Kinematics in One Dimension: Position, Velocity, and Acceleration

1) Describing Motion Without Forces

Position and displacement (and why distance is different)

In one-dimensional motion, we describe an object’s location along a line using a coordinate position x (for example, along a hallway, a road, or an elevator shaft). You choose an origin (x = 0) and a positive direction (often “to the right” or “up”).

• Displacement is the change in position: Δx = x_f − x_i. It can be positive, negative, or zero.
• Distance traveled is the total path length along the line. It is always nonnegative and depends on the route (including back-and-forth motion).

Example (back-and-forth): Suppose you start at x = 2 m, walk to x = 9 m, then return to x = 5 m. Displacement: Δx = 5 − 2 = +3 m. Distance: 7 m out + 4 m back = 11 m. The displacement is not the same as the distance because direction matters for displacement.

Average vs instantaneous velocity

Velocity describes how position changes with time and includes direction.

• Average velocity over a time interval: v_avg = Δx / Δt.
• Instantaneous velocity at a moment: v = dx/dt (the rate of change of position at that instant).

Misconception alert: speed vs velocity. Speed is how fast you move (a magnitude, nonnegative). Velocity can be negative if you move in the negative direction. In 1D, “negative velocity” simply means motion opposite your chosen positive direction.

Example (average velocity vs average speed): Using the earlier walk: total time Δt = 10 s. Average velocity: v_avg = +3/10 = +0.3 m/s. Average speed: distance/time = 11/10 = 1.1 m/s. Different numbers because one uses displacement and direction, the other uses total distance.

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Average vs instantaneous acceleration

Acceleration describes how velocity changes with time.

• Average acceleration: a_avg = Δv / Δt.
• Instantaneous acceleration: a = dv/dt.

Misconception alert: “acceleration means speeding up.” Not always. Acceleration tells you how velocity changes, not whether speed increases. If velocity and acceleration have the same sign, speed increases; if they have opposite signs, speed decreases.

2) Connecting Motion to Graphs: x–t, v–t, a–t

Position–time graph (x–t): slope is velocity

An x–t graph shows position as a function of time.

• Slope of the secant line between two times gives average velocity.
• Slope of the tangent line at a time gives instantaneous velocity.

Guided example: You read an x–t graph and see that from t = 2 s to t = 6 s, position goes from x = 1 m to x = 9 m. Average velocity: v_avg = (9 − 1)/(6 − 2) = 8/4 = 2 m/s. If the graph is a straight line in that interval, then velocity is constant at 2 m/s throughout.

Common mix-up: A “high” position on an x–t graph does not mean “high velocity.” Velocity depends on slope, not height.

Velocity–time graph (v–t): slope is acceleration, area is displacement

A v–t graph shows velocity as a function of time.

• Slope gives acceleration: a = dv/dt.
• Area under the curve (signed area) gives displacement: Δx = ∫ v dt.

Guided example (area as displacement): Suppose velocity is constant at v = +3 m/s from t = 0 to t = 4 s. The area under the v–t graph is a rectangle: Δx = vΔt = 3×4 = 12 m.

Guided example (signed area): If v = +2 m/s for 0–3 s and then v = −1 m/s for 3–5 s, displacement is (+2)(3) + (−1)(2) = 6 − 2 = +4 m. Distance traveled would be 6 + 2 = 8 m (use absolute value of velocity when accumulating distance).

Acceleration–time graph (a–t): area is change in velocity

An a–t graph shows acceleration as a function of time.

• Area under the curve gives change in velocity: Δv = ∫ a dt.

Guided example: If a = −2 m/s^2 from t = 0 to t = 3 s, then Δv = aΔt = (−2)(3) = −6 m/s. If the initial velocity was +5 m/s, the final velocity is v_f = 5 − 6 = −1 m/s (it reversed direction by the end).

