Question

A particle moves along a circle with speed given by v = 3t2 m/s, where t is time in seconds. Calculate ratio of time taken for traversing second complete revolution to first complete revolution.

Tell correct answer pls fast

Answer 1

Answer:

this is the correct answer

Answer 2

Given:

A particle moves along a circle with speed given by v = 3t² m/s, where t is time in seconds.

To find:

Ratio of time taken for traversing second complete revolution to first complete revolution.

Calculation:

• Let time for 1st revolution be t ,
• Time for 2nd revolution be t to t_(2):

For 1st complete revolution:

$\therefore \: v = 3 {t}^{2}$

$\implies \: \dfrac{ds}{dt} = 3 {t}^{2}$

$\implies \: ds = 3 {t}^{2} \: dt$

$\displaystyle \implies \: \int_{0}^{s} ds = 3 \int_{0}^{t} {t}^{2} \: dt$

$\displaystyle \implies \: s = {t}^{3}$

$\boxed{ \displaystyle \implies \: t = {s}^{ \frac{1}{3} } }$

For 2nd complete revolution:

$\therefore \: v = 3 {t}^{2}$

$\implies \: \dfrac{ds}{dt} = 3 {t}^{2}$

$\implies \: ds = 3 {t}^{2} \: dt$

$\displaystyle \implies \: \int_{s}^{2s} ds = 3 \int_{t}^{t_{2}} {t}^{2} \: dt$

$\displaystyle \implies \: 2s - s = { (t_{2}) }^{3} - {t}^{3}$

$\displaystyle \implies \: s= { (t_{2}) }^{3} - {t}^{3}$

$\displaystyle \implies \: s= { (t_{2}) }^{3} - { ({s}^{ \frac{1}{3} }) }^{3}$

$\displaystyle \implies \: s= { (t_{2}) }^{3} - s$

$\displaystyle \implies \: { (t_{2}) }^{3} = 2s$

$\displaystyle \implies \: t_{2} = {(2s)}^{ \frac{1}{3} }$

$\displaystyle \implies \: t_{2}-t = {(2s)}^{ \frac{1}{3} }-{s}^{\frac{1}{3}}$

$\displaystyle \implies \: t_{2}-t = {s}^{ \frac{1}{3} }({2}^{\frac{1}{3}}-1)$

So, required ratio :

$\therefore \: (t_{2}-t) : t = {(s)}^{ \frac{1}{3} }(2^{\frac{1}{3}}-1) : {s}^{ \frac{1}{3} }$

$\implies \: (t_{2}-t) : t = ({2}^{ \frac{1}{3} } -1 ): 1$

So, final answer is:

$\boxed{ \bold{\: (t_{2} -t): t = ({2}^{ \frac{1}{3} } -1) : 1}}$