annoying physics problem?

Question

okay a girl watches a train pass by. the cars of each train is 8.6 m long. the first one passes her in 1.5 s and the second in 1.1 s. Find the constant acceleration. Please explain how you approached this problem?

Answer 1

if we assume that acceleration is constant throughout, the correct answer should be 1.604 m/s/s. If this is wrong, then I don't know sh!t. If it's correct then you already know how to do it. Or email me for any comment.

I'm no genius. Just tried a little common sense to attack the problem. Ok, I will explain: It's obvious that we have 2 average velocities: one for car1 and one for car2...these velocities are also the instantaneous velocities at the midpoints of each the measurements, respectively. That means these can be thought of as "speedometer readings" at time 0.75 sec for the first car and time 0.55 sec for the second car, independently. so the time between the two "speedometer readings" must be 0.75 + 0.55 = 1.3 sec
If it took 1.3 secs for the "speedometer needle" to climb up from 5.733m/s to 7.818m/s then you have what you need to calculate acceleration.
Thus, accel. = (7.818 - 5.733)/1.3 = positive 1.604 m/s/s

Is that ok, friend? You gotta think like a ninja. that's the secret.

Answer 2

Let u, v, w be the speeds at time 0, 1.5 and 2.6 respectively.
Distance traveled = average velocity x time
8.6 = [v +u] t1 / 2
[v +u] = 2 x 8.6 / 1.5= 11.47 --------A
[w +v] = 2 x 8.6 /1.1=15.63--------------B
Subtracting A from B
[w - u]= 4.16
Acceleration a = [w - u] / [time to change from u to w]
a = 4.16 / 2.6 = 1.6 m/s^2
=====================================
Or
The average speed for the first car U = 8.6 / 1.5 m/s
For the second one V = 8.6 / 1.1 m/s.
Using V^2 - U^2 = 2as
a = [V^2 - U^2] / 2 * 8.6 = 1.64 m/s^2

Answer 3

The average velocity of the first car is 8.6/1.5 = 5.73 m/sec; the average velocity of the second car is 8.6/1.1 = 7.82 m/sec. For constant acceleration, the average velocity occurs at the mid-point of the measurements, or car 1 @0.75 sec and car 2 at (1.5+1.1)/2 = 1.3 sec; The time difference is 1.3 - 0.75 = 0.55 sec, The velocity change is 7.82 - 5.73 = 2.09 m/sec. The acceleration is change in velocity divided by time, or 2.09/0.55 = 3.8 m/sec^2