Graph translation checklist (avoid mixing graphs)

• If x(t) is a straight line, v is constant and a = 0.
• If x(t) curves upward (slope increasing), v is increasing and a is positive.
• If v(t) is a horizontal line, a = 0.
• If v(t) crosses zero, the object changes direction at that time (in 1D).

3) Constant-Acceleration Relationships (Kinematics Equations)

When these equations apply

The following relationships are valid only when acceleration is constant over the time interval you are analyzing. Constant acceleration means a does not change with time (it can be zero, positive, or negative).

Deriving the key relationships from definitions

Start with the definition of constant acceleration:

a = Δv/Δt becomes a = (v − v_0)/t if we set t = 0 at the start of the interval.

Rearrange:

v = v_0 + at

Now connect velocity to position. For constant acceleration, velocity changes linearly, so the average velocity over the interval is:

v_avg = (v_0 + v)/2

Displacement is average velocity times time:

Δx = v_avg t = ((v_0 + v)/2)t

Substitute v = v_0 + at to get a position-time formula:

Δx = v_0 t + (1/2)at^2

Finally, eliminate time to connect v and x. From v = v_0 + at, we have t = (v − v_0)/a (when a ≠ 0). Substitute into Δx = ((v_0 + v)/2)t:

Δx = ((v_0 + v)/2)((v − v_0)/a) = (v^2 − v_0^2)/(2a)

Rearrange:

v^2 = v_0^2 + 2aΔx

Constant-acceleration toolkit (choose based on knowns)

• v = v_0 + at (connects velocity and time)
• Δx = v_0 t + (1/2)at^2 (connects position change and time)
• v^2 = v_0^2 + 2aΔx (connects velocity and displacement; no time)
• Δx = ((v_0 + v)/2)t (uses average velocity for constant a)

Practical step: Before plugging numbers, write a sign convention sentence such as: “Up is positive” or “East is positive.” Then assign signs to v_0, v, a, and Δx based on that choice.

4) Practice in Real Contexts (with Sign Conventions)

A) Braking car (slowing down is acceleration too)

Scenario: A car moving east at v_0 = +20 m/s brakes with constant acceleration a = −5 m/s^2. Find (i) time to stop, (ii) stopping distance.

Step 1: Choose direction. Take east as positive. Then braking means acceleration is negative.

Step 2: Time to stop. Stopping means v = 0. Use v = v_0 + at:

0 = 20 + (−5)t → t = 4 s

Step 3: Stopping distance. Use Δx = v_0 t + (1/2)at^2:

Δx = 20(4) + (1/2)(−5)(4^2) = 80 − 40 = 40 m

Interpretation: The displacement is positive (still moving east while slowing). Acceleration was negative, yet the car was moving in the positive direction—this is the “slowing down” case.

B) Elevator motion (piecewise kinematics)

Elevator trips often have stages: speeding up, cruising, slowing down. Each stage can be modeled with constant acceleration (possibly zero) over that stage.

Scenario: An elevator starts from rest and accelerates upward at a = +1.2 m/s^2 for 3 s, then moves at constant velocity for 5 s, then decelerates (accelerates downward) at a = −1.2 m/s^2 until it stops.

Step 1: End of first stage (accelerating). Up is positive, v_0 = 0. Use v = v_0 + at:

v_1 = 0 + (1.2)(3) = 3.6 m/s

Displacement in stage 1: Δx_1 = (1/2)at^2 = 0.5(1.2)(9) = 5.4 m

Step 2: Cruise stage. a = 0, velocity constant at 3.6 m/s for 5 s:

Δx_2 = v t = 3.6(5) = 18 m

Step 3: Decelerating to rest. Initial velocity for this stage is v_0 = 3.6 m/s, final v = 0, acceleration a = −1.2 m/s^2. Time to stop:

0 = 3.6 + (−1.2)t → t = 3 s

Displacement in stage 3:

Δx_3 = v_0 t + (1/2)at^2 = 3.6(3) + 0.5(−1.2)(9) = 10.8 − 5.4 = 5.4 m

Total displacement: Δx_total = 5.4 + 18 + 5.4 = 28.8 m

Graph insight: The v–t graph is a ramp up (linear increase), then flat, then ramp down to zero. The area under that graph equals the total displacement.

