Motion Along a Line: Position, Velocity, Speed, and Acceleration

1) Core quantities in one-dimensional motion (with units)

In motion along a line, we describe where an object is, how fast it is moving, and how its motion is changing using three related functions of time t.

• Position: s(t) (units: length, e.g., meters m or feet ft). It can be positive or negative depending on the chosen origin and direction.
• Velocity: v(t)=s'(t) (units: length/time, e.g., m/s). Velocity is signed: it tells both speed and direction.
• Acceleration: a(t)=v'(t)=s''(t) (units: length/time², e.g., m/s^2). Acceleration measures how velocity changes.

Unit check: If s is in meters and t is in seconds, then s'(t) must be in m/s, and s''(t) must be in m/s^2. Any final statement about motion should include units.

Instantaneous vs average (interpretation)

At a specific time t, v(t) is the instantaneous rate of change of position: it is the slope of the tangent line to the position graph s at that time. Similarly, a(t) is the slope of the tangent line to the velocity graph v.

2) Velocity vs speed; detecting direction changes

Velocity is signed; speed is nonnegative.

• Speed: |v(t)| (units: m/s). It ignores direction.
• Direction of motion (relative to the chosen positive direction): determined by the sign of v(t).

When does the object change direction? A direction change occurs at a time t=c where v(c)=0 and v(t) changes sign across c (from positive to negative or vice versa). If v(c)=0 but the sign does not change, the object does not reverse direction there.

Continue in our app.

• Listen to the audio with the screen off.
• Earn a certificate upon completion.
• Over 5000 courses for you to explore!

Or continue reading below...

Download the app

Step-by-step: checking for a direction change

1. Find times where v(t)=0 (and any times where v is undefined, if applicable).
2. Test the sign of v(t) just before and just after each candidate time.
3. Conclude “changes direction” only if the sign flips.

Reasonableness check: If the problem statement says “moving left” (negative direction) during an interval, your computed v(t) should be negative there.

3) Using signs of v(t) and a(t) to describe motion qualitatively

The sign of v tells direction; the sign of a tells whether velocity is increasing or decreasing. Combining them tells whether the object is speeding up or slowing down.

Speeding up vs slowing down

An object is:

• Speeding up when v(t) and a(t) have the same sign (speed increases).
• Slowing down when v(t) and a(t) have opposite signs (speed decreases).

Unit-consistent statement example: “On 2<t<5 s, v(t)<0 m/s and a(t)<0 m/s², so the object moves in the negative direction and speeds up.”

Worked example: intervals of speeding up/slowing down

Let s(t)=t^3-6t^2+9t (meters), for 0≤t≤5 seconds.

Step 1: Compute velocity and acceleration.

v(t)=s'(t)=3t^2-12t+9  (m/s) = 3(t-1)(t-3) (m/s) a(t)=v'(t)=6t-12 (m/s^2) = 6(t-2) (m/s^2)

Step 2: Find sign-change points. Velocity is zero at t=1 and t=3. Acceleration is zero at t=2.

Step 3: Make a sign chart on intervals.

• 0<t<1: pick t=0.5. v(0.5)=3(−0.5)(−2.5)>0, a(0.5)=6(−1.5)<0 ⇒ moving positive, slowing down.
• 1<t<2: pick t=1.5. v(1.5)=3(0.5)(−1.5)<0, a(1.5)=6(−0.5)<0 ⇒ moving negative, speeding up.
• 2<t<3: pick t=2.5. v(2.5)=3(1.5)(−0.5)<0, a(2.5)=6(0.5)>0 ⇒ moving negative, slowing down.
• 3<t<5: pick t=4. v(4)=3(3)(1)>0, a(4)=6(2)>0 ⇒ moving positive, speeding up.

Direction changes: at t=1 (from v>0 to v<0) and at t=3 (from v<0 to v>0).

Reasonableness check: Since a(t) becomes positive after t=2, velocity should be increasing after t=2. Our sign chart shows v goes from negative (between 2 and 3) to positive (after 3), consistent with “increasing.”

