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Physics · Ch 2 — Motion in a Straight Line

Velocity-Time Graphs and Graphical Analysis

2.7

Velocity-Time Graphs and Graphical Analysis

A velocity-time (v-t) graph plots the instantaneous velocity of a body along the vertical axis against time along the horizontal axis. It is, in a definite sense, more information-rich than a position-time graph for our present purposes, because it lets us read off both acceleration (from its slope) and displacement (from the area beneath it) directly.

Slope of a v-t graph gives acceleration. Since instantaneous acceleration is defined as a=dv/dta = dv/dt (Section 2.5), and the derivative at a point is the slope of the tangent to the graph at that point, the slope of a v-t graph at any instant is exactly the instantaneous acceleration at that instant. For uniformly accelerated motion, aa is constant, so the v-t graph is a straight line whose (constant) slope is that constant acceleration — steeper lines mean larger acceleration, and a line sloping downward corresponds to a negative (deceleration/retarding) acceleration.

Area under a v-t graph gives displacement. This is the more subtle — and more powerful — reading. Because v=dx/dtv = dx/dt, we have dx=v dtdx = v\,dt, and integrating both sides between times t1t_1 and t2t_2,

x2−x1=∫t1t2v dtx_2 - x_1 = \int_{t_1}^{t_2} v\,dt

The integral on the right-hand side is, by the fundamental geometric meaning of a definite integral, exactly the area enclosed between the v-t curve and the time axis over the interval [t1,t2][t_1, t_2] (counted as negative area for any stretch where vv is negative). So:

displacement=area under the v-t graph\text{displacement} = \text{area under the } v\text{-}t \text{ graph}

Applying this to uniformly accelerated motion. For a body with constant acceleration aa, starting with velocity uu at t=0t = 0 and reaching velocity vv at time TT, the v-t graph is a straight line rising (or falling) from uu to vv. The region under this line, above the time axis, between t=0t=0 and t=Tt=T, is a trapezium with parallel sides of length uu and vv and width TT. Its area is

s=12(u+v) Ts = \frac{1}{2}(u+v)\,T

which is exactly the displacement covered in time TT. This graphical route reproduces the average-of-initial-and-final-velocity kinematic relation used later in Section 2.8, entirely from the shape of the graph, with no separate calculus required. The accompanying figure shows exactly this trapezoidal shaded region. …

Figure 1Velocity-time graph for uniformly accelerated motion (area = displacement)

What this figure shows. A Cartesian graph with time t (s) on the horizontal axis and velocity v (m/s) on the vertical axis. A straight line rises from the point (0, u) on the vertical axis to a point (T, v_T) at time T, reflecting constant (uniform) acceleration equal to the line's slope. The trapezoidal region bounded by this line above, the time axis below, the vertical line t = 0 on the left, and the vertical line t = T on the right is shaded, with a caption 'shaded area = displacement covered in time T = (1/2)(u + v_T) x T'. The slope of the line is separately annotated as 'slope = (v_T - u)/T = a …