Particle position?? ?

Question

A particle's position on the x axis is given by x(t)= -6t^2 + 2t, where x is in meters and t is in seconds. Find the initial velocity of the particle and its maximum displacement.

(i got the derivative of the formula and kinda think the initial velocity is 2m/s if that helps anything)

Thanks in advance!!!!

Answer 1

x=-6*t^2+2*t  "position"

x'=-12t+2     "velocity"

when t=0, x'=2 m/s

x has its maximum +value when d(x)/dt=0=x'

0=-12t+2

solve for t

t=-2/-12 = 1/6 s

Xmax=-6*(1/6)^2+2*(1/6) = 0.166 m

Answer 2

Velocity is derivative of displacement with.respect.to (w.r.t henceforth) time so V= -12t+2

At t=0 V= 2m/s ( you found this correct).

I hope you know enough calculus for the other part, To find the maximum or minimum for any function you find its derivative and put it equal to zero.

For max displacement derivative of displacement w.r.t Time =0

this would give you -12t+2=0 that is at t=1/6 seconds the displacement would be maximum.

How did i knew it would be maximum?

Well you find the derivative again and put this value of t in it  if result is negative then this value of t would give you maximum if its positive it would give you minimum value. This is true for any function

So just plug in this value of t in the displacement equation above you will get the result.

Answer 3

if you differentiate x(t) you will get v(t)

v(t)=-12t+2

the initial velocity occurs when t=0, so v(0)=2

the max displacement occurs when x(t) is a maximum, which occurs when its derivative is equal to zero

we already know the derivative of x(t) is -12t+2 which is zero when

t=1/6, so t=1/6 is when the max displacement occurs, now just substitute t=1/6 into x(t) for the value of that displacement