A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s2. Its maximum cruising speed is 90 mi/h. (Round your answers to three decimal places.)
(a) What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?
(b) Suppose that the train starts from rest and must come to a complete stop in 15 minutes. What is the maximum distance it can travel under these conditions?
(c) Find the minimum time that the train takes to travel between two consecutive stations that are 45 miles apart.
(d) The trip from one station to the next takes at minimum 37.5 minutes. How far apart are the stations?
90 mi/hr = 132 ft/s
(a) first how long and how far to reach 132 ft/s
v = 0 + 4 t
132 = 4 t
t = 33 s to reach 132
d = .5 a t^2 = 2 (33)^2 = 2178 ft to reach cruising speed
15*60 = 900 seconds total
900 - 33 = 867 seconds left
how far at cruising speed
132 ft/s * 867 s = 114,444ft
total distance = 114,444 + 2178
= 116,622 ft
= 22 miles
I did put these answers in but they're marked wrong
Thank you!
What is a solid figure?
Bcz it is wrong lol
(a) What did the bullet train say when it reached its cruising speed? "Choo choo, I'm on a roll!" The maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes can be calculated using the formula s = ut + 0.5at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. Let's assume the initial velocity is 0 and the acceleration is 4 ft/s^2. Converting 90 mi/h to ft/s gives us 132 ft/s (rounded to the nearest whole number), and converting 15 minutes to seconds gives us 900 seconds. Plugging in these values, we get s = 0(900) + 0.5(4)(900)^2 = 0 + 0.5(4)(810,000) = 0 + 1,620,000 = 1,620,000 feet (rounded to three decimal places). So, the maximum distance the train can travel is 1,620,000 feet.
(b) Why did the bullet train want to come to a complete stop? It wanted to show its friends how "train-ed" it is! To calculate the maximum distance the train can travel from rest and come to a complete stop in 15 minutes, we can use the same formula as in part (a). However, this time we need to find the acceleration needed to stop the train in 15 minutes. Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we get 0 = 0 + a(15 minutes), which simplifies to a = 0. Dividing the distance formula s = ut + 0.5at^2 by 2, we get s = 0.5at^2, since the initial velocity is 0. Plugging in the values, we get s = 0.5(0)(15 minutes)^2 = 0 feet. So, the maximum distance the train can travel while coming to a complete stop in 15 minutes is 0 feet.
(c) Why did the bullet train bring a stopwatch to the station? It wanted to "watch" the time "fly" by! To find the minimum time that the train takes to travel between two consecutive stations that are 45 miles apart, we need to calculate the time it takes for the train to accelerate to its cruising speed and decelerate back to rest. Let's assume the cruising speed is 90 mi/h. Converting 45 miles to feet, we get 45 * 5280 = 237,600 feet. Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can find the time it takes for the train to accelerate from 0 to 90 mi/h. Since the final velocity of 90 mi/h is the cruising speed, and the initial velocity is 0, the formula becomes 90 mi/h = 0 + 4 ft/s^2 * t. Converting 90 mi/h to ft/s, we get 132 ft/s (rounded to the nearest whole number). Solving for t, we get t = 132 ft/s / 4 ft/s^2 = 33 seconds (rounded to the nearest whole number). The deceleration time will be the same, so the total time is 2 * 33 = 66 seconds. Converting this to minutes, we get 66 seconds / 60 seconds per minute = 1.1 minutes (rounded to one decimal place). Therefore, the minimum time that the train takes to travel between two consecutive stations that are 45 miles apart is 1.1 minutes.
(d) Why was the bullet train not happy during the trip from one station to the next? It felt like it was going "station"-ary! To find the distance between two consecutive stations when the trip takes a minimum of 37.5 minutes, we need to determine the average speed of the train during this time. Using the formula s = ut + 0.5at^2 and assuming the initial velocity is 0, the acceleration is 4 ft/s^2, and the time is 37.5 minutes converted to seconds (37.5 minutes * 60 seconds/minute), we can solve for s. Plugging in the values, we get s = 0 + 0.5(4 ft/s^2)(37.5 minutes * 60 seconds/minute)^2 = 0 + 0.5(4)(37.5^2)(60^2) = 0 + 0.5(4)(1,406,250) = 0 + 2,812,500 = 2,812,500 feet (rounded to the nearest whole number). Converting this distance to miles, we get 2,812,500 feet / 5280 feet/mile = 532.197 miles (rounded to three decimal places). Therefore, the stations are approximately 532.197 miles apart.
b) the answer is in (a)
twice 2178 feet
I think you can figure the rest out