A train traveling at 55.6 m s[sup]-1[/sup] covers a distance in 26.6 s, which gives 1480 m. Easy …
Now we start the train from stand still and it covers the same distance in 44.1 s; and we know it’s at the 55.6 m s[sup]-1[/sup] speed at the end of the run, but we don’t know for how long.
Is this enough information, and if not, what else do I need to figure acceleration?
(No, it’s not homework, it’s a video game)
Not enough information. If acceleration is constant, then you just have a = v/t, or 55.6/44.1 = 1.26 m/s[sup]2[/sup]. You can then figure the distance using d = 0.5at[sup]2[/sup]. That does not give the 1480 m in the original problem, though. So acceleration must not be constant, and if it’s not constant then it could be any complicated function you might imagine.
If you’re playing a train simulator, then that could certainly explain it. Very few vehicles have constant acceleration and trains are no exception.
Yes, it’s enough. You have initial velocity (zero), final velocity, and how many seconds to get up to final velocity. Just divide v1-v0 by the duration to get acceleration.
(edit: with the assumption of constant acceleration, that is)
The acceleration is assumed to be constant, however the train reaches top speed at some unknown point, I’m trying to work with:
1480 m = a t[sub]1[/sub][sup]2[/sup] + 55.6 m s[sup]-1[/sup] t[sub]2[/sub] and
t[sub]1[/sub] + t[sub]2[/sub] = 44.1 s
My problem is the time which the train is accelerating is unknown, only it’s less than 44.1 s
Heh, those equations look similar to the ones I got on the envelope in front of me right now.
What I did was plot v(t) with the assumption that the acceleration was constant until you reached v[sub]max[/sub].
This gives us a plot consisting of a triangle and a rectangle where the area of the triangle represents the distance traveled while accelerating and the area of the rectangle the distance traveled while at v[sub]max[/sub].
So, the distance traveled while accelerating is 0.5v[sub]max[/sub]*t[sub]1[/sub] and for the distance at v[sub]max[/sub] we have v[sub]max[/sub]*t[sub]2[/sub].
Giving us the equations:
0.5v[sub]max[/sub]*t[sub]1[/sub]+v[sub]max[/sub]*t[sub]2[/sub] = 1480 m
and
t[sub]1[/sub] + t[sub]2[/sub] = 44.1 s
Two equations, two unknowns, easily solved, giving us t[sub]1[/sub]=34.963 and t[sub]2[/sub]=9.137 s
Then we can get the acceleration by solving:
v[sub]max[/sub]=a*t[sub]1[/sub]
<=> a=1.590 m s[sup]-2[/sup]
Perfect, thank you, I have that diagram too … average speed during acceleration is half max speed … an excellent insight.
Also easily solved with excel.
final_speed = 55.6
Distance_traveled = final_speed * (44.1 - T1 ) + 1/2 * final_speed * T1
Data, what if ,
set distance_traveled to be 1480 by adjusting T1