Lesson 9 of 9 · 15 min

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Must-know facts

16 facts

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

Common traps

Where marks are lost

Taking a negative acceleration to always mean the body is slowing down.

Slowing down depends on the relative signs of v and a. A body moving in the negative direction with negative acceleration is speeding up.

Putting x = v₀t + ½at² equal to the total distance when the body reverses direction during the interval.

The equation gives displacement. Find the time at which v = 0, then add the magnitudes of the displacements before and after the turn.

Using g = +10 m s⁻² for the upward part of a throw and −10 for the downward part.

Gravity always points down. Fix one positive direction for the whole flight and keep a = −g (if up is positive) from launch to landing.

Substituting speeds in km/h into equations with g in m s⁻² or distances in metres.

Convert every speed to m s⁻¹ first (× 5/18) before using any kinematic equation.

Using the kinematic equations when x or v is given as a function of t with a changing acceleration (for example x = 6t² − t³).

Check whether a is constant. If a depends on t, use v = dx/dt and a = dv/dt, and find turning points where v = 0.

Assuming zero velocity implies zero acceleration.

At the highest point of a vertical throw the velocity is zero but the acceleration is still g downward; that is why the body falls back.

Reading the area under a v-t graph below the time axis as positive when asked for displacement.

For displacement, areas below the axis count as negative. For distance, add all areas as positive.

Averaging initial and final speeds to get the average speed when the acceleration is not constant, or when two phases last for unequal times.

(v₀ + v)/2 works only for one phase of constant acceleration. In general average speed = total path length ÷ total time.

Formulas

13 to know

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.

Key terms

12 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.

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