C) Dropping objects (careful sign convention)

Near Earth’s surface, an object in free fall has approximately constant acceleration downward. The key is to be consistent with your sign convention.

Option 1: Up is positive. Then gravitational acceleration is a = −g (negative). If you drop an object from rest, v_0 = 0 and it will quickly have negative velocity.

Scenario: Drop a ball from a balcony 10 m above the ground. Take the balcony as x_0 = 0 and up as positive. The ground is at x = −10 m. Use a = −9.8 m/s^2. Find impact speed.

Step 1: Identify knowns. v_0 = 0, Δx = −10 m, a = −9.8 m/s^2.

Step 2: Use the no-time equation. v^2 = v_0^2 + 2aΔx:

v^2 = 0 + 2(−9.8)(−10) = 196 → v = −14 m/s (choose negative because motion is downward)

Interpretation: The speed is 14 m/s. The velocity is −14 m/s because it points downward in this sign convention.

Common pitfall: Plugging Δx = +10 m while also using a = −9.8 m/s^2 would produce a sign mismatch. Decide what direction is positive first, then assign signs to Δx and a accordingly.

Misconceptions to Actively Check

• “If velocity is zero, acceleration must be zero.” Not true. At the top of a tossed ball’s motion, v = 0 momentarily but a is still nonzero (downward).
• “Acceleration always points in the direction of motion.” Not true. During braking, velocity and acceleration point in opposite directions.
• “A steeper x–t graph means higher acceleration.” Steeper x–t means higher velocity. Acceleration relates to how the slope changes (curvature), not the slope itself.
• “A high value on a v–t graph means high position.” v–t height is velocity, not position. Position comes from the area under the v–t curve.

Mixed Exercises (Choose Equation or Graph Method)

1) Displacement vs distance

A runner starts at x = 0, runs to x = +60 m, then back to x = +20 m. (a) Find displacement. (b) Find distance traveled.

2) Reading an x–t graph (slope)

An object’s position changes linearly from x = −2 m at t = 1 s to x = +10 m at t = 5 s. (a) Find the velocity. (b) Is acceleration zero or nonzero in this interval?

3) Reading a v–t graph (area and sign)

Velocity is +4 m/s from t = 0 to 2 s, then −2 m/s from t = 2 s to 6 s. (a) Find displacement. (b) Find distance traveled. (c) At what time does the object change direction?

4) Reading an a–t graph (area)

An object starts with v_0 = +1 m/s. Acceleration is +3 m/s^2 for 2 s, then 0 for 3 s, then −1 m/s^2 for 4 s. Find the final velocity using areas under the a–t graph.

5) Constant-acceleration choice (braking)

A bike moves at v_0 = +8 m/s and slows with constant acceleration a = −2 m/s^2. (a) How long until it stops? (b) How far does it travel while stopping? Use two different equations and verify they agree.

6) Elevator: piecewise modeling

An elevator goes down (take up as positive). It starts from rest, accelerates downward at magnitude 1.0 m/s^2 for 4 s, then continues at constant velocity for 6 s. (a) What is its velocity after the first stage (include sign)? (b) What is total displacement after 10 s?

7) Free fall with sign convention

Take up as positive. A ball is thrown upward from x = 0 with v_0 = +12 m/s and a = −9.8 m/s^2. (a) Find the time when v = 0. (b) Find the maximum height (displacement at that time). (c) Without finding total time, find the speed when it returns to x = 0 (use v^2 relation).

8) Graph-method vs equation-method (choose wisely)

You are given a v–t graph that is a straight line from v = +6 m/s at t = 0 to v = −2 m/s at t = 4 s. (a) Find acceleration from the slope. (b) Find displacement from the area. (c) Find the time when the object changes direction.