FIGURE EX2.11 shows the velocity graph of a particle moving along the x-axis. Its initial position is x0 = 2.0 m at t0 = 0 s. At t = 2.0 s, what are the particle’s (a) position, (b) velocity, and (c) acceleration?

The initial position of the particle $x_0=2m$

Initial time is$t_0=0s$

The displacement of a particle $∆x=x_f-x_0$between $t_0$and $t_f$is the area under the curve from $t_0=0$to $t_f=2s$. In this case, the area is equal to sum of triangle and rectangle as follows.

Thus, the area under velocity - time curve is

$∆x=$area of the triangle and rectangle $t=0s$and $t_f=2s$.

localid="1648454664157" role="math" $$\begin{gathered} ∆x=\frac{1}{2}\times 2s\times 4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.+2s\times 2m/s \\ =8m \end{gathered}$$

Now,

The position of a particle at $t=2s$is given by

$x_f=x_0+$area under the velocity curve $v_s$between $t_0$and $t_f$

localid="1648454683214" role="math" $$\begin{gathered} x_f=2m+8m \\ =10m \end{gathered}$$

Therefore, the position of a particle at $t=2s$is localid="1648454696315" role="math" $10m$.

From the given velocity-versus-time graph it can be observed that $t=2s$corresponds to $v_s=2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Therefore, the velocity of a particle at$t=2s$is$2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Acceleration is given by the slope of the velocity-versus-time graph.

Consider two points on the graph $\left(t_1,v_1\right)=\left(0,6\right)$and $\left(t_2,v_2\right)=\left(3,0\right)$.

The acceleration of a particle at$t=2s$is

$$\begin{gathered} a=\frac{v_2-v_1}{t_2-t_1} \\ =\frac{0-6}{3-0} \\ =-2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Therefore an acceleration of a particle at$t=2s$is$-2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.