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

Answer:

a) the answer is t=4 seconds

b) acceleration is zero

c) displacement= 142 m

Explanation:

Given the position of the particle

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

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

$v=72=\frac{ds}{st}=6t^2-24$

now we solve for time $t$

$6t^2=72+24=96\\\\\implies t^2=\frac{96}{6}\\\therefore t=4 s$

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

acceleration is the time derivative of velocity i.e

$a=\frac{dv}{dt}=\frac{d}{dt}(30)=0$

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

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