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Q104P

Expert-verifiedFound in: Page 39

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: A proton moves along the ** **axis** **according to the equation x = 50t + 10t ^{2}**

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

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

**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:

${v}_{avg}=\frac{\u2206x}{\u2206t}$ … (i)

Here, $\u2206x$ is the displacement and $\u2206t$ 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" $x\left(3.0\text{s}\right)=50\text{m/s\xd73.0 s+10m/s\xd7}{\left(3.0\text{s}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=150\text{m+90m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=240\text{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

$x\left(0\text{s}\right)=50\text{m/s\xd70 s+10m/s\xd7}{\left(0\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=0\text{m}$

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

${v}_{avg}=\frac{x\left(3.0\text{s}\right)-x\left(0\text{s}\right)}{{t}_{2}-{t}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{240\text{m-0 m}}{3.0\text{s -0 s}}\phantom{\rule{0ex}{0ex}}=80\text{m/s}$

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

Using equation (ii), the instantaneous velocity is,

$v=\frac{d\left(50t+10{t}^{2}\right)}{dt}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}v=50+20t$

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

role="math" localid="1650539082653" $v=50+20\left(3.0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=50+60\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=110\text{m/s}$

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

Using equation (iii), the instantaneous acceleration is,

$a=\frac{d\left(50+20t\right)}{dt}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=20{\text{m/s}}^{2}$

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,

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

So, the average velocity is,

${v}_{avg}=\frac{240\text{m}}{3.0\text{s}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=80\text{m/s}$

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,

$\text{slope=}\frac{350-120}{4-1.9}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\approx 110\text{m/s}$

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

$\text{Slope =}\frac{132-60}{4-0.5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=20.57{\text{m/s}}^{2}$

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

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