Answer:
Please refer to the attachment.......
we are given
position function as
we know that acceleration is second derivative of position
so, firstly, we will find first derivative
now, we can simplify it
now, we can find derivative again
now, we can plug t=2
and we get
so,
the value of the person’s acceleration a at t=2s is -4..........Answer
Answer:
c. (x + 3)
Step-by-step explanation:
using factor theorem
if x - 3 is a factor then p(a) = 0
p(a)= x^3 - 3x^2 - 4x + 12
a.(x-3)
p(3) = (3)^3 - 3(3)^2 - 4(3) + 12
= 27 - 27 - 12 + 12
= 0
therefore x-3 is a factor
b.(x + 2)
p(-2) = (-2)^3 - 3(-2)^2 - 4(-2) + 12
= -8 -12 + 8 + 12
,= 0
therefore x + 2 is a factor
c.(x + 3)
p(-3) = (-3)^3 - 3(-3)^2 - 4(-3) + 12
= -27 -27 + 12 + 12
= -30
therefore x + 3 is not a factor
d.(x-2)
p(2) = (2)^3 - 3(2)^2 - 4(2) + 12
= 8 -12 - 8 + 12
= 0
therefore x - 2 is a factor
I’ll say it is a large difference
I can’t read it it is upside down