Respuesta :
Answer with Explanations:
We are given:
a(t)=4*t^2-2............................(1)
where t= time in seconds, and a(t) = acceleration as a function of time.
and
x(0)=-2 .................................(2)
x(2) = -20 ............................(3)
where x(t) = distance travelled as a function of time.
Need to find x(4).
Solution:
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 ................(4)
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 .............(5) 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 ......................(6)
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 ....................................(7)
Thus substituting (6) and (7) in (5), we get
x(t) = (t^4)/3 - t^2 - 29t/3 + 2 ..............(8)
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)
The answer is "28.87 m" and the further calculation can be defined as follows:
Given:
[tex]\to a=(4t^2-2)\ \frac{m}{s}\\\\ [/tex]
When
[tex] \to t=0 \ \ \ \ \ \ \ v= 2\ m\\\\ \to t=2 \ \ \ \ \ \ \ v= 20\ m\\\\ \to t=4\ \ \ \ \ \ \ v=?[/tex]
To find:
value=?
Solution:
[tex]V(t)=\int a(t)\ dt=\int (4t^2-2)\ dt=\frac{4}{3}t^3-2t+C_1\\\\ x(t)=\int V(t)\ dt =\int (\frac{4}{3}t^3-2t+C_1)\ dt=\frac{1}{3}t^4-t^2+C_1t+C_2\\\\ x(0)=-2=\frac{1}{3} 0^4-0^2+C_1(0)+C_2\to C_2=-2\ m\\\\ x(2)=-20=\frac{1}{3} 2^4-2^2+C_1(2)-2\to C_1=-\frac{29}{3}\ \frac{m}{s}\\\\ x(4) = \frac{1}{3}4^4-4^2 -\frac{29}{3}4-2 = 28.87 \ m \\\\[/tex]
The particle will be at 28.87 m at the right of the origin.
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