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A body moves in a straight line under the retardation a which is given by a= -bv^2 . if the initial velocity is 2m/s the distance covered in t =2 s is?

A body moves in a straight line under the retardation a which is given by a= -bv^2 . if the initial velocity is 2m/s the distance covered in t =2 s is?

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2 Answers

Nasrin
18 Points
6 years ago
a=-bv^2, dv/dt =-bv^2 dv/v^2(limit 2 to v) =-b dt (limit 0 to 2)now integrating both sides we get, - 1/v+1/2=-2b, v=2/(1+4b)V=ds/dt,ds/dt=2/(1+4b),ds(limit 0 to s) =2/(1+4b) [limit 0 to 2] again integrating both sides s=ln(1+4b)/2b[limit 0 to 2],s=1/b[ln (1+4b)]So distance covered s=1/b[ln(1+4b)]
A08
11 Points
6 years ago
Retardation is given by the expression
a=−bv2
=>dv/dt =−bv2
Assuming that velocity changes by an amount dv in infinitesimal time interval dt. Above expression can be rewritten as
dv/v2 =−b dt
Integrating both sides with respective variable  we get
∫ (dv)/v2 =−∫ b dt
=>−1/v=−bt+C
where C is a constant of integration. To be ascertained from initial conditions. At t=0 we have v=2 ms-1
We get
−1/2=C
Velocity becomes
−1/v=−bt−1/2
=>1/v=(1+2bt)/2
=>v=2/(1+2bt)
=>dS/dt=2/(1+2bt)
Assuming that displacement dS takes place in infinitesimal time interval dt. Above expression can be rewritten as
dS=2/(1+2bt)dt
Integrating both sides with respective variables we get
∫ dS=∫ 2/(1+2bt)dt
=>S=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 displacement becomes.
S=ln(|1+2bt|)/b
Therefore, distance covered in time t=2 s
S(2)=[ln(1+4b)] /b

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