Question

8 A body moves in a straight line under the retardation 'a' which is given by a = bv. If the initial velocity is 2 m/s, the distance covered in time t = 2 seconds is (1) blog, (4) (2) / log, (1 + 46) (3) blog, (1 - 46 14) / 10g, (46-1) ime relatin The starting from​

Answer 1

$\huge{ \underline{ \underline{ \mathbb{ \purple{SOLUTION:-}}}}}$

Retardation is given by the expression

$\bigstar \: \tt{a=−bv^{2} } \\ \\ \tt{⇒ \dfrac{dv}{dt}=−bv^2}$

Assuming that velocity changes by an amount dv in infinitesimal time interval dt. Above expression can be rewritten as

$\rightarrow \tt{ \dfrac{dv}{v^2}=−bdt}$

Integrating both sides with respective variable we get

$\tt{∫ \dfrac{dv}{v^2}=−∫b dt} \\ \\ \tt{⇒− \dfrac{1}{v}=−bt+C}$

where C is a constant of integration. To be ascertained from initial conditions.

$\underline{ \bold{At t=0 we \: have v=2 ms^{−1}}}$

We get

• −12=C

Expression for velocity becomes

$\tt{− \dfrac{1}{v}=−bt− \dfrac{1}{2}}$

$\tt{⇒ \dfrac{1}{v}= \dfrac{1+2bt}{2}}$

$\tt{⇒ \: v= \dfrac{2}{1+2bt}}$

$\tt{⇒ \dfrac{dS}{dt}= \dfrac{2}{1+2bt}}$

Assuming that displacement dS takes place in infinitesimal time interval dt. Above expression can be rewritten as

$\tt{dS= \dfrac{2}{1+2bt}dt}$

Integrating both sides with respective variables we get

$\tt{∫ dS=∫ \dfrac{ 2}{1+2bt}dt}$

$\tt{⇒S= \dfrac{ln(|1+2bt|)}{b}+C1}$

• where C1 is a constant of integration.

At t=0, S=0. Inserting in above we get C1=0 and expression for displacement becomes.

$\tt{S= \dfrac{ln(|1+2bt|)}{b}}$

Therefore, distance covered in time t=2 s

$\boxed{ \underline{ \frak{ \pink{S(2) = \dfrac{ln(1+4b)}{b}}}}}$