Velocity-Time Graphs | Learn Physics with Erkan Koymen

Definition

A velocity-time graph (v-t graph) represents the velocity of an object as a function of time.

The vertical axis (y-axis) shows the instantaneous velocity
The horizontal axis (x-axis) shows the elapsed time (t) in seconds (s)

Reading a Velocity-Time Graph

To determine the kinematic properties of an object at any instant from a v-t graph:

Vertical Position (y-value): Gives the instantaneous velocity and the direction of motion.

Above the time axis (v > 0): The object is moving in the positive direction (forward).
Below the time axis (v < 0): The object is moving in the negative direction (backward).
On the time axis (v = 0): The object is momentarily at rest or actively reversing its direction of motion.

Horizontal Position (x-value): Indicates the elapsed time (t).

**Figure 3.6.1 –** Reading instantaneous velocity and motion direction from coordinates on a v-t graph.

Slope of a Velocity-Time Graph

The slope of a line on a velocity-time graph equals the acceleration (a) of the object:

\text{Slope}=\frac{\text{Rise}}{\text{Run}}=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}=a

Units of Slope: m/s²

A steeper line indicates a greater magnitude of acceleration (velocity changes at a faster rate).
A straight line indicates that the object experiences constant (uniform) acceleration.
A horizontal line indicates zero acceleration (a = 0), meaning the velocity remains constant.

**Figure 3.6.2** – A linear velocity-time graph showing constant acceleration calculated via slope

Positive, Negative, and Zero Slope

Positive Slope (a > 0): The line rises to the right. Velocity is increasing in the positive direction (speeding up forward) or becoming less negative (slowing down backward).
Zero Slope (a = 0): The line is completely horizontal. Velocity does not change over time (constant speed in a straight line).
Negative Slope (a < 0): The line falls to the right. Velocity is decreasing in the positive direction (slowing down forward) or becoming more negative (speeding up backward).

**Figure 3.6.3 –** Physical interpretation of positive slope $(a>0)$ , zero slope $(a=0)$ , and negative slope $(a<0)$ .

Average vs. Instantaneous Acceleration

When an object accelerates non-uniformly, the velocity-time graph is curved rather than straight:

Average Acceleration (a_avg): The slope of the secant line connecting two distinct points (t₁, v₁) and (t₂, v₂) on the curve:

a_{\text{avg}} = \frac{v_2 – v_1}{t_2 – t_1} = \frac{\Delta v}{\Delta t}

Instantaneous Acceleration (a): The slope of the tangent line touching the curve at a single specific instant in time (t).

**Figure 3.6.4** – Secant line representing average acceleration $a_{avg}$ versus tangent line representing instantaneous acceleration $a$ on a non-linear $v-t$ curve.

Area Under a Velocity-Time Graph (Displacement)

1. Why Area Equals Displacement?

The fundamental relationship between velocity, time, and displacement is:

\Delta x = v \cdot \Delta t

On a velocity-time graph, multiplying the vertical axis (velocity in m/s) by the horizontal axis (time in s) yields meters:

\text{Units of Area} = \left(\frac{\text{m}}{\text{s}}\right) \times \text{s} = \text{m}

Therefore, the geometric area bounded between the graph line and the horizontal time axis represents the object’s displacement $\Delta x$ .

2. Calculating Displacement for Uniform Acceleration (Geometric Decomposition)

When an object moves with constant acceleration from an initial velocity $v_i$ to a final velocity $v_f$ over a time interval $\Delta t$ , the region under the line forms a trapezoid.

As shown in Figure 3.6.5, this total region can be broken down into two simple geometric shapes:

A bottom rectangle representing the displacement if the object had continued at its initial velocity:

\text{Area}_{\text{Rectangle}} = v_i \Delta t

A top triangle representing the additional displacement gained due to constant acceleration (a):

\text{Area}_{\text{Triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(\Delta t)(\Delta v) = \frac{1}{2} a (\Delta t)^2

Adding these two areas gives the standard kinematic equation for displacement:

\text{Displacement } (\Delta x) = \text{Area}_{\text{Rectangle}} + \text{Area}_{\text{Triangle}} = v_i \Delta t + \frac{1}{2} a (\Delta t)^2

**Figure 3.6.5:** Finding total displacement $(\Delta x)$ under a linear velocity-time graph by decomposing the region into a constant-velocity rectangle $(\text{Area}_1)$ and an accelerated-motion triangle $(\text{Area}_2)$

Motion with Direction Changes and Total Analysis

When an object changes its direction of motion, its velocity changes sign by crossing the horizontal time axis (v = 0). To analyze this complete motion, the velocity-time graph is divided into signed geometric regions.

