Interpret and sketch position-time, velocity-time and acceleration-time graphs, relating the slope of a graph to a rate of change and the area under a velocity-time graph to displacement.

What this topic is asking

Motion graphs are tested heavily on the Regents, in Part B-1 multiple choice and in Part B-2 where you may have to plot points, draw a best-fit line, and read its slope. This dot point asks you to interpret and sketch the three graph types, position-time, velocity-time and acceleration-time, and to use the two key tools: the slope of a graph (a rate of change) and the area under a velocity-time graph (a displacement).

Position-time graphs

On a graph of position against time, the slope is the velocity, because slope is rise over run, which here is change in position over change in time. A straight line means constant velocity: a steep line is fast, a gentle line is slow, a horizontal line means the object is at rest, and a line sloping the other way means motion in the opposite direction. A curved position-time graph means the velocity is changing, so the object is accelerating; the instantaneous velocity at any point is the slope of the tangent there.

Velocity-time graphs

The velocity-time graph is the most useful of the three, because it shows both acceleration (from the slope) and displacement (from the area). For a straight line from rest, the area is a triangle, $\tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} t v$; for constant velocity it is a rectangle, $v \times t$. Area below the time axis counts as negative displacement (motion the other way).

Acceleration-time graphs

On an acceleration-time graph, a horizontal line means constant acceleration, and the area between the line and the time axis is the change in velocity ($\Delta v = a \times t$). These appear less often on the Regents than the other two but follow the same logic: the area accumulates the quantity that the vertical axis is the rate of.

Drawing and using a best-fit line

A frequent Part B-2 task gives you a data table (for example, distance against time for a rolling cart) and asks you to plot the points and draw a best-fit line. The line should pass through the trend of the points, not connect them dot to dot, with roughly as many points above as below. You then read a value from the line, or calculate its slope using two points on the line (not two raw data points). The slope is interpreted as the physical rate, for example a velocity on a distance-time graph.

Reference Tables note

The slope-and-area relationships of motion graphs are not printed on the Reference Tables, so they are recall items. The tables do print the kinematic equations, and a velocity-time graph is simply a picture of those equations: the slope reproduces $\bar{a} = \dfrac{\Delta v}{\Delta t}$, and the triangular area reproduces $d = \tfrac{1}{2}(v_i + v_f)t$.

Try this

Q1. State what the slope of a position-time graph represents and what the slope of a velocity-time graph represents. [2 points]

• Cue. Slope of position-time = velocity; slope of velocity-time = acceleration.

Q2. A velocity-time graph shows a horizontal line at $6.0$ m/s for $3.0$ s. Calculate the displacement. [2 points]

• Cue. Area = $v \times t = (6.0)(3.0) = 18$ m.