MCAT content library / Chemistry & physics / Motion: connect vectors, units, and graphs
Use the same cart to see why a zero velocity can occur with a nonzero acceleration—and why zero displacement does not mean zero distance traveled.
About 18 minutes, including the worked examples and checks.
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On this page
- Choose the direction before interpreting the sign
- Components preserve information that a magnitude loses
- Worked record: a cart that turns around
- Read slopes and signed areas as different operations
- Acceleration concerns velocity, not speed alone
- Use units to reject an impossible calculation
- Check your reasoning
- Continue learning
- Sources
Choose the direction before interpreting the sign
A cart can be to the right of its starting point while moving left. Its velocity can be zero at an instant when its acceleration is not. These statements describe different features of motion, so one graph cannot be read as if it plotted all three. AAMC includes displacement, velocity, acceleration, vectors, and units in the motion scope.1
Choose an origin, a positive direction, and a reference frame. For the straight track below, let east be positive and let the starting position be x = 0. Position tells where the cart is relative to that origin. Displacement is the final position minus the initial position: Δx = xf − xi. The distance traveled instead adds the lengths of the segments along the route.
Components preserve information that a magnitude loses
A vector includes direction as well as magnitude. Add perpendicular components separately before finding the magnitude. For a displacement with eastward component Δx and northward component Δy, its magnitude is √(Δx² + Δy²). If a vector of magnitude d makes an angle θ measured from the positive x-axis, its components are d cos θ and d sin θ.3
For example, an original route travels 6 m east and then 8 m north in 20 s. The displacement components are (+6, +8) m and its magnitude is 10 m; the route length is 14 m. The average velocity has components (+0.30, +0.40) m/s and magnitude 0.50 m/s. The average speed is 14/20 = 0.70 m/s. Adding the component lengths would confuse the route with the direct displacement.
Changing the coordinate convention changes component signs, not the physical route. A west-positive axis would assign a negative component to the same eastward motion. A sign becomes useful only after its direction has been defined.
Worked record: a cart that turns around
The following graphs and values are original illustrative motion. At t = 0, the cart moves east at 2.0 m/s. Its acceleration remains 1.0 m/s² westward for the next 4.0 s. Thus a = −1.0 m/s², v = v0 + at, and x = x0 + v0t + ½at². These constant-acceleration relationships connect the same motion quantities represented in the graphs.6
| Time (s) | Position x (m) | Velocity v (m/s) |
|---|---|---|
| 0 | 0 | +2.0 |
| 1.0 | 1.5 | +1.0 |
| 2.0 | 2.0 | 0 |
| 3.0 | 1.5 | −1.0 |
| 4.0 | 0 | −2.0 |
During the first two seconds, position increases while velocity falls toward zero. After two seconds, position decreases and velocity becomes negative. The turn occurs where velocity changes sign, at t = 2.0 s. Returning to x = 0 at t = 4.0 s is a different event: the cart is moving west at that time.
Read slopes and signed areas as different operations
Average velocity over a time interval is Δx/Δt: the slope of the line connecting two points on a position graph. Instantaneous velocity describes the slope at one instant. On a smooth position curve, that is the tangent slope. A horizontal tangent means zero velocity at that instant, not necessarily a prolonged stop.4
From 0 to 1 s, the average velocity is (1.5 − 0)/1 = +1.5 m/s. The velocity at the end of that interval is +1.0 m/s. Both are correct; they answer different questions. From 0 to 4 s, the displacement is zero and the average velocity is zero, even though the cart was moving almost the entire time.
A velocity graph provides displacement through its signed area. The positive triangle from 0 to 2 s has area ½ × 2 s × 2 m/s = +2 m. The negative triangle from 2 to 4 s contributes −2 m. Adding them gives zero displacement. Adding their magnitudes gives 4 m traveled, so average speed over all four seconds is 1.0 m/s. The area method follows the same constant-acceleration relationships used to construct the record.6
Acceleration concerns velocity, not speed alone
Average acceleration is Δv/Δt; instantaneous acceleration is the local slope of the velocity graph. Here the slope is (−2 − 2)/4 = −1.0 m/s² throughout. The cart therefore has the same acceleration while slowing down, turning, and speeding up westward.5
When velocity and acceleration point in opposite directions, speed decreases. When they point in the same direction, speed increases. Calling every negative acceleration “slowing down” would fail after the turn. Likewise, setting acceleration to zero because velocity momentarily reaches zero would confuse the value of a graph with its slope.
Use units to reject an impossible calculation
Every term added in a position equation must have dimensions of length. The product v0t has units (m/s)s = m; the product at² has units (m/s²)s² = m. The product at has units m/s, so it cannot be added directly to a position. Units also require converting measurements to a consistent scale before arithmetic: 25 cm/s is 0.25 m/s.2
A dimensional check can reject a formula, but it cannot determine a numerical coefficient or establish that an assumption holds. Both at² and ½at² have dimensions of length. Constant-acceleration reasoning, rather than units alone, supplies the factor ½ in the position relation.
