Acceleration Problem: Solving for Distance, Velocity & Time

[sproles134]

Homework Statement

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)?

Homework Equations

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?

 

[Borek]

From your calculations distance covered in the third second is smaller than the distance covered in the second second, even if the train accelerates, so there is something wrong.

Your third equation will do, you just have to correctly deal with x₀ term. 998 cm is for t=0, 1000 cm is for t=1, can you use this information to calculate both a & V?