1. Direction and Signed Area

Area Above the Time Axis (v > 0): The object moves forward in the positive direction, producing a positive displacement $(+\Delta x)$ .
Area Below the Time Axis (v < 0): The object moves backward in the negative direction, producing a negative displacement $(-\Delta x)$ .

2. Net Displacement vs. Total Distance Traveled

Net Total Displacement $(\Delta x_{\text{total}})$ : The vector sum of all individual displacements, accounting for directional signs:

\Delta x_{\text{total}} = \text{Area}_{\text{above}} – \text{Area}_{\text{below}}

Total Distance Traveled $(d_{\text{total}})$ : The total scalar ground covered, calculated by summing the absolute magnitudes of all areas:

d_{\text{total}} = |\text{Area}_{\text{above}}| + |\text{Area}_{\text{below}}|

Worked Example: Multi-Segment Motion

Consider an object undergoing four distinct phases of linear motion over a 10-second interval, as shown in Figure 3.6.6.

**Figure 3.6.3** – Multi-segment velocity-time graph showing constant positive acceleration $(\text{Seg A})$ , constant velocity $(\text{Seg B})$ , deceleration to a momentary stop $(\text{Seg C})$ , and direction reversal with acceleration in the negative direction $(\text{Seg D})$ .

Step-by-Step Segment Analysis:

Segment A (0 to 3 s):

Acceleration: Constant positive acceleration speeding up from rest: $v$

a_A = \frac{6 – 0}{3 – 0} = +2.0 \text{ m/s}^2

Displacement: Triangular area under the curve:

\Delta x_A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(3\text{ s})(6\text{ m/s}) = +9.0\text{ m}

Segment B (3 to 6 s):

Acceleration: Constant positive acceleration speeding up from rest:

a_B = \frac{6 – 6}{6 – 3} = 0\text{ m/s}^2

Displacement: Rectangular area under the line:

\Delta x_B = \text{base} \times \text{height} = (3\text{ s})(6\text{ m/s}) = +18.0\text{ m}

Segment C (6 to 8 s):

Acceleration: Constant deceleration slowing down until momentarily stopping at t = 8 s:

a_C = \frac{0 – 6}{8 – 6} = -3.0\text{ m/s}^2

Displacement: Triangular area bringing the object to rest:

\Delta x_C = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(2\text{ s})(6\text{ m/s}) = +6.0\text{ m}

Segment D (8 to 10 s):

Acceleration: Negative acceleration speeding up in the reverse direction:

a_D = \frac{-4 – 0}{10 – 8} = -2.0\text{ m/s}^2

Displacement: Triangular area located below the time axis:

\Delta x_D = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(2\text{ s})(-4\text{ m/s}) = -4.0\text{ m}

Total Motion Summary:

Net Total Displacement:

\Delta x_{\text{total}} = (+9.0\text{ m}) + (+18.0\text{ m}) + (+6.0\text{ m}) + (-4.0\text{ m}) = +29.0\text{ m}

Total Distance Traveled:

d_{\text{total}} = |+9.0\text{ m}| + |+18.0\text{ m}| + |+6.0\text{ m}| + |-4.0\text{ m}| = 37.0\text{ m}

Key Points

Instantaneous velocity (v) is read directly from the vertical coordinate (y-value) at any time (t).
Acceleration (a) is determined by calculating the slope of the line $(\frac{\Delta v}{\Delta t})$ .
A straight line indicates constant (uniform) acceleration; a horizontal line indicates zero acceleration (a = 0).
Displacement $(\Delta x)$ equals the total geometric area bounded between the function line and the time axis.
Regions above the time axis represent positive displacement $(+\Delta x)$ , while regions below represent negative displacement $(-\Delta x)$ .
Crossing the time axis (v = 0) signifies an active reversal in the object’s direction of motion.

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