The total time taken is 271.0 seconds
Explanation:
The object in this problem is moving by uniformly accelerated motion until it reaches a top velocity of
v = 297.733 ft/s
So in the first part of the motion, we can use suvat equation:
[tex]v = u+at[/tex]
where
v is the final velocity
u = 0 is the initial velocity
[tex]a=2.93 ft/s^2[/tex] is the acceleration
Solving for t, we find the time taken for the object to reach top speed:
[tex]t=\frac{v-u}{a}=\frac{297.733-0}{2.93}=101.6 s[/tex]
So the distance covered by the object in the first part of the motion (the accelerated part) is
[tex]s=ut+\frac{1}{2}at^2=0+\frac{1}{2}(2.93)(101.6)^2=15,122.6 ft[/tex]
The total distance to cover is
d = 65,557.6 ft
So, the distance left is
s' = d - s = 65,557.6 - 15,122.6 = 50,435 ft
This second distance is covered by the object at a constant speed of
v = 297.733 ft/s
So it takes a time of
[tex]t'=\frac{s'}{v}=\frac{50,435}{297.733}=169.4 s[/tex]
Therefore, the total time taken is
T = t + t' = 101.6 + 169.4 = 271.0 s
Learn more about accelerated motion:
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