(1 point) A Carroll student decided to depart from Earth after his graduation to
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Question
(1 point) A Carroll student decided to depart from Earth after his graduation to find work on Mars. Before building a shuttle, he conducted careful calculations. A model for the velocity of the shuttle, from liftoff at t = 0 s until the solid rocket boosters were jettisoned at t = 57.1 s, is given by r(t) = 0.00 1405000t3-0.08 11 35P+ 19.73t+ 1.79 (in feet per second). Using this model, estimate the global maximum value and global minimum value of the ACCELERATION of the shuttle between liftoff and the jettisoning of the boosters.Explanation / Answer
we know that dertivative of velocity is acceleration
so we find v'(t) that is a(t)
given
v(t)=0.001405000t^3-0.081135t^2+19.73t+1.79
so v'(t)=3*0.001405000t^2-2*0.081135t^2+19.73+0 [ using power rule x^n= nx^n-1]
so v'(t)= 0.004215 t^2 - 0.16227 t + 19.73
this is equal to acceleration
that is a(t)=0.004215 t^2 - 0.16227 t + 19.73
now for maxima or minima . we set a'(t)=0
a(t)=0.004215 t^2 - 0.16227 t + 19.73
now for maxima ,minima of acceleration , we derivate the a(t)
so a'(t)=2*0.004215 t - 0.16227 +0
a'(t)= 0.00843 t-0.16227
now for crititcal value , we set a'(t)=0
so
a'(t)=0
0.00843 t-0.16227 =0
so t= 0.16227/0.00843
t=19.2491 or 19.25
now for global maximum or minimum , we will check value of a(t) [ note the value of acceleration not its derivative]
at these three point
t=0
t=19.25
t=57.1
SO
t=0 sec
a(t)=0.004215 t^2 - 0.16227 t + 19.73 = 0+0+19.73
so a(t)=19.73
(2) at t= 19.25
a(t)= 0.004215 t^2 - 0.16227 t + 19.73
a(t)= 0.004215 (19.25)^2 - 0.16227 (19.25) + 19.73
a(t)=18.1682 or 18.17
(3) at t=57.1 sec
a(t)= 0.004215 t^2 - 0.16227 t + 19.73
a(t)= 0.004215 (57.1)^2 - 0.16227 (57.1)+ 19.73
a(t)=24.20
hence we see that
global maximum is 24.20
global minimum is 18.17
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