Question

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).

Short answer

1. The average velocity of the proton during the first $3.0\text{s}$of its motion is $80\text{m/s}$.
2. The instantaneous velocity of the proton at $t=3.0\text{s}$is $110\text{m/s}$.
3. The instantaneous acceleration of the proton at $t=3.0\text{s}$ is $20{\text{m/s}}^2$
4. From graph,$x\left(3.0\text{s}\right)=240\text{m,}x\left(0\text{s}\right)=0\text{m}$,Therefore, $v_{avg}=80\text{m/s}$
5. The instantaneous velocity at $t=3.0\text{s}$ is $110\text{m/s}$.
6. The instantaneous acceleration from the graph is $20.57{\text{m/s}}^2$.

Step-by-step solution

To understand concept

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

Understanding the average velocity and instantaneous velocity.

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)

(a) Determination of the average velocity of the proton during the first 3.0s of its motion

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 $80\text{m/s}$.

(b) Determination of the instantaneous velocity of the proton at  t=3.0 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 $110\text{m/s}$.

(c) Determination of the instantaneous acceleration of the proton at  t=3.0 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 $20{\text{m/s}}^2$.

(d) Determination of average velocity from the graph. 

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,

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

(e) Indicate the instantaneous velocity of the proton at t=3.0 s  on the plot  

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.

(f) Plot the graph of V vs t

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