Aparticle moves along a straight line and its position at time t is given by s(t)=2t3−27t2+84t where s is measured in feet and t in seconds. a) find the velocity (in ft/sec) of the particle at time t=0: 84 b) the particle stops moving twice, once when t=a and again when t=b where a

a)

b)

c)

d) Δs=4428ft

Explanation:

From the exercise we know the equation of position

a) To calculate the velocity we need to derivate the equation of position

So, v(0) is:

b) To find the two times where the particle stops A and B we need to solve the quadratic equation:

Solving for t

and

c)

d) Δs=4428ft-0ft=4428ft