Question

$A \: body \: moves \: in \: a \: \\ straight \: line \: under \: the \: \\ retardation \: 'a' \: which \: is \: given \\ \: by \: a = b {v}^{2}. If \: the \: initial \: \\ velocity \: is \: 2m /s , \: then \: what \: is \: \\ the \: distance \: covered \: in \: \\ time \: t=2 \: seconds?$

Answer 1

Given : u = 2 m/s at t =0 s

a = - bv^2 m/s^2 _____{retardation}

t = 2 s

As we know that

a= −bv^2

=> dv/dt = - bv^2

=> dv/v^2 = -bdt

Integrating both sides , we get

=> - 1/v = - bt + C ________(1)

As we have v at t=0 or u = 2 m/s

So, - 1/2 = - b(0) + C

Or, - 1/2 = C

So, now putting the value of C in (1)

We get ;

- 1/v = - bt - 1/2

Or, 1/v = bt + 1/2

Or, v = 1/(bt + 1/2)

We also know that v = ds/dt

So, ds/dt = 1/(bt + 1/2)

Or, ds = dt/(bt + 1/2)

Or, ds = 2dt/(2bt + 1)

Integrating both sides we get ;

Or, s =ln(2bt + 1)/b + C' ________(2)

Also, at t = 0 , s = 0 m

So, to get C' we put t and s as 0 in (2)

=> 0 = ln (0 + 1)/b + C'

=> 0 = C' ____{as ln1 = 0 }

So, s = ln(2bt + 1)/b ________(3)

Now to get displacement at t = 2s

We put t = 2 in (3)

s at 2 s = ln(2b(2) + 1)/b

= ln(4b + 1)/b

So, the distance covered in time 2s is

s = 1/b . ln(4b + 1)

Answer 2

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