For a new record, preserve three distinctions: position versus change in position, velocity versus speed, and a graph’s height versus its slope or area. State the coordinate convention, identify the interval, and attach units to the result. Those choices make a proposed motion internally checkable before any force explanation is added.
ACTIVE PRACTICE · PHYSICS GRAPHS
Find the Turn, Not Just the Origin
For the original cart, choose east as positive: x(0) = 0, v(0) = +2.0 m/s, and constant acceleration a = −1.0 m/s² over 4 seconds.
At t = s, position x = m, velocity v = m/s, and acceleration a = −1.00 m/s².
At the start, the cart moves east.
The turn occurs when velocity changes sign, not when position returns to zero. Acceleration is nonzero at the instant v = 0.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
A different cart follows x = −2.5 + 3t − ½t², with x in metres and t in seconds, over 0–4 s. Its velocity is v = 3 − t m/s and its acceleration is −1 m/s². At t = 1 s it crosses x = 0. When does it reverse direction, and what is its acceleration then?
A. At 1 s; acceleration is 0 m/s².
Review this choice. Crossing the origin is a position event. At 1 s, v = +2 m/s, so the cart continues east; the stated acceleration remains −1 m/s².
B. At 1 s; acceleration is −1 m/s².
Review this choice. The acceleration is correctly retained, but x = 0 does not identify the turn. The cart is still moving east at 1 s.
C. At 3 s; acceleration is 0 m/s².
Review this choice. Velocity changes sign at 3 s, but a zero velocity does not imply a zero slope of the velocity graph. The acceleration stays −1 m/s².
D. At 3 s; acceleration is −1 m/s².
Correct. Before 3 s, v is positive; after 3 s, it is negative. The velocity passes through zero while its slope, the acceleration, remains −1 m/s².
The principle to carry forward
Locate a reversal with a change in the sign of velocity. Crossing an origin and having zero acceleration are separate conditions.
Check 2
A cart travels 4.0 m east at 2.0 m/s, rests for 2.0 s, and returns 4.0 m west at 1.0 m/s. Take east as positive. What are its average velocity and average speed over the complete trip, including the rest?
A. Average velocity 0 m/s; average speed 0 m/s.
Review this choice. The displacement is zero, but the traveled distance is 8.0 m. Speed does not cancel between opposite directions.
B. Average velocity 0 m/s; average speed 1.0 m/s.
Correct. The elapsed time is 2.0 + 2.0 + 4.0 = 8.0 s. Zero displacement gives zero average velocity; 8.0 m/8.0 s gives an average speed of 1.0 m/s.
C. Average velocity 0 m/s; average speed 4/3 m/s.
Review this choice. This speed uses only the six seconds of motion. The requested complete interval also includes the two-second rest.
D. Average velocity 1.0 m/s; average speed 1.0 m/s.
Review this choice. The average speed is correct, but average velocity uses net displacement, which is zero. Total route length cannot replace displacement in that calculation.
The principle to carry forward
Pauses add elapsed time without adding distance. Opposite displacements cancel in average velocity, but not in average speed.
Check 3
Let x0 be a position, v0 a velocity, a an acceleration, and t a time. Which candidate expression is dimensionally consistent with a final position? Numerical coefficients are dimensionless.
A. x₀ + v₀t + ½at
Review this choice. The first two terms have length units, but at has velocity units. Quantities with different dimensions cannot be added this way.
B. x₀ + v₀/t + ½at²
Review this choice. The term v₀/t has acceleration units, not length. The other two terms do not repair that mismatch.
C. x₀ + v₀t + ½at²
Correct. Each term has length units: x₀ is length, v₀t is length, and at² is length. This passes the requested unit check; using it as a motion law additionally requires the appropriate constant-acceleration assumptions.
D. x₀ + v₀²t² + ½at²
Review this choice. The middle term has units (m²/s²)s² = m². An area cannot be added to the length terms.
The principle to carry forward
Consistency of dimensions is necessary, but not sufficient, for a physical equation to describe a particular motion.
Continue learning
From forces to fluid flow
Lesson 1 of 6.
- Next in From forces to fluid flow: Net force and friction: build the right free-body diagram
- Return to this path’s overview
See every step in From forces to fluid flow
From forces to fluid flow
Select the physical model that explains pressure, speed and flow resistance.
- Motion: translate between vectors, units, and graphs · You are here
- Net force and friction: build the right free-body diagram
- Work, energy, and power: choose the useful conservation model
- Pressure, buoyancy, and the role of fluid interfaces
- Ideal fluid flow: use continuity before Bernoulli
- Viscous flow: explain pressure loss and radius dependence
Related lessons
- Net force and friction: build the right free-body diagram
- Work, energy, and power: choose the useful conservation model
- Reading graphs: axes, scales, and the comparison that matters
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Sources
1. AAMC: C/P Content Category 4A
2. OpenStax University Physics 1: Dimensional Analysis
3. OpenStax University Physics 1: Coordinate Systems and Components of a Vector
4. OpenStax University Physics 1: Instantaneous Velocity and Speed
5. OpenStax University Physics 1: Average and Instantaneous Acceleration
6. OpenStax University Physics 1: Motion with Constant Acceleration