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Topic 2 of 20

One-Dimensional Kinematics

Kinematics describes how objects move without asking why. In one dimension, motion along a straight line is fully characterized by position, velocity, and acceleration. When acceleration is constant — as in free fall near Earth's surface — a set of four kinematic equations connects these quantities and makes every 1D motion problem solvable.

Key Concepts

Position & Displacement

Position

x

locates an object on a number line. Displacement

\Delta x = x_f - x_i

is the change in position — a vector (has sign/direction). Distance is the total path length — always positive.

Average Velocity

Average velocity over an interval:

\bar{v} = \Delta x / \Delta t

. It is a vector (positive or negative) and equals the slope of a position-vs-time graph over that interval.

Instantaneous Velocity

The velocity at a single instant:

v = dx/dt

, the slope of the tangent to the

x(t)

curve. Speed is the magnitude

|v|

and is always non-negative.

Acceleration

Rate of change of velocity:

a = \Delta v / \Delta t = dv/dt

. Positive acceleration does not necessarily mean speeding up — it depends on the sign of the velocity.

Constant Acceleration

When acceleration is constant, the four kinematic equations apply. The

v

-vs-

t

graph is a straight line with slope

a

; the

x

-vs-

t

graph is a parabola.

Free Fall

Near Earth's surface, a freely falling object (no air resistance) has constant downward acceleration

g \approx 9.8 \text{ m/s}^2

. Taking upward as positive,

a = -g

Key Equations

Average Velocity

\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}

Average rate of change of position over a time interval.

Velocity (const. acceleration)

v = v_0 + at

Velocity as a linear function of time when acceleration is constant.

Position (const. acceleration)

x = x_0 + v_0 t + \tfrac{1}{2}at^2

Position as a quadratic function of time for constant acceleration.

Velocity–Position Relation

v^2 = v_0^2 + 2a\,\Delta x

Eliminates time; useful when time is not given or needed.

Average Velocity (const. accel.)

\bar{v} = \frac{v_0 + v}{2}

For constant acceleration only, average velocity equals the mean of initial and final velocities.

Worked Example

Braking Distance

Problem

A car traveling at 30 m/s brakes with constant deceleration of 6 m/s². How far does it travel before stopping?

Solution

Identify known quantities: $v_0 = 30$ m/s, $v = 0$ (stops), $a = -6$ m/s².

Use the velocity–position relation since time is not asked for:

v^2 = v_0^2 + 2a\,\Delta x

Solve for $\Delta x$ :

\Delta x = \frac{v^2 - v_0^2}{2a} = \frac{0 - (30)^2}{2(-6)} = \frac{-900}{-12}

\Delta x = 75 \text{ m}

Answer The car travels 75 m before stopping.

Practice

Exercises

7 problems

1 of 7

A car starts from rest and accelerates uniformly at $4\,\text{m/s}^2$ . How far does it travel in $6$ seconds?

Your answer

2 of 7

An object is dropped from rest. Taking $g = 9.8\,\text{m/s}^2$ , how far does it fall in $5$ seconds?

Your answer

3 of 7

A car moving at $30\,\text{m/s}$ brakes with uniform deceleration of $5\,\text{m/s}^2$ . How far does it travel before stopping?

Your answer

4 of 7

A cyclist traveling at $12\,\text{m/s}$ decelerates uniformly at $3\,\text{m/s}^2$ . How many seconds does it take to come to a complete stop?

Your answer

5 of 7

A sprinter starts from rest and reaches $10\,\text{m/s}$ over a distance of $20\,\text{m}$ . What is their average acceleration?

Your answer

m/s²

6 of 7

A ball is thrown vertically upward at $20\,\text{m/s}$ ( $g = 9.8\,\text{m/s}^2$ ). What is the maximum height it reaches above the launch point? Give your answer to one decimal place.

Your answer

7 of 7

A ball is thrown vertically upward from the ground at $14.7\,\text{m/s}$ ( $g = 9.8\,\text{m/s}^2$ ). How long does it take to return to its starting height?

Your answer

Key Takeaways

Displacement is a vector; distance is scalar. Average velocity = displacement / time.
The four kinematic equations apply only when acceleration is constant.
Free fall is a constant-acceleration problem with $a = -g = -9.8 \text{ m/s}^2$ (upward positive).
The $v$ -vs- $t$ graph slope gives acceleration; its area gives displacement.
Choose the kinematic equation that contains the unknown and all known quantities to minimize algebra.