Question

10 POINTS!!! FULL ANSWER WITH FULL STEP BY STEP SOLUTION PLEASE

Answer 1

A
The velocity is found by differentiating the given cubic.
x'(t) = 2t^3 - 15t^2 + 24t      This is the distance
x'(t) = 6t^2 - 20t + 24           This is the velocity (first derivative) 
x"(t) = 12t - 20                      This is the acceleration (second derivative)
x'(2) = 6(2)^2 - 20*2 + 24 
x'(2) = 24 - 40 + 24
x'(2) = 48 - 40
x'(2) = 8 units per second. 

x"(2) = 12(2) - 20
x"(2) = 24 - 20
x"(2) = 4 units per second^2


B
at t = 2 the acceleration is positive. That's a pretty good indication that the velocity is increasing.

C
When x(t) is minus, then the particle is moving left.
At t = 0 the object is not moving (x(t) = 0)
Any value less than 0 and the object is moving left. Try a couple of values.
x(-1) = 2(-1)^3 - 15(-1)^2 +24(-1) Whatever that is is minus.
t < 0 will make the object move left.

Answer 2

A) The velocity and acceleration are the first and second derivatives of x(t), respectively.
  x'(t) = 6t² -30t +24
  x''(t) = 12t -30

At t=2, the velocity is x'(2) = -12 . . . . units/time unit
At t=2, the acceleration is x''(2) = -6 . . . . units/time unit²


b) Acceleration has the same sign as velocity at t=2, so P is speeding up. When acceleration is in the same direction as velocity, the magnitude of velocity (speed) increases.


c) P is moving left where its velocity is negative, when t is in the interval (1, 4).