9.
12. 12Stopping distance varies as the square of initial speed.
13. 13Reaction time from a dropped ruler: t = √(2d/g).
14. 14Free-fall acceleration does not depend on the mass of the body (air resistance neglected).
15. 15Uniform motion: straight x-t line and a v-t line parallel to the time axis.
16. 16Speeding up means v and a have the same sign; slowing down means they have opposite signs.

Where marks go

Common traps

Reference

Formulas

Average velocity

v̄ = Δx/Δt = (x₂ − x₁)/(t₂ − t₁)

Displacement over time; sign follows the chosen positive direction.

Average speed

average speed = total path length / total time

Never negative.

Instantaneous velocity

v = dx/dt

Slope of the tangent to the x-t graph.

Instantaneous acceleration

a = dv/dt = d²x/dt²

Slope of the v-t graph; unit m s⁻².

Velocity-time relation

v = v₀ + at

Constant acceleration only.

Position-time relation

x = v₀t + ½at²

x is displacement from the position at t = 0.

Velocity-position relation

v² = v₀² + 2ax

Constant acceleration; x is displacement.

Displacement with average velocity

x = (v₀ + v)t/2

Constant acceleration only.

Free fall from rest

y = ½gt², v = gt, v² = 2gy

y measured downward from the release point, g ≈ 9.8 m s⁻².

Maximum height of a vertical throw

H = u²/2g, time to top = u/g

Air resistance neglected.

Stopping distance

d = v₀²/2a

a is the magnitude of the constant retardation.

Reaction time from a falling ruler

t = √(2d/g)

d is the distance the ruler falls before it is caught.

Displacement in the nth second

sₙ = v₀ + a(2n − 1)/2

Derived from x = v₀t + ½at² as x(n) − x(n − 1), with t in seconds; constant acceleration, no reversal of direction within that second. A derived result, not an equation NCERT lists.

Glossary

Key terms

Point object

A body whose size can be ignored because it is small compared with the distance it moves.

Displacement

The change in position, with sign; depends only on where the motion starts and ends.

Path length

The total length of the route actually covered; always non-negative.

Average velocity

Displacement divided by the time interval in which it occurs.

Instantaneous velocity

The rate of change of position at a particular instant, dx/dt.

Speed

The magnitude of velocity at an instant; average speed is path length over time.

Acceleration

How fast velocity changes with time, dv/dt.

Uniform motion

Motion with constant velocity, so equal displacements in equal time intervals.

Uniformly accelerated motion

Motion in which velocity changes by equal amounts in equal time intervals.

Free fall

Motion under gravity alone, with air resistance neglected, at acceleration g downward.

Stopping distance

The distance a moving vehicle covers after the brakes are applied until it comes to rest.

Reaction time

The time a person takes to respond after noticing something.