5,5.08,5.8,8.05,8.5 :))))
3/5n - 4/5 = 1/5n....multiply everything by 5 to get rid of the fractions
3n - 4 = n
-4 = n - 3n
-4 = -2n
-4/-2 = n
2 = n <==
(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when
and
, and because the velocity function is continuous, you need only check the sign of
for values on the intervals (0, 3) and (3, 6).
We have, for instance
and
, which means the particle is moving the positive direction for
, or the interval (3, 6).
(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:

which follows from the definition of absolute value. In particular, if
is negative, then
.
The total distance traveled is then 4 ft.
(g) Acceleration is the rate of change of velocity, so
is the derivative of
:

Compute the acceleration at
seconds:

(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)
Using quadratic formula
x = [-18 +-sq root (324 -4 *3*15)] / 2*3
x = [-18 +-sq root (144)]/6
x1 = (-18 + 12) / 6 = -1
x2 = (-18 -12) / 6 = -30/6 = -5