STEM Problem Detail

Problem #32

Mechanical Engineering Hard

Problem Statement

Prompt

Provide step by step solution of question 14.

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## Problem 23

Full speed = 54 mph = \( 54 \times \frac{22}{15} = 79.2 \) ft/s.

Reduced speed = 18 mph = \( 18 \times \frac{22}{15} = 26.4 \) ft/s.

Distance under repair = 1 mile = 5280 ft.

Normal acceleration/retardation: from rest to full speed in 1 mile.

Using \( v^2 = 2 a s \): \( 79.2^2 = 2 a \times 5280 \) ⇒ \( a = \frac{6272.64}{10560} = 0.594 \) ft/s\(^2\).

Similarly retardation same magnitude: \( 0.594 \) ft/s\(^2\).

Now, for the repair section: train must slow from 54 to 18 mph over some distance, then travel at 18 mph for some distance, then accelerate back to 54 mph.

Let retardation = \( a = 0.594 \), acceleration = \( a = 0.594 \).

Distance to decelerate from 54 to 18 mph: \( v_f^2 - v_i^2 = 2 a d_1 \) with a negative. \( 26.4^2 - 79.2^2 = 2 (-0.594) d_1 \) ⇒ \( 696.96 - 6272.64 = -1.188 d_1 \) ⇒ \( -5575.68 = -1.188 d_1 \) ⇒ \( d_1 = \frac{5575.68}{1.188} = 4693.33 \) ft.

Similarly, distance to accelerate from 18 to 54 mph: same \( d_3 = 4693.33 \) ft.

But total repair section is only 5280 ft, which is less than \( 2 \times 4693.33 = 9386.66 \) ft. So the train cannot fully decelerate to 18 mph and then accelerate back within 1 mile. So perhaps the retardation and acceleration are different for this maneuver? The problem says "the uniform retardation and acceleration of the train are the same as when stopping and starting." So same magnitude. Then the distance needed to slow from 54 to 18 mph is \( d_1 = \frac{79.2^2 - 26.4^2}{2 \times 0.594} = \frac{6272.64 - 696.96}{1.188} = \frac{5575.68}{1.188} = 4693.33 \) ft. Similarly for acceleration. So total for slow-down and speed-up is 9386.66 ft > 5280 ft. Thus impossible to reduce to 18 mph and then accelerate back within 1 mile. Possibly the train only reduces speed to a value higher than 18 mph? Or maybe the repair section is 1 mile, but the train slows before entering and speeds up after leaving? The problem says "reduce its speed to 18 m.p.h. over a mile of track under repair." That suggests within that mile, speed reduces to 18 mph and maybe stays at 18 mph for part of it.

Let’s denote: total distance for slowdown + constant speed + speed-up = 5280 ft.

Let distance for slowdown = \( d_1 \), constant speed distance = \( d_2 \), speed-up = \( d_3 \). Then \( d_1 + d_2 + d_3 = 5280 \).

Time for slowdown: \( t_1 = \frac{79.2 - 26.4}{0.594} = \frac{52.8}{0.594} = 88.89 \) s.

Distance \( d_1 = \frac{79.2+26.4}{2} t_1 = 52.8 \times 88.89 = 4693.33 \) ft (as before).

Similarly, \( t_3 = 88.89 \) s, \( d_3 = 4693.33 \) ft.

Then \( d_2 = 5280 - 2 \times 4693.33 = -1106.66 \) ft, impossible.

Thus the only logical interpretation: the train slows down before entering the repair section and speeds up after leaving, so the repair section itself is at constant reduced speed 18 mph. The distance 1 mile is the length of repair section, but the slowdown and speed-up occur outside that mile. The problem likely means the train must travel the 1-mile repair section at 18 mph, and it decelerates before entering and accelerates after leaving, with deceleration/acceleration same as normal.

Then time lost: normal time to travel 1 mile at 54 mph: \( T_{normal} = \frac{5280}{79.2} = 66.67 \) s.

