Question

A car travels a distance on a road of 12.42 miles at a slope of 6.4%. The car has a maximum speed of 203 mph. Assume the car will accelerate at an average of 0.091 g (2.93 ft/s^2). How long will it take the car to reach the end of the road?

Answer 1

Answer:

4 mins 31 seconds

Step-by-step explanation:

Maximum speed is 203 mph equivalent to

$203*\frac {5280}{3600}=297.73 ft/s$

Time taken to attain maximum speed, $t=\frac {v}{a}$ where v is velocity and a is acceleration hence

$t=\frac {297.73}{2.93}=101.6155 s$

The distance traveled when at top speed

$d=\frac {v^{2}}{2a}=\frac {(297.73)^{2}}{2*2.93}=15127.16 \approx 15127.2 ft$

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

$t=\frac {50450.44}{297.73}=169.45 s$

Total time=101.6155 s+169.45 s=271.06 s

271.06/60=4 mins 31 seconds

Answer 2

The car  will take 1805.3 s to reach the end of the road

Further explanation

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration

• At constant velocity motion:

the speed of vo = v = constant

acceleration = a = 0

Δx = vt or x = xo + vt

An equation of constant velocity motion

$\large {\boxed {\bold {d = v \times\: t}}}$

d = distance = m

v = speed = m / s

t = time = seconds

• An equation of uniformly accelerated motion

$\large {\boxed {\bold {x=xo+vo.t+\frac {1} {2} at ^ 2}}}$

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

a = Δv / Δ t

$\displaystyle a = \frac {v2-v1} {t2-t1}$

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

• 1. the car accelerates

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

$\displaystyle a = \frac {v2-v1} {t2-t1}$

$\displaystyle 0.1875 = \frac {297.7-0} {t}$

$\displaystyle t = 1587.7 ~ s$

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

• 2. a car with a constant speed

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

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Keywords: constant velocity motion, uniformly accelerated motion,  distance, speed, acceleration

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