solution:
Increasing the magnification and decreasing the field view
we are given:
a(t)=4*t^2-)
where t= time in seconds, and a(t) = acceleration as a function of time.
and
x(0)=-2 )
x(2) = -20 )
where x(t) = distance travelled as a function of time.
need to find x(4).
from (1), we express x(t) by integrating, twice.
velocity = v(t) = integral of (1) with respect to t
v(t) = 4t^3/3 - 2t + k1 )
where k1 is a constant, to be determined.
integrate (4) to find the displacement x(t) = integral of (4).
x(t) = integral of v(t) with respect to t
= (t^4)/3 - t^2 + (k1)t + k2 ) where k2 is another constant to be determined.
from (2) and (3)
we set up a system of two equations, with k1 and k2 as unknowns.
x(0) = 0 - 0 + 0 + k2 = -2 => k2 = 2 )
substitute (6) in (3)
x(2) = (2^4)/3 - (2^2) + k1(2) -2 = -20
16/3 -4 + 2k1 -2 = -20
2k1 = -20-16/3 +4 +2 = -58/3
=>
k1 = -29/3 )
thus substituting (6) and (7) in (5), we get
x(t) = (t^4)/3 - t^2 - 29t/3 + 2 )
which, by putting t=4 in (8)
x(4) = (4^4)/3 - (4^2 - 29*4/3 +2
= 86/3, or
= 28 2/3, or
= 28.67 (to two places of decimal)
