Question

The acceleration 'a' for a particle depends on displacement s
as a = 5 + s. At t=0,5 = 0 and
velocity v = 5. Then the velocity V, corresponding to displacement s is given by
(a) v = 5+s
(b) V= under root(5+s ).
(c) v = under root(s square+10s)
(d) v = s - 5​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

• The acceleration 'a' for a particle depends on displacement s as a = 5 + s.

• At t = 0 , s = 0 and velocity v = 5

TO DETERMINE

The velocity v corresponding to displacement s

CALCULATION

$\sf{Acceleration \: = \: a = 5+ s}$

$\displaystyle \sf{v \frac{dv}{ds} = 5 + s \: \: }$

$\implies \: \displaystyle \sf{v dv = 5ds + s. ds\: \: }$

On integration

$\displaystyle \sf{ \frac{ {v}^{2} }{2} = 5s + \frac{ {s}^{2} }{2} + \frac{c}{2} } \: \: \: \: ( \: where \: \frac{c}{2} \: is \: constant \: )$

$\implies \: \displaystyle \sf{ {v}^{2} = 10s + {s}^{2} + c } \: \: \: \: ....(1)$

Now At t = 0 , s = 0 and velocity v = 5

So

$\sf{ \: 25 = 0 + 0 + c \: }$

$\implies \: \sf{ c = 25\: \: }$

From Equation (1)

$\: \displaystyle \sf{ {v}^{2} = 10s + {s}^{2} + 25 } \: \: \: \:$

$\implies \: \: \displaystyle \sf{ {v}^{2} = {s}^{2} + 10s + 25 } \: \: \: \:$

$\implies \: \: \displaystyle \sf{ {v}^{2} = {s}^{2} + 2 \times s \times 5 + 25 } \: \: \: \:$

$\implies \: \: \displaystyle \sf{ {v}^{2} = {(s + 5)}^{2} } \: \: \: \:$

$\implies \: \: \displaystyle \sf{v = s + 5 } \: \: \: \:$

$\implies \: \: \displaystyle \sf{v =5 + s } \: \: \: \:$

RESULT

Hence the velocity is given by

$\boxed{ \: \: \displaystyle \sf{v =5 + s } \: \: \: \: }$

Answer 2

The correct option is A

Here $a=s+5$

So, $\frac{dv}{dt}=s+5$

or $\frac{dv}{ds} \frac{ds}{dt} =s+5$

or $\frac{dv}{ds}v=s+5(as \frac{ds}{dt} =v)$

Or  $vdv=(s+5)ds$

Integrating both sides, $\int\limits vdv=\int\limits (s+5)ds$

or $\frac{v^2}{2}=\fracs^2}{2}+5s+C$ where C is integrating constant

when $S=0, v=5, C=\frac{25}{2}$

so, $\frac{v^2}{2}=\frac{s^2}{2}+5s+\frac{25}{2} or v^2=s^2+10s+25$

or $v^2=(s+5)^2$

Or $v=s+5$

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