Answer:
4 mins 31 seconds
Step-by-step explanation:
Maximum speed is 203 mph equivalent to
[tex]203*\frac {5280}{3600}=297.73 ft/s[/tex]
Time taken to attain maximum speed, [tex]t=\frac {v}{a}[/tex] where v is velocity and a is acceleration hence
[tex]t=\frac {297.73}{2.93}=101.6155 s[/tex]
The distance traveled when at top speed
[tex]d=\frac {v^{2}}{2a}=\frac {(297.73)^{2}}{2*2.93}=15127.16 \approx 15127.2 ft[/tex]
Total distance is provided as 12.42 miles converted to ft is 12.42*5280=65577.6 ft
Total distance-distance moved to attain maximum speed will be
65577.6 ft-15127.2 ft=50450.44 ft
Time taken during maximum speed
[tex]t=\frac {50450.44}{297.73}=169.45 s[/tex]
Total time=101.6155 s+169.45 s=271.06 s
271.06/60=4 mins 31 seconds
Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration
the speed of vo = v = constant
acceleration = a = 0
Δx = vt or x = xo + vt
An equation of constant velocity motion
[tex] \large {\boxed {\bold {d = v \times\: t}}} [/tex]
d = distance = m
v = speed = m / s
t = time = seconds
[tex] \large {\boxed {\bold {x=xo+vo.t+\frac {1} {2} at ^ 2}}} [/tex]
v = vo + at
vt² = vo² + 2a (x-xo)
x = distance on t
vo / vi = initial speed
vt / vf = speed on t / final speed
a = acceleration
Acceleration is a change in speed within a certain time interval
[tex] \displaystyle a = \frac {v2-v1} {t2-t1} [/tex]
Because no images are included, we assume the car will move at a constant speed of 203 mph to reach a distance of 12.42 miles and move at an acceleration of 2.93 ft / s² from rest
slope 6.4% = acceleration 6.4% of the actual acceleration
acceleration = 6.4% x 2.93 ft / s² = 0.1875 ft / s²
time required
1 mile = 5280 ft
1 h = 3600 s
V2 = vf = 203 mph = 297.7 ft / s
[tex] \displaystyle a = \frac {v2-v1} {t2-t1} [/tex]
[tex] \displaystyle 0.1875 = \frac {297.7-0} {t} [/tex]
[tex] \displaystyle t = 1587.7 ~ s [/tex]
Distance traveled
Vt² = vo² + 2a (x-xo)
297.7 ft / s = 0 + 2.0.1875 ft / s². x
x1 = 793.86 ft
Total distance
12.42 miles = 12.42 x 5280 = 65577.6 ft
Remaining distance traveled with a constant speed of 297.7 ft / s:
x2 = 65577.6 ft - 793.86 ft
x2 = 64783.74 ft
time required
x2 = d = v x t
t2 = d : v
t2 = 64783.74 : 297.7 ft / s
t2 = 217.6 s
Total time
t1 + t2 = 1587.7 + 217.6
t total = 1805.3 s
The distance of the elevator Â
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resultant velocity Â
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the average velocity Â
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Keywords: constant velocity motion, uniformly accelerated motion, Â distance, speed, acceleration
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