Question

a particle moves along the x axis with an acceleration of a=18t, where a has units if m/s2. if the particle at time t=0 is at the origin with a velocity of -12 m/s, what js its position at t=4.0s?

Answer 1

Answer:

 Position at t= 4 seconds is 144 m

Explanation:

 It is given that acceleration, a = 18 t, where t is the time.

 We know that Velocity, $v = \int { a} \, dt$

  Substituting value of a,

           Velocity, $v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c$

 We know that at t = 0, v = -12 m/s

         So, $9*0^2+c=-12\\ \\ c=-12m/s$

So velocity, $v = (9t^2-12)m/s$

  We also know that displacement, $x = \int { v} \, dt$

     Substituting value of v,  

        Displacement, $x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c$

  We know that at t = 0, particle is at origin, x =0.

               So,  $0=3*0^3-12*0+c\\ \\ c=0$

   Displacement, $x = 3t^3-12t$

At t = 4 seconds

   $x = 3*4^3-12*4=192--48=144m$

Position at t= 4 seconds is 144 m