91Ó°ÊÓ

The equation for the motion of the proton is, $x=50t+10t^2$

Average velocity is the ratio of total displacement to the total time interval. Instantaneous velocity is the velocity of a moving object at a specific moment.

The expression for the average velocity is given as follows:

$v_{avg}=\frac{∆x}{∆t}$ … (i)

Here, $∆x$is the displacement and $∆t$ is the time duration.

The expression for the instantaneous velocity is given as follows:

$v=\frac{dx}{dt}$ … (ii)

The expression for the instantaneous acceleration is given as follows:

$a=\frac{dv}{dt}$ … (iii)

Position of the proton at $t=3.0\text{s}$is,

localid="1660821352042" $$\begin{gathered} x\left(3.0\text{s}\right)=50\text{³¾/²õ×3.0²õ+10³¾/²õ×}{\left(3.0\text{s}\right)}^2 \\ =150\text{m+90m} \\ =240\text{m} \end{gathered}$$

Position of the proton at $t=0\text{s}$ is,

$$\begin{gathered} x\left(0\text{s}\right)=50\text{³¾/²õ×0²õ+10³¾/²õ×}{\left(0\right)}^2 \\ =0\text{m} \end{gathered}$$

Using equation (i), the average velocity is calculated as follows:

$$\begin{gathered} v_{avg}=\frac{x\left(3.0\text{s}\right)-x\left(0\text{s}\right)}{t_2-t_1} \\ =\frac{240\text{m-0m}}{3.0\text{s-0s}} \\ =80\text{m/s} \end{gathered}$$

Thus, the average velocity of the proton is $80\text{m/s}$.

Using equation (ii), the instantaneous velocity is,

$$\begin{gathered} v=\frac{d\left(50t+10t^2\right)}{dt} \\ v=50+20t \end{gathered}$$

The instantaneous velocity at $t=3.0\text{s}$is,

role="math" localid="1650539082653" $$\begin{gathered} v=50+20\left(3.0\right) \\ =50+60 \\ =110\text{m/s} \end{gathered}$$

Thus, the instantaneous velocity at time$t=3.0\text{s}$ is $110\text{m/s}$.

Using equation (iii), the instantaneous acceleration is,

$$\begin{gathered} a=\frac{d\left(50+20t\right)}{dt} \\ =20{\text{m/s}}^2 \end{gathered}$$

Thus, the instantaneous acceleration at $t=3.0\text{s}$ is $20{\text{m/s}}^2$.

The graph x vs t is plotted below.

From the graph,

The positions at 3.0 s and 0 s are,

$x\left(3.0\text{s}\right)=240\text{mandx}\left(0\text{s}\right)\text{=0m}$

So, the average velocity is,

$$\begin{gathered} v_{avg}=\frac{240\text{m}}{3.0\text{s}} \\ =80\text{m/s} \end{gathered}$$

The instantaneous velocity at$t=3.0\text{s}$is the slope of the triangle drawn at that point in the above graph and it is,

$$\begin{gathered} \text{slope=}\frac{350-120}{4-1.9} \\ ≈110\text{m/s} \end{gathered}$$

Thus, the instantaneous velocity at 3.0 s is 110 m/s.

The graph of$v$vs$t$is as below,

From the graph, the instantaneous acceleration is the slope of the graph

$$\begin{gathered} \text{Slope=}\frac{132-60}{4-0.5} \\ =20.57{\text{m/s}}^2 \end{gathered}$$

Thus, the instantaneous acceleration from the graph is$20.57{\text{m/s}}^2$.