physics 1234567

Question

So I had a problem in class where you were traveling by train to get from Columbia to atlanta, it took 20 minutes to get from Columbia to Atlanta. The train gets accelerated for a few minutes, reaches terminal velocity, move with velocity for a few minutes, and then gets decelerated for the last few minutes. calculate the time during which the train was moving with its terminal velocity. assume terminal velocity was 1200 km per hour, distance from columbia to atlanta is 300 kilometers. and the time the train accelerated and decelerated are the same.

So i got this shown to me and i can get the right answer one way but am confused in the process.... i took 20 min and made it 1/3 hour and then took the 300 km and diveded by 1200km/hr and got 1/4, then proceeded to 1/3-1/4 and got 1/12, then multiplied that by 60 and got 5 min. then the time with terminal velocity is equal too the total-20 minus the acceleration time plus the acceleration time so 20-(5+5) = 10 min is the answer.......so my question is why can you just divide the 300 km total distance by the terminal velocity of 1200 and get that 1/4 value is that a formula of some sort that is known....cause with that im doing distance divided by velocity and getting part of a time, is that a formula if so which one

Answer 1

Is your solution a general result?...what if the distance was greater? See if it works in this case: the total time is 40 minutes and the distance is 600 km, times accelerating and decelerating are the same, I've played around with your method a few times, and it looks like it works - haven't proven that it is a general result that will hold in any case.

You are correct in changing the time in minutes to fraction of an hour, because the units must be consistent.

I always recommend drawing a picture, since that helps clarify the problem. In this case, your picture will be a line segment divided into 3 parts - distances:

Part 1 - where the train is accelerating, part 2 where it is moving at constant velocity, and part 3 where it is decelerating.

Since the acceleration/deceleration are constant, we can calculate the average velocity during these times: v_ave = (v2 - v1) / 2. In both cases, one velocity is the terminal velocity, v_t, the other is zero, so the average velocity when accelerating and decelerating are v_ave = (v_t / 2)

Since we are told the times accelerating and decelerating are the same, the distances traveled are the same. We have therefore two unknowns - the distance at constant velocity and the distance when the velocity is changing and two times the time a constant speed and the time accelerating/decelerating.

time equation:

t1 + t2 + t3 = 2*t1 + t2 = 20 min (I converted from minutes to hrs just before putting in numbers to avoid using fractions).

Distance equation:

D1 + D2 + D3 = 2*D1 + D2 = 300 km

The time and distance are related by D = V*t, so the distance equation becomes:

2*(V1*t1 )+ (V2*t2) = 300 km

It was previously established that V1 = V_t / 2 and solving for t1, get: t1 = (20 - t2) / 2 and V2 - V_t, substituting these into the equation,

2*[(V_t/2)*(20 - t2)/2] + V_t*t2 = 10*V_t - (V_t * t2)/2 + (V_t * t2) = 10*V_t + (V_t * t2)/2 = 300 km

this is where I converted minutes to hours...multiply by 2, then solve for t2

V_t / 6 + (V_t * t2)/2 = 300 km --> V_t / 3 + (V_t * t2) = 600 km --> t2 = [600 km - (V_t / 3)] / V_t = 1/6 hours or 10 minutes.