Question: A proton moves along theaxis according to the equation x = 50t + 10t² , where xis in meters and tis in seconds. Calculate (a) the average velocity of the proton during the first 3.0 sof its motion, (b) the instantaneous velocity of the proton at t = 3.0 s, and (c) the instantaneous acceleration of the proton at t = 3.0 s . (d) Graph xversus tand indicate how the answer to (a) can be obtained from the plot. (e) Indicate the answer to (b) on the graph. (f) Plotvversus tand indicate on it the answer to (c).

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" $$\begin{gathered} x\left(3.0\text{s}\right)=50\text{m/s×3.0s+10m/s×}{\left(3.0\text{s}\right)}^2 \\ =150\text{m+90m} \\ =240\text{m} \end{gathered}$$

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

$$\begin{gathered} x\left(0\text{s}\right)=50\text{m/s×0s+10m/s×}{\left(0\right)}^2 \\ =0\text{m} \end{gathered}$$

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,

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

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$20.57{\text{m/s}}^2$.