Physics Lab

Velocity-Time Graphs

A velocity-time (vv-tt) graph is one of the most useful diagnostic tools in kinematics. Its slope is the acceleration; the area beneath it is the displacement.

Concept

FeaturePhysical meaning
Slope at a pointInstantaneous acceleration
Area under the curve (above tt-axis)Displacement (positive)
Area below the tt-axisNegative displacement
Horizontal line at v=0v = 0Object at rest
Horizontal line at v0v \ne 0Uniform velocity
Straight line, non-zero slopeUniform acceleration
CurveVariable acceleration

Why Area = Displacement

x(t)=x0+0tvdt.x(t) = x_0 + \int_0^t v\,dt'. The integral is the geometric area between the curve and the tt-axis.

Sign Conventions

  • v>0v > 0: motion in chosen positive direction.
  • v<0v < 0: motion in negative direction.
  • Net displacement = (positive area) − (negative area).

Velocity vs. Speed Graphs

A vv-tt graph shows signed velocity. A v|v|-tt (speed) graph only shows magnitude — slope of v|v| doesn't equal a|a| when sign changes.

Worked Example

Q: A car's vv-tt graph:

  • 0t50 \le t \le 5 s: v=4tv = 4t (accelerates uniformly from rest to 20 m/s).
  • 5t105 \le t \le 10 s: v=20v = 20 m/s (uniform).
  • 10t1510 \le t \le 15 s: vv decreases linearly to 0.

Find (a) acceleration in each phase, (b) total distance.

Solution:

Phase 1: a1=20/5=4a_1 = 20/5 = 4 m/s².

Phase 2: a2=0a_2 = 0.

Phase 3: a3=(020)/(1510)=4a_3 = (0 - 20)/(15-10) = -4 m/s².

Distances (areas under graph):

  • Phase 1: 12(5)(20)=50\tfrac{1}{2}(5)(20) = 50 m.
  • Phase 2: 5×20=1005 \times 20 = 100 m.
  • Phase 3: 12(5)(20)=50\tfrac{1}{2}(5)(20) = 50 m.

Total: 200200 m.

Reading Multiple Features

  • Speeding up or slowing down? Sign of slope vs sign of velocity. Same sign → speeding up; opposite → slowing down.
  • Stationary? vv on the tt-axis (i.e. v=0v = 0).
  • Direction reversal? vv crosses the tt-axis.

Common Confusions

  • "Area on a vv-tt graph gives distance." — It gives displacement (which is signed). For distance, take absolute values of each piece.
  • "If vv increases, the object speeds up." — Only if v>0v > 0; if v<0v < 0, "increasing" (less negative) actually means slowing down.
  • "Steeper slope = faster." — Steeper slope on vv-tt = greater acceleration, not necessarily greater speed.

Key Takeaways

  • Slope of vv-tt = acceleration.
  • Area under vv-tt = displacement.
  • Distance = sum of |areas| of each piece.
  • Direction reverses where vv crosses zero.
  • Position-time and velocity-time graphs are complementary diagnostic tools.

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