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Physics · Ch 2 — Motion in a Straight Line

Kinematic Equations for Uniformly Accelerated Motion

2.8

Kinematic Equations for Uniformly Accelerated Motion

For the common and important special case of uniformly accelerated motion — motion in which the acceleration aa has one constant value throughout the interval considered — three standard relations connect initial velocity uu, final velocity vv, displacement ss, acceleration aa and time tt. As required by the WBCHSE syllabus, we derive all three both graphically (from the v-t graph of Section 2.7) and by calculus (by integrating the defining relations a=dv/dta = dv/dt and v=dx/dtv = dx/dt).

First equation: v=u+atv = u + at

Graphical derivation. On a v-t graph, a body with constant acceleration aa has a straight-line graph of constant slope aa, starting at v=uv = u when t=0t = 0. The slope of a straight line equals (rise)/(run), so

a=v−ut−0⟹v=u+ata = \frac{v - u}{t - 0} \quad\Longrightarrow\quad v = u + at

Calculus derivation. Since aa is constant, a=dv/dta = dv/dt can be directly integrated with respect to time from 00 to tt (with velocity going from uu to vv):

∫uvdv=∫0ta dt⟹v−u=at⟹v=u+at\int_u^v dv = \int_0^t a\,dt \quad\Longrightarrow\quad v - u = at \quad\Longrightarrow\quad v = u + at

Second equation: s=ut+12at2s = ut + \tfrac{1}{2}at^2

Graphical derivation. From Section 2.7, displacement equals the area under the v-t graph — here a trapezium with parallel sides uu and v=u+atv = u+at, and width tt:

s=12(u+v) t=12(u+u+at)t=ut+12at2s = \frac{1}{2}(u+v)\,t = \frac{1}{2}\big(u + u + at\big)t = ut + \frac{1}{2}at^2

Calculus derivation. Since v=dx/dtv = dx/dt and, from the first equation, v(t)=u+atv(t) = u + at, we integrate position from x0x_0 (taken as the origin, s=x−x0s = x - x_0) as tt runs from 00 to tt:

s=∫0tv dt=∫0t(u+at) dt=ut+12at2s = \int_0^t v\,dt = \int_0^t (u + at)\,dt = ut + \frac{1}{2}at^2

This is also, as noted in Section 2.6, exactly why the position-time graph of uniformly accelerated motion is a parabola: ss is a quadratic function of tt.

Third equation: v2=u2+2asv^2 = u^2 + 2as

Graphical derivation. Eliminate tt between the first two equations: from v=u+atv = u + at we get t=(v−u)/at = (v-u)/a; substituting into s=ut+12at2s = ut + \tfrac{1}{2}at^2 and simplifying algebraically gives v2=u2+2asv^2 = u^2 + 2as.

Calculus derivation (chain rule). Using a=dv/dta = dv/dt and the chain rule dvdt=dvdxdxdt=vdvdx\dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}, we can write the constant acceleration as a=v dv/dxa = v\,dv/dx, i.e. a dx=v dva\,dx = v\,dv. Integrating both sides — position from 00 to ss, and velocity from uu to vv:

∫0sa dx=∫uvv dv⟹as=v2−u22⟹v2=u2+2as\int_0^s a\,dx = \int_u^v v\,dv \quad\Longrightarrow\quad as = \frac{v^2-u^2}{2} \quad\Longrightarrow\quad v^2 = u^2 + 2as

(This chain-rule derivation is worked out in full generality in Numerical 4.) …

Table 1Worked position, velocity values from the kinematic equations (u = 0, a = 4 m/s²)
t (s)v = u + at (m/s)s = ut + ½at² (m)
000
142
288
31218