Question

Particles A, B, and C move along the x-axis. Particle C has an initial velocity of 10 m/s. In FIGURE P2.37, the graph for A is a position-versus-time graph; the graph for B is a velocity-versus time graph; the graph for C is an acceleration-versus-time graph. Find each particle’s velocity at t = 7.0 s.

Short answer

The velocity of particle A at $t=7s$is $-10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The velocity of particle B at $t=7s$is $-20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The velocity of particle c at$t=7s$is$95\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Step-by-step solution

Particle A

The position-versus-time graph is given for particle A.

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

The graph for particle A is a straight line from $t=5s$to $t=8s$.

Consider points $\left(t_1,x_1\right)=\left(5,0\right)$and $\left(t_2,x_2\right)=\left(8,-30\right)$

The velocity of particle A at $t=7s$is

$$\begin{gathered} v=\frac{x_2-x_1}{t_2-t_1} \\ =\frac{-30-0}{8-5} \\ =\frac{-30}{3} \\ =-10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Therefore, the velocity of particle A at$t=7s$is$-10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Particle B

The velocity-versus-time graph is given for particle B.

From the graph of a velocity-versus-time, it can be observed that the velocity of particle B is$-20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Particle C

The acceleration-versus-time graph is given for particle C.

The velocity of a particle $∆v=v_f-v_0$between $t_0$and $t_f$is the area under the curve from $t_0$to $t_f$.

Thus, the area under the acceleration - time curve is

$∆v=$area of the rectangle between $t=0s$and role="math" localid="1648480759533" $t=2s+$area of the triangle between $t=2s$and role="math" localid="1648480812177" $t=5s-$area of the triangle between $t=5s$and $t=7s$

$$\begin{gathered} ∆v=\left(2s\right)\left(30\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right)+\frac{1}{2}\left(3s\right)\left(30\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right)-\frac{1}{2}\left(2s\right)\left(20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right) \\ =60+45-20 \\ =60+25 \\ =85\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Now,

The velocity of a particle at $t=7s$is

$v_f=v_0+$area under the acceleration-time curve

$$\begin{gathered} v_f=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.+85\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ =95\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Therefore, the velocity of particle C is $95\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.