Agustin G. answered • 03/22/24
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"""
[Note: \int represents the integral sign.]
Recall,
(1) a(t) = dv / dt,
(2) v(t) = ds / dt.
It then follows that
v(t) = \int a(t) dt from (1) and
s(t) = \int v(t) dt from (2).
Thus,
v(t) = \int a(t) dt,
= \int [ 36t + 8 ] dt,
= ((1 / 2) * 36 * t^2) + (8 * t) + v_0,
= 18t^2 + 8t + v_0,
where v_0 is the velocity at t = 0 s. From the information
given in the problem, we see that v_0 = 1 m/s. So,
v(t) = 18t^2 + 8t + 1.
We can proceed with finding s(t):
s(t) = \int v(t) dt,
= \int [ 18t^2 + 8t + 1 ] dt,
= ((1 / 3) * 18 * t^3) + ((1 / 2) * 8 * t^2) + t + s_0,
= 6t^3 + 4t^2 + t + s_0,
where s_0 is the position at t = 0 s, namely s_0 = 3 m
from the information in the problem. So,
s(t) = 6t^3 + 4t^2 + t + 3.
This is now position as a function of time, so at t = 8 s,
s(8) = 6(8)^3 + 4(8)^2 + (8) + 3,
= 6(512) + 4(64) + 11,
= 3072 + 256 + 11,
= 3339 m.
"""
