A freight train traverses 998 cm in one second and 1,000 cm in the next second; the acceleration is constant.
a) What is the distance traversed in the third second?
b) Find the acceleration in cm/s^2
c) How long had the (initially stationary) train been moving at the beginning of the first second? (498.5 seconds. Hint: First show that the time from rest to the midpoint of the second second is 500 seconds)
d) For a real train whose acceleration might have varied, if the given data were the same, could the answer have been appreciably different to part (a)? to part (c)?
V= final velocity ; V0 = initial velocity ; V' = average velocity ; a = acceleration ; t = time ; X = final point reached ; X0 = initial point
1) V = V0 + at
2) V^2 = V0^2 + 2a(X-X0)
3) X = X0 + V0t + 1/2at^2
4) V' = V + V0 / 2
The Attempt at a Solution
I have the answer for a, which is 999 cm under my calculations. I took 1000 cm + 998 cm / 2 and then multiplied it by 3 seconds, to get 2997 cm traversed in three seconds. Then I subtracted 1998 cm from 2997 cm to get the amount of distance traversed in the third second to 999 cm.
For part b, I used equation 1 above, plugged in all the numbers and had an acceleration of 249.75 cm/s^2. However, I am not sure if this is correct.
For part C, I used equation #1 above. I used V = X - X0 / t - t0 to find the velocity in the first second, which is 998 cm/s. Then plugged all the numbers in equation 1 and got the answer 3.99 seconds (which is wrong). I must have the wrong acceleration or the wrong formula used.
Can anyone help?