Question

The displacement X of a particle moving in one dimension ,the actuon of a constant force is related to time t by the eqn
t=√x +3 where x is in metre and t in sec . calculate a ) The displacement of the particle when its velocity is 0
b) the work done by the force in first 6 s​

Answer 1

$\huge\mathfrak{solution}$

Given :

• t= √x +3

To find :

• a) Displacement of the particle when v=0.

• b) The work done by the force in first 6 seconds .

Answer:

Here .

• t= √x+3

=> t-3=√x

now, square both sides we get .

=> (t-3)^2 = x --------- (*)

Now , we have to find velocity .

As we know that :

$\boxed{ \bold{v \: = \frac{dx}{dt} }}$

$\implies \: v \: = \frac{d}{dt} {(t - 3)}^{2} \\ \\ \implies \: v = 2(t - 3) \: \: \: \boxed{ \bold{ \frac{d}{dx} {x}^{n} = n {x}^{n - 1} }}$

a) In this situation , it is given that we have to take v= 0

$\implies \: 0 = 2(t - 3) \\ \\ \implies \: t - 3 = 0 \\ \\ \implies \: t = 3$

We get t=3s, put in in equation (*) to find displacement:

$\implies \: x = {(3 - 3)}^{2} \\ \\ \implies \: x = 0$

Hence , when v=0 then displacement is also 0

b) Now , find velocity when t = 0s and t=6s

$\star \: v_{t = 0} = 2(0 - 3) = - 6 \: \frac{m}{s} \\ \\ \star \: v_{t = 6} \: = 2(6 - 3) = 2(3) = 6 \frac{m}{s}$

Now , Use Work-Energy Theorem:

•W= ∆KE

$\implies \: W = \frac{1}{2} m( {v_f}^{2} - {v_i ^{2} } ) \\ \\ \implies \: W = \frac{1}{2} m( {6}^{2} - ( { - 6)}^{2} ) = \frac{1}{2} m (36 - 36) \\ \\ \implies \: W = \frac{1}{2} m(0) = 0$

Work done by the force in first 6sec is zero ..

Answer 2

$</p><p>\huge{\pink{\tt{Answer \: - }}}$

$</p><p>{\blue{\underline{\tt{Displacement \: = \: 0 \: Metres .}}}}$

$</p><p>\red{\tt{\underline{Work\: Done \: = 0. }}}$

T = √X + 3

√X = T - 3

$\huge{\fbox{\fbox{\therefore{\mathfrak {\green{X = T^2 - 6T + 9 }}}}}}$

$\huge{\fbox{\fbox{\mathfrak {\red{V = dx / dt }}}}}$

V = 0

=> T = 3 Seconds

$\huge{\fbox{\fbox{\mathfrak {\orange{Displacement= 3^2 + 9 - 6×3 = 0 \:Metres. }}}}}</p><p>$

WORK DONE :

$\huge{\fbox{\fbox{\mathfrak {\red{V = u + at }}}}}$

T = 0 Seconds

$\huge{\fbox{\fbox{\therefore{\mathfrak {\green{v =-6 \: m \ s.}}}}}}$

$\huge{\fbox{\fbox{\mathfrak {\red{a = dv / dt = 2}}}}}$

$\huge{\fbox{\fbox{\therefore{\mathfrak {\green{Final \: Velocity =-6 \: metre/sec.}}}}}}$

$\huge{\fbox{\fbox{\mathfrak {\red{Work \: Done = 1/2 × m ( 6 ) ^2 - 1/2 × m ( - 6 ) ^2 = 0 }}}}}$

$</p><p>\huge{\pink{\tt{Hope \: This \: Helps!!!!}}}$