The distance x of a particle moving in 1 dimension under the action of a constant force is related to time t by equation t = root over x + 3 where x is in metres and t is in seconds . Find the displacement when it's velocity is zero.

I don't undrstand the root is over x + 3 or just x, so i will solve both cases.

t = √x + 3

on squaring

√x = t - 3

x = t^{2} - 6t + 9

Now differentiate x wrt t, dx/dt is velocity

dx/dt = 2t - 6

velocity v = 2t - 6

v = 0

2t - 6 = 0

that gives t = 3

at t = 3 ,

x = 3 - 3

0 displacement

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