Question

t = √x + 3

Find the value of displacement when velocity is zero​

Answer 1

★ Answer :

t = √x + 3

√x = t - 3

→ Squaring both sides

$\rm{\implies (\sqrt{x})^2 = (t - 3)^2} \\ \\ \bf{We \: know \: that,} \\ \\ \Large{\star{\boxed{\tt{(a - b)^2 = a^2 + b^2 - 2ab}}}} \\ \\ \rm{\implies x = t^2 + 9 - 2(1t)(3)} \\ \\ \rm{\implies x = t^2 - 6t + 9 .......(1)} \\ \\ \bf{Now,}$

$\rm{\implies v = \frac{dx}{dt} = \frac{d(t^2 - 6t + 9)}{dt}}$

$\rm{\implies v = \frac{d(t^2)}{dt} - \frac{6 \: d(t)}{dt} + \frac{d(9)}{dt}}$

$\rm{\implies v = 2t - 6(1) + 0} \\ \\ \rm{\implies v = 2t - 6}$

$\bf{v = 0 \: \: \: \: (Given)} \\ \\ \sf{\implies 0 = 2t - 6} \\\ \\ \rm{\implies t = 3} \\ \\ \Large{\star{\boxed{\sf{t = 3}}}}$

$\rule{200}{2}$

Put value of t in equation (1).

$\rm{\implies x = (3)^2 + 9 - 6(3)} \\ \\ \rm{\implies x = 9 + 9 - 18} \\ \\ \rm{\implies x = 18 - 18} \\ \\ \rm{\implies x = 0} \\ \\ \Large{\star{\boxed{\tt{x = 0}}}}$

Answer 2

t = √x+3

√x = t-3

x = t² - 6t + 9  

dx/dt is the velocity

dx/dt = 2t -6

v = 0

2t-6 = 0  

which will give us 3 so t = 3

at t = 3

x = 3-3

so there will be 0 displacement.