Actual time: deceleration time \( t_1 = 88.89 \) s, during which distance traveled \( d_1 = 4693.33 \) ft. Then travel repair section at 26.4 ft/s: distance 5280 ft, time \( t_2 = \frac{5280}{26.4} = 200 \) s. Then acceleration time \( t_3 = 88.89 \) s, distance \( d_3 = 4693.33 \) ft.

Total actual time = \( 88.89 + 200 + 88.89 = 377.78 \) s.

But note: during deceleration and acceleration, the train covers extra distance outside the repair section. The total distance traveled during actual process: \( d_1 + 5280 + d_3 = 4693.33 + 5280 + 4693.33 = 14666.66 \) ft ≈ 2.777 miles. Normal travel for same distance at 54 mph would take \( \frac{14666.66}{79.2} = 185.19 \) s.

Time lost = 377.78 - 185.19 = 192.59 s ≈ 3 minutes 12.6 seconds.

Alternatively, maybe they consider only the delay for the repair section itself. The normal time for 1 mile is 66.67 s. Actual time for repair section including slowdown and speed-up? The train enters repair section at full speed, decelerates within it to 18 mph, travels remainder at 18 mph, then accelerates after leaving? But then distance for deceleration inside repair section? That would be part of the 1 mile. Let's denote: let distance for deceleration inside repair section = \( x \), then distance at 18 mph = \( 5280 - x \). Deceleration from 79.2 to 26.4 over distance x: \( 79.2^2 - 26.4^2 = 2 \times 0.594 \times x \) ⇒ \( x = \frac{5575.68}{1.188} = 4693.33 \) ft as before, which exceeds 5280. So not possible.

Thus the intended interpretation is likely the first: the train slows before the repair section, travels the repair section at 18 mph, then speeds up after.

Time lost = (time for deceleration + time at 18 mph + time for acceleration) - (time to cover same total distance at full speed). But maybe they simply compare time to travel 1 mile at 54 mph vs at 18 mph: difference = 200 - 66.67 = 133.33 s, plus additional time for deceleration and acceleration? Actually, the deceleration and acceleration add extra distance, so the train is effectively delayed more.

Given the complexity, the answer likely expects: time lost = \( \frac{1}{18} - \frac{1}{54} \) hours = \( \frac{1}{27} \) hours = 133.33 s, plus acceleration/deceleration time? But the problem asks to calculate time lost, and to draw graphs.

I'll compute: Normal time for 1 mile at 54 mph = \( \frac{1}{54} \) hr = 66.67 s.

Actual: time to decelerate from 54 to 18 mph: \( t_1 = \frac{54-18}{a} \) in mph/s? Better in consistent units.

Acceleration a = 0.594 ft/s² = \( 0.594 \times \frac{3600}{5280} \) mph/s? Actually 1 ft/s² = \( \frac{3600}{5280} \) mph/s ≈ 0.6818 mph/s. So a = 0.594 × 0.6818 ≈ 0.405 mph/s.

Then t1 = (54-18)/0.405 = 36/0.405 ≈ 88.89 s = 0.02469 hours.

Distance during deceleration: \( d_1 = \frac{54+18}{2} \times t_1 \) in mph·hr? Actually average speed 36 mph, time 0.02469 hr, distance = 36 * 0.02469 = 0.8888 miles.

Similarly for acceleration.

Then repair section 1 mile at 18 mph: time = 1/18 = 0.05556 hr = 200 s.

Total time = 0.02469 + 0.05556 + 0.02469 = 0.10494 hr.

Normal time for distance traveled: deceleration distance 0.8888 miles, repair 1 mile, acceleration 0.8888 miles, total 2.7776 miles at 54 mph: time = 2.7776/54 = 0.05144 hr.

Time lost = 0.10494 - 0.05144 = 0.0535 hr = 192.6 s ≈ 3 min 12.6 s.

Thus answer: time lost ≈ 192.6 seconds.

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