Question

A particle is moving along the x axis such that its position is given by x² =3t²-16t+8. What are the velocity and acceleration of the particle at t = 4s? At what time is the particle stationary?​

Answer 1

Below

Step-by-step explanation:

Usually, position is given by x(t) =.......

if x^2 = 3t^2 - 16t+ 8 then

x = ( 3t^2 - 16t + 8)^1/2 ( I do not think this is what you want....but it IS what was posted)

Velocity will be given by the derivative : (3t-8) / ( 3t^2-16t+ 8)^1/2

at t = 4 this equals an imaginary number....

so I think your question equation is incorrect

The answer above this one used x(t) = 3t^2-16t+8 as the position function and is likely the correct way to do it

then velocity = dx/dt = 6t-16 at t = 4 this is 8

the particle is stationary when this = 0 at t = 16/6 = 2.67 s (NOT 4s as posted in the other 'verified' answer !!!!!)

Acceleration = dv/dt = 6

at t= 4 the velocity would be 8 units/s

and the accel would be 6 units/s^2

and when v= 0 would occur at t = 2.67 s