a particle moves along a straigth line so that its acceleration at any given time t?

Question

is a(t)=2t -12 (in units of ft and sec)
If the initial velocity of the particle is 27 ft/ sec, at what time t during the interval 0 less than or equal to t and 8 is greater than or equal to t (in other words 0

Answer 1

a=2t-12, so, v, which is the integral of a w.r.t., is

v = t^2-12t+c, where c is the constant of integration.

You are given that the initial velocity is 27f.p.s, meaning that

when t=0, then v=27, so substitute these values to give:

27=0^2-12*0+c, so,

c=27, and

v=t^2-12t+27.

Particle changes direction when t=0, so

t^2-12t+27=0,so

[t-3][t-9]=0, so

t=3 OR t=9 but t=9 is inadmissible, so

the particle changes direction after 3 seconds.

Answer 2

v = t^2 - 12t + 27
t^2 - 12t + 27 = 0
(t - 9)(t - 3) = 0
t = 3, 9
9 lies outside the permitted interval , so
t = 3 sec.