Question

Plz es question ka ans bta do​

Answer 1

hey mate!

Solve the distance x of a particle moving in one dimension

Answer:

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

Answer 2

$t = \sqrt{x+3} \\x = t^{2} -3\\\frac{dx}{dt} = \frac{d}{dt} (t^{2}-3)\\v=2t=0\\t=0$

Applying t=0 in equation above,

$x = t^{2} -3= 0^{2}-3 = -3\\Displacement = 3units$