One-Dimensional Motion

Position, velocity, and acceleration in one dimension

One-Dimensional Motion

Introduction to Kinematics

Kinematics is the study of motion without considering the forces that cause it. We describe motion using:

Position ( $x$ )
Velocity ( $v$ )
Acceleration ( $a$ )
Time ( $t$ )

Position and Displacement

Position

Position ( $x$ ) is the location of an object relative to a reference point (origin).

Measured in meters (m)
Can be positive or negative
Example: $x = 5$ m means 5 meters to the right of the origin

Displacement

Displacement ( $\Delta x$ ) is the change in position.

$\Delta x = x_f - x_i$

Where:

$x_f$ = final position
$x_i$ = initial position

Note: Displacement is different from distance traveled!

Displacement: straight-line change (can be negative)
Distance: total path length (always positive)

Velocity

Average Velocity

Average velocity is displacement divided by time interval.

$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$

Units: m/s
Can be positive (moving right/up) or negative (moving left/down)
Vector quantity (has direction)

Instantaneous Velocity

Instantaneous velocity is the velocity at a specific instant.

$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$

This is the slope of the position-time graph at a point
On a graph: tangent line slope

Speed

Speed is the magnitude of velocity (always positive).

$\text{speed} = |v|$

Acceleration

Average Acceleration

Average acceleration is the change in velocity divided by time.

$a_{avg} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}$

Units: m/s²
Can be positive or negative
Positive $a$ : speeding up in positive direction OR slowing down in negative direction

Instantaneous Acceleration

Instantaneous acceleration is the acceleration at a specific instant.

$a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

This is the slope of the velocity-time graph at a point

Kinematic Equations (Constant Acceleration)

When acceleration is constant, we can use these powerful equations:

The Big Four Kinematic Equations

$v = v_0 + at$

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

$v^2 = v_0^2 + 2a(x - x_0)$

$x = x_0 + \frac{1}{2}(v_0 + v)t$

Where:

$x_0$ = initial position
$x$ = final position
$v_0$ = initial velocity
$v$ = final velocity
$a$ = acceleration (constant)
$t$ = time

Choosing the Right Equation

| Missing variable | Use equation | |-----------------|--------------| | $x$ | $v = v_0 + at$ | | $v$ | $x = x_0 + v_0 t + \frac{1}{2}at^2$ | | $a$ | $x = x_0 + \frac{1}{2}(v_0 + v)t$ | | $t$ | $v^2 = v_0^2 + 2a(x - x_0)$ |

Free Fall

Free fall is motion under gravity alone (no air resistance).

$a = -g = -9.8 \text{ m/s}^2$

Negative because gravity pulls downward
Same equations apply, but use $a = -g$
Objects thrown upward: velocity decreases at rate $g$ until $v = 0$ at peak

Key Facts About Free Fall

At the peak of motion, $v = 0$ but $a = -g$
Time up = time down (for same height)
Landing speed = launching speed (magnitude)

Sign Conventions

Always establish a coordinate system first!

Common convention:

Positive direction: right, up
Negative direction: left, down

Then:

Velocity in positive direction → $v > 0$
Velocity in negative direction → $v < 0$
Acceleration in positive direction → $a > 0$
Acceleration in negative direction → $a < 0$

Problem-Solving Strategy

Draw a diagram with coordinate system
List known variables ( $x_0$ , $v_0$ , $v$ , $a$ , $t$ )
Identify unknown you need to find
Choose equation that relates knowns and unknown
Solve algebraically before plugging in numbers
Check units and reasonableness

📚 Practice Problems

1Problem 1medium

❓ Question:

A car accelerates uniformly from rest to 25 m/s in 5.0 seconds. (a) What is the car's acceleration? (b) How far does the car travel during this time? (c) What is the car's average velocity during this interval?

💡 Show Solution

Solution:

Given: v₀ = 0 m/s (starts from rest), v = 25 m/s, t = 5.0 s

(a) Acceleration: Using v = v₀ + at 25 = 0 + a(5.0) a = 25/5.0 = 5.0 m/s²

(b) Distance traveled: Using Δx = v₀t + ½at² Δx = 0 + ½(5.0)(5.0)² Δx = ½(5.0)(25) Δx = 62.5 m or 63 m

Alternatively, using v² = v₀² + 2aΔx: (25)² = 0 + 2(5.0)Δx 625 = 10Δx Δx = 62.5 m ✓

Or: v_avg = Δx/t = 62.5/5.0 = 12.5 m/s ✓

2Problem 2easy

❓ Question:

A car starts from rest and accelerates uniformly at $2.0$ m/s² for $5.0$ seconds. How far does it travel?

💡 Show Solution

Given information:

Initial velocity: $v_0 = 0$ m/s (starts from rest)
Acceleration: $a = 2.0$ m/s²
Time: $t = 5.0$ s
Initial position: $x_0 = 0$ (assume)

Find: Distance traveled ( $x$ )

Choose equation: We have $v_0$ , $a$ , and $t$ , and need $x$ . Missing $v$ .

Use: $x = x_0 + v_0 t + \frac{1}{2}at^2$

Solve: $x = 0 + (0)(5.0) + \frac{1}{2}(2.0)(5.0)^2$ $x = 0 + 0 + \frac{1}{2}(2.0)(25)$ $x = \frac{1}{2}(50)$ $x = 25 \text{ m}$

Answer: The car travels 25 meters.

Check: Units are correct (m). A car accelerating for 5 seconds should travel a reasonable distance, and 25 m ≈ 82 feet seems right.

