Question

The position coordinate of a particle which is confined to move along a straight line is given by s =2t3−24t+6, where s is measured in meters from a convenient origin and t is in seconds. Determine (a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at t = 0, (b) the acceleration of the particle when v = 30 m/s, and (c) the net displacement of the particle during the interval t = 1 s to t = 4 s.

Answer 1

a) the answer is t=4 seconds

b) acceleration is zero

c) displacement= 142 m

Step-by-step explanation:

Given the position of the particle

`s=2t^3-24t+6`

a) the time required when velocity
`v=72 m/s`

`v=72=(ds)/(st)=6t^2-24`

now we solve for time
`t`

`6t^2=72+24=96\\\\\implies t^2=(96)/(6)\\\therefore t=4 s`

b) acceleration when
`v=30 m/s`

acceleration is the time derivative of velocity i.e

`a=(dv)/(dt)=(d)/(dt)(30)=0`

c) the net displacement of the particle during the interval t = 1 s to t = 4 s is

`s_4-s_1=2t^3-24+6|_(t=4)-2t^3-24+6|_(t=1)\\s_4-s_1=126-(-16)=142 m`