A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s2. Its maximum cruising speed is 120 mi/h.
A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s2. Its maximum cruising speed is 120 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 60 miles apart
(d) The trip from one station to the next takes at minimum 37.5 minutes. How far apart are the stations?
1 Answer | Add Yours
a) If it accelerate at the rate of 10ft/s^2, starting from rest at the time t=0,
Its speed is `10t` ft/s
The distance is the anti derivative of the velocity
`d= 10t^2/2=5t^2 ft.`
The maximum speed is 120 mi/h ie 176 ft/s
The train reaches its maximum speed at `t=176/10=17.6s`
The distance during the acceleration period is `d=5*17.6^2=1548.8ft`
Then the train travels for 15mn ( a quarter of an hour) at the speed of 120mi/h.
The distance is `120/4=30mi.`
1548.8 ft is 0.293mi
Therefore the maximum total distance is 30.293mi.
We’ve answered 319,808 questions. We can answer yours, too.Ask a question