3Problem 3medium

❓ Question:

💡 Show Solution

Solution:

Given: v₀ = 0 m/s (starts from rest), v = 25 m/s, t = 5.0 s

(a) Acceleration: Using v = v₀ + at 25 = 0 + a(5.0) a = 25/5.0 = 5.0 m/s²

(b) Distance traveled: Using Δx = v₀t + ½at² Δx = 0 + ½(5.0)(5.0)² Δx = ½(5.0)(25) Δx = 62.5 m or 63 m

Alternatively, using v² = v₀² + 2aΔx: (25)² = 0 + 2(5.0)Δx 625 = 10Δx Δx = 62.5 m ✓

Or: v_avg = Δx/t = 62.5/5.0 = 12.5 m/s ✓

4Problem 4hard

❓ Question:

A ball is thrown straight upward with an initial velocity of 20 m/s. Taking upward as positive and g = 10 m/s², find: (a) the maximum height reached, (b) the time to reach maximum height, (c) the total time in the air before returning to the starting point, and (d) the velocity when it returns to the starting height.

💡 Show Solution

Solution:

Given: v₀ = +20 m/s, a = -10 m/s² (gravity acts downward)

(a) Maximum height: At max height, v = 0 Using v² = v₀² + 2aΔy 0 = (20)² + 2(-10)Δy 0 = 400 - 20Δy Δy = 20 m

(b) Time to max height: Using v = v₀ + at 0 = 20 + (-10)t t = 2.0 s

(c) Total time in air: The motion is symmetric for projectiles. Time up = Time down Total time = 2 × 2.0 s = 4.0 s

Alternatively, using Δy = v₀t + ½at² with Δy = 0 (returns to start): 0 = 20t + ½(-10)t² 0 = 20t - 5t² 0 = t(20 - 5t) t = 0 or t = 4.0 s → 4.0 s

(d) Velocity upon return: Using v = v₀ + at v = 20 + (-10)(4.0) v = 20 - 40 = -20 m/s (downward)

The speed is the same, but direction reversed.

5Problem 5hard

❓ Question:

💡 Show Solution

Solution:

Given: v₀ = +20 m/s, a = -10 m/s² (gravity acts downward)

(a) Maximum height: At max height, v = 0 Using v² = v₀² + 2aΔy 0 = (20)² + 2(-10)Δy 0 = 400 - 20Δy Δy = 20 m

(b) Time to max height: Using v = v₀ + at 0 = 20 + (-10)t t = 2.0 s

(c) Total time in air: The motion is symmetric for projectiles. Time up = Time down Total time = 2 × 2.0 s = 4.0 s

Alternatively, using Δy = v₀t + ½at² with Δy = 0 (returns to start): 0 = 20t + ½(-10)t² 0 = 20t - 5t² 0 = t(20 - 5t) t = 0 or t = 4.0 s → 4.0 s

(d) Velocity upon return: Using v = v₀ + at v = 20 + (-10)(4.0) v = 20 - 40 = -20 m/s (downward)

The speed is the same, but direction reversed.

6Problem 6medium

❓ Question:

A ball is thrown vertically upward with an initial velocity of $20$ m/s. How high does it go? (Use $g = 10$ m/s²)

💡 Show Solution

Set up coordinate system:

Positive direction: upward
Origin: ground level

Given information:

Initial velocity: $v_0 = +20$ m/s (upward)
Acceleration: $a = -g = -10$ m/s² (gravity pulls down)
At maximum height: $v = 0$ m/s
Initial position: $y_0 = 0$ (ground)

Find: Maximum height ( $y$ )

Choose equation: We have $v_0$ , $v$ , and $a$ , need $y$ . Missing $t$ .

Use: $v^2 = v_0^2 + 2a(y - y_0)$

Solve: $0^2 = (20)^2 + 2(-10)(y - 0)$ $0 = 400 - 20y$ $20y = 400$ $y = 20 \text{ m}$

Answer: The ball reaches a maximum height of 20 meters.

Physical insight: At the peak, velocity is zero but acceleration is still $-10$ m/s² (gravity never stops pulling down!).

7Problem 7hard

❓ Question:

A train traveling at $30$ m/s begins to decelerate uniformly at $-1.5$ m/s². How long does it take to stop, and how far does it travel during this time?

💡 Show Solution

Given information:

Initial velocity: $v_0 = 30$ m/s
Final velocity: $v = 0$ m/s (stops)
Acceleration: $a = -1.5$ m/s² (negative = slowing down)
Initial position: $x_0 = 0$ (assume)

Find:

Time to stop ( $t$ )
Distance traveled ( $x$ )

Part 1: Find time

Use: $v = v_0 + at$

$0 = 30 + (-1.5)t$ $1.5t = 30$ $t = 20 \text{ s}$

Part 2: Find distance

Method 1 - Use the time we just found: $x = x_0 + v_0 t + \frac{1}{2}at^2$ $x = 0 + (30)(20) + \frac{1}{2}(-1.5)(20)^2$ $x = 600 + \frac{1}{2}(-1.5)(400)$ $x = 600 - 300$ $x = 300 \text{ m}$

Method 2 - Without using time (faster!): $v^2 = v_0^2 + 2a(x - x_0)$ $0^2 = (30)^2 + 2(-1.5)(x - 0)$ $0 = 900 - 3x$ $3x = 900$ $x = 300 \text{ m}$

Answers:

Time to stop: 20 seconds
Distance traveled: 300 meters

Check: Average velocity = $\frac{30 + 0}{2} = 15$ m/s. Distance = $(15)(20) = 300$ m ✓

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