Q104P - Fundamentals Of Physics

Question

Question: A proton moves along the  axis according to the equation x = 50t + 10t² , where x is in meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3.0 s of 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 x versus t and indicate how the answer to (a) can be obtained from the plot. (e) Indicate the answer to (b) on the graph. (f) Plot v versus t and indicate on it the answer to (c).

Step-by-Step Solution

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 $20.57{\text{m/s}}^2$ .

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 . 0 s of its motion

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

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

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

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

 

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

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

 

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,

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

 

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, 

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

 

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$20.57 {\text{m/s}}^2$  .