Question

The distance x of a particle moving in one dimension, under the action of a constant force is related to time is given by the equation t = 2x +1+3 . where x is in metres and in seconds. Find the displacement of the particle when its velocity is zero. PLEASE SOLVE ​

Answer 1

0 m

Step-by-step explanation:

The question is incomplete, the complete question is attached.

Solution:

Velocity is the rate of change of displacement with respect to time. It is the ratio of change in displacement to change in time.

Give that the displacement x is given by the equation:

`t=√(x)+3\\\\√(x) =t-3\\\\squaring\ both\ sides\ to\ get:\\\\(√(x)) ^2 =(t-3)^2\\\\x=t^2-6t+9\\\\The\ velocity\ v=(dx)/(dt) ;differentiating \ through\ with\ respect\ to\ t:\\\\(dx)/(dt) =(d)/(dt) (t^2-6t+9)\\\\v=2t-6\\\\At\ v=0:\\\\0=2t-6\\\\2t=6\\\\t=3\ s\\\\substitute\ t=3\ in \ x=t^2-6t+9\ to\ find\ displacement:\\\\x=3^2-6(3)+9\\\\x=0\ m`