Question

find avg velocity of given data between 1s to 3s : ​

Answer 1

Explanation:

plz refer to the attachment

Answer 2

Answer:

• The average velocity (v) of the data is 20 i + 6 j

Given:

1. Given Equation $\displaystyle \sf \overrightarrow{\sf r} = 5\;t^{2}\;\hat{i}\;+\;6\;t\;\hat{j}\;+\;7\;\hat{k}$

Explanation:

$\rule{300}{1.5}$

Firstly finding the Distance travelled at t = 1 sec.

$\displaystyle \sf \overrightarrow{\sf r} = 5\;t^{2}\;\hat{i}\;+\;6\;t\;\hat{j}\;+\;7\;\hat{k}$

Substituting,

$\displaystyle \Rightarrow\sf r_{1} = 5 \times(1)^{2}\;\hat{i}\;+\;6\times(1)\;\hat{j}\;+\;7\;\hat{k}\\\\\\\Rightarrow\sf r_{1}=5\;\hat{i}\;+\;6\;\hat{j}\;+\;7\;\hat{k}\\\\\\\Rightarrow\sf r_{1}=5\;\hat{i}\;+\;6\;\hat{j}\;+\;7\;\hat{k}\quad\dfrac{\quad}{}\;(1)$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Finding the Distance travelled at t=3 sec.

$\displaystyle \sf \overrightarrow{\sf r} = 5\;t^{2}\;\hat{i}\;+\;6\;t\;\hat{j}\;+\;7\;\hat{k}$

Substituting,

$\displaystyle \Rightarrow\sf r_{2} = 5 \times(3)^{2}\;\hat{i}\;+\;6\times(3)\;\hat{j}\;+\;7\;\hat{k}\\\\\\\Rightarrow\sf r_{2}=5\times 9\;\hat{i}\;+\;6\times3\;\hat{j}\;+\;7\;\hat{k}\\\\\\\Rightarrow\sf r_{2}=45\;\hat{i}\;+\;18\;\hat{j}\;+\;7\;\hat{k}\\\\\\\Rightarrow\sf r_{2}=45\;\hat{i}\;+\;18\;\hat{j}\;+\;7\;\hat{k}\quad\dfrac{\quad}{}\;(2)$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the Formula we know,

$\Rightarrow\sf Average\;velocity=\dfrac{Final\;disp.-Initial\;Disp.}{Total\;time\;taken}$

i.e. $\sf v=\dfrac{r_{2}-r_{1}}{t_{2}-t_{1}}$

Substituting the values,

$\displaystyle\Rightarrow\sf v=\dfrac{45\;\hat{i}\;+\;18\;\hat{j}\;+\;7\;\hat{k}-(5\;\hat{i}\;+\;6\;\hat{j}\;+\;7\;\hat{k})}{3-1}\\\\\\\Rightarrow\sf v=\dfrac{45\;\hat{i}\;+\;18\;\hat{j}\;+\;7\;\hat{k}-5\;\hat{i}\;-\;6\;\hat{j}\;-\;7\;\hat{k}}{3-1}\\\\\\\Rightarrow\sf v=\dfrac{(45-5)\;\hat{i}\;+\;(18-6)\;\hat{j}\;+\;(7-7)\;\hat{k}}{3-1}\\\\\\\Rightarrow\sf v=\dfrac{40\;\hat{i}\;+\;12\;\hat{j}\;+\;0\;\hat{k}}{2}\\\\\\\Rightarrow\sf v=\dfrac{2\times(20\;\hat{i}\;+\;6\;\hat{j})}{2}$

$\displaystyle\Rightarrow\sf v=20\;\hat{i}\;+\;6\;\hat{j}\\\\\\\Rightarrow\sf \large{\underline{\boxed{\red{\sf v=20\;\hat{i}\;+\;6\;\hat{j}\;m/s}}}}$

∴ The average velocity (v) of the data is 20 i + 6 j.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Finding the magnitude of the velocity.

$\displaystyle\Rightarrow\sf |v|=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

Substituting the values,

$\displaystyle\Rightarrow\sf |v|=\sqrt{(20)^{2}+(6)^{2}}\\\\\\\Rightarrow\sf |v|=\sqrt{400+36}\\\\\\\Rightarrow\sf |v|=\sqrt{436}\\\\\\\Rightarrow\sf |v|=20.88\\\\\\\Rightarrow \large{\underline{\boxed{\red{\sf |v|=20.88\;m/s}}}}$

∴ The magnitude of average velocity is 20.88 m/s.

$\rule{300}{1.5}$