4) Connecting position behavior to velocity, and position shape to acceleration

Position increasing/decreasing and velocity

• If v(t)>0 on an interval, then s(t) is increasing on that interval.
• If v(t)<0 on an interval, then s(t) is decreasing on that interval.

Interpretation: the sign of the slope of the s-graph tells whether the object’s position is rising or falling as time passes.

Concavity of position and acceleration

• If a(t)=s''(t)>0 on an interval, the position graph is concave up there (velocity increasing).
• If a(t)=s''(t)<0 on an interval, the position graph is concave down there (velocity decreasing).

Practical interpretation: concavity of s is not “speeding up” by itself; it indicates whether velocity is increasing or decreasing. Speeding up depends on both v and a.

5) Practice tasks and procedures (with unit-consistent final statements)

Task A: Find intervals of speeding up and slowing down

Procedure

1. Compute v(t)=s'(t) and a(t)=v'(t).
2. Find critical times where v(t)=0 and where a(t)=0 (and any undefined points).
3. Make a sign chart for v and a on the resulting intervals.
4. Speeding up where v and a have the same sign; slowing down where they have opposite signs.

Final statement template (include units): “The object speeds up on (a,b) s and slows down on (c,d) s. On these intervals, v has units m/s and a has units m/s^2.”

Task B: Total distance vs displacement

Displacement over [t_1,t_2] is the net change in position:

Displacement = s(t2) - s(t1)   (units: m)

Total distance traveled accounts for direction changes and is the integral of speed:

Total distance = ∫[t1,t2] |v(t)| dt   (units: m)

In practice with a formula for s(t), you can compute total distance by splitting the interval at direction-change times (where v=0 and changes sign), then adding absolute changes in position.

Worked example: distance vs displacement

Using the earlier example s(t)=t^3-6t^2+9t meters on [0,5].

Step 1: Find direction-change times. We found v(t)=3(t-1)(t-3), so direction changes at t=1 and t=3.

Step 2: Compute positions.

s(0)=0 m s(1)=1-6+9=4 m s(3)=27-54+27=0 m s(5)=125-150+45=20 m

Displacement: s(5)-s(0)=20-0=20 m.

Total distance:

• From 0 to 1: distance |s(1)-s(0)|=|4-0|=4 m
• From 1 to 3: distance |s(3)-s(1)|=|0-4|=4 m
• From 3 to 5: distance |s(5)-s(3)|=|20-0|=20 m

Total distance =4+4+20=28 m.

Reasonableness check: Total distance (28 m) is larger than displacement (20 m) because the object reverses direction (it goes forward, then backward, then forward again).

Task C: Interpreting graphs of s(t), v(t), and a(t)

You may be given a graph instead of formulas. The key is to translate slopes and areas into motion language.

From a position graph s(t)

• Velocity is the slope of the tangent line to s. Positive slope means moving positive; negative slope means moving negative.
• Speed corresponds to the steepness (magnitude of slope): steeper means faster.
• Acceleration relates to concavity of s: concave up implies a>0, concave down implies a<0.

From a velocity graph v(t)

• Direction: above the time axis means v>0 (positive direction); below means v<0 (negative direction).
• Acceleration is the slope of the v-graph.
• Displacement over an interval is the signed area under v.
• Total distance is the area under |v| (area below the axis counts positive after taking absolute value).

From an acceleration graph a(t)

• Velocity change over an interval is the signed area under a: v(t2)-v(t1)=∫ a(t) dt.
• If a>0, velocity is increasing; if a<0, velocity is decreasing.

Task D: A quick “motion description” checklist

• State the interval of time with units (seconds).
• State direction using the sign of v (positive/negative direction).
• State speeding up/slowing down using signs of v and a.
• Include units in any computed values (m, m/s, m/s^2).
• Do a reasonableness check: does the sign of velocity match the described direction? does total distance ≥ |displacement|?