Motion with constant acceleration

Motion with constant acceleration occurs in everyday life whenever an object is dropped: the object moves downward with the constant acceleration $9.81 {\rm m s^{-2}}$, under the influence of gravity.

Fig. 8 shows the graphs of displacement versus time and velocity versus time for a body moving with constant acceleration. It can be seen that the displacement-time graph consists of a curved-line whose gradient (slope) is increasing in time. This line can be represented algebraically as

$$x = x_0 + v_0 t + \frac{1}{2} a t^2.$$ (19)

Here, $x_0$ is the displacement at time $t=0$: this quantity can be determined from the graph as the intercept of the curved-line with the $x$-axis. Likewise, $v_0$ is the body's instantaneous velocity at time $t=0$.

Figure 8: Graphs of displacement versus time and velocity versus time for a body moving with constant acceleration

The velocity-time graph consists of a straight-line which can be represented algebraically as

$$v = \frac{dx}{dt}= v_0 + a t.$$ (20)

The quantity $v_0$ is determined from the graph as the intercept of the straight-line with the $x$-axis. The quantity $a$ is the constant acceleration: this can be determined graphically as the gradient of the straight-line (i.e., the ratio ${\mit\Delta}v/{\mit\Delta} t$, as shown). Note that $dv/dt=a$, as expected.

Equations (19) and (20) can be rearranged to give the following set of three useful formulae which characterize motion with constant acceleration:

$$s = v_0 t + \frac{1}{2} a t^2,$$ (21)

$$v = v_0 + a t,$$ (22)

$$v^2 = v_0^{ 2} + 2 a s.$$ (23)

Here, $s=x-x_0$ is the net distance traveled after $t$ seconds.

Fig. 9 shows a displacement versus time graph for a slightly more complicated case of accelerated motion. The body in question accelerates to the right [since the gradient (slope) of the graph is increasing in time] between times $A$ and $B$. The body then moves to the right (since $x$ is increasing in time) with a constant velocity (since the graph is a straight line) between times $B$ and $C$. Finally, the body decelerates [since the gradient (slope) of the graph is decreasing in time] between times $C$ and $D$.

Figure 9: Graph of displacement versus time