# I need help with a distance problem involving constant acceleration

1. Sep 10, 2008

### science_rules

1. The problem statement, all variables and given/known data
A train starts at a railway station and accelerates, at a constant rate, for 7 seconds, to a velocity of 115 kilometers per hour east. Set t = 0 at the instant the train starts to accelerate.

a) determine the acceleration of the train (in km/hr/sec.) at t = 3.5 sec.
b) determine the distance, in meters, the train travels during the first 3.5 seconds of the acceleration period.
c) determine the distance, in meters, the train travels during the second 3.5 seconds of the acceleration period.

2. Relevant equations
To find the acceleration at t = 3.5 sec., I write: accel. = change in veloc. divided by change in time = 115 km/hr. - 0 km/hr.
_____________________ = 115 km/hr.
7 sec. - 0 s __________ = 16.4 km/hr/sec. constant
7 sec.

Then, to determine distance traveled in the first 3.5 seconds, i use the equation:
x = initial x + initial veloc. X time + 1/2 accel. X time squared

x = 0m + 0m/squ X 3.5 sec. + 1/2 X 4.5 m/sec squared X (3.5 sec.) squared

4.5 m is from converting the 16.4 km/hr/sec to meters.

I think I use the same equation to find the distance the train travels in the second 3.5 seconds.

Im confused about which numbers to use in the equation.

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 10, 2008

### science_rules

oops i messed up on the top part there. i meant to write:
acceler. = 115 km/hr/sec. divided by 7 seconds = 16.4 km/hr/sec. constant

3. Sep 10, 2008

### Solaxys

1 a.
Since this is constant acceleration, just use
(Vf - Vo)/(7)

1 b. Use equation:
x = vt + 1/2at^2
x = (0)(3.5) + 0.5(a)(3.5)^2

1 c. Find velocity at 3.5 seconds use acceleration, Vo = 0, and t = 3.5
and equation: vf = vo + at.
And then use that as initial velocity, and repeat 1b.

4. Sep 10, 2008

### science_rules

5. Sep 10, 2008

### science_rules

oh, i have one more quick question-do i need to use the distance calculated from the first 3.5 sec, and plug that into the initial position of the equation for the second 3 seconds? hope that's clear enough.

6. Sep 10, 2008

### science_rules

second *3.5 seconds

7. Sep 10, 2008

### science_rules

i got 15.75 m for initial velocity and plugged that in to x = x initial + veloc. initial X 3.5s X 1/2 (4.5 m/s/s) (3.5)^2
i got 27. 5 meters for the distance of the first 3.5 seconds, but when i try to calculate for the second 3.5 seconds, it seems way too much 82.6 m-maybe because i shoulnt plug in 27.5m for the inital position the second time. it remains at 0? i just dont know.

8. Sep 10, 2008

### Solaxys

yes dont sub, the question asks what is the distance from 3.5s.
So initial = 0.

9. Sep 10, 2008

### science_rules

so, in that case, i think this would be right:
the distance for the first 3.5 seconds would be
x = x initial + (v initial X 3.5 s) + 1/2 a (t)^2
= 0 m + (0 m/s X 3.5s) + 1/2 (4.5m/s/s)(3.5s)^2 = 27.5 m

the distance for the second 3.5 seconds is:

x = x initial + (v initial X 3.5 s) + 1/2 a (t)^2
= 0m + (4.5m/s/s X 3.5 s) + 1/2 X 4.5 m/s/s X (3.5s)^2
= 43.31 m