Question

FIGURE EX2.5 shows the position graph of a particle. a. Draw the particle’s velocity graph for the interval 0 s … t … 4 s. b. Does this particle have a turning point or points? If so, at what time or times?

Short answer

Part (a)

The particle’s velocity graph for the interval $0\le t\le 4$is shown below

Part (b)

The particle does not have turning points.

Step-by-step solution

Given information

The position-time graph of the particle is shown below

Part (a)

The velocity is the slope of the position-time graph.

Consider points $\left(t_2,x_2\right)=\left(1,10\right)$and $\left(t_1,x_1\right)=\left(0,0\right)$.

The velocity of the particle between $t=0s$and $t=1s$is

$$\begin{gathered} v=\frac{x_2-x_1}{t_2-t_1} \\ =\frac{10-0}{1-0} \\ =10m/s \end{gathered}$$

Consider points $\left(t_3,x_3\right)=\left(3,10\right)$and $\left(t_2,x_2\right)=\left(1,10\right)$.

The velocity of the particle between $t=1s$and $t=3s$is

$$\begin{gathered} v=\frac{x_3-x_2}{t_3-t_2} \\ =\frac{10-10}{3-1} \\ =0m/s \end{gathered}$$

Consider points $\left(t_4,x_4\right)=\left(4,20\right)$and $\left(t_3,x_3\right)=\left(3,10\right)$$\left(t_4,x_4\right)=\left(4,20\right)$.

The velocity of the particle between $t=3s$and$t=4s$

$$\begin{gathered} v=\frac{x_4-x_3}{t_4-t_3} \\ =\frac{20-10}{4-3} \\ =10m/s \end{gathered}$$

With the help of the above-obtained information, we can draw a graph of velocity versus time as follows

Part (b)

The turning point is a point in motion where a particle reverses direction. Here, there are no turning points as the particle is not changing its direction.