physics ch.2 question?

[bahamababe4ever]

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.4m/s^2 for 15 s. It then proceeds at constant speed for 1100 m before slowing down at 2.2 m/s^2 until it stops at the station.
What is the distance between the stations?
How much time does it take the train to go between the stations?

[poolmule22]

Using 1D coord. system (x-axis) with +x pointing ->
Note: units are enclosed within [ ] brackets...

Break problem into 3 sections... 1 (acceleration), 2 (constant speed), 3 (negative acceleration).

Will utilize the following 3 equations:

(1) acceleration [m/s^2] = a, where: a =acceleration

(2) velocity [m/s] = a*t + v_0, where: t=time, v_0=initial velocity; function of time -> v(t)

(3) position [m] = a/2*t^2 + v_0*t + p_0, where: p_0=initial position; function of time -> p(t)

----------Section 1:

Known:
t=15 [s]
a=1.4 [m/s^2]

so... we know all the information to find equations (1), (2), and (3).

This yields...

(1) = a = 1.4 [m/s^2]
(2) = v(t) = a*t + v_0 = 1.4*(15) + 0 = 21 [m/s]; v_0 = 0 due to the fact that the train starts from rest and, thus, no intial velocity
(3) = p(t) = a/2*t^2 + v_0*t + p_0 = .7*(15)^2 + 0*(15) + 0 = 157.5 [m]; p_0 = 0 due to the fact that the train was at the first station where position can be set to 0

From this we can see...

-train traveled 157.5 meters in the 21 second acceleration.
-reached 21 meters/second after the 21 second acceleration.

END SECTION 1

----------Section 2:

Known:

train travels 1100 meters at constant speed...

So... all we need to find is the time it takes for the train to travel this distance... this is very simple...

Distance traveled = velocity * time, solving for time we find...

time = distance traveled / velocity.

we know the distance travleled (1100 [m]) and train velocity (21 [m/s] after the train accelerates for the first 15 seconds in section 1)

so we find that...

1100 [m] / 21 [m\s] = 52.38 [s]

From this we can see...



END SECTION 2

----------Section 3:

Known:

Train decelerates from 21 [m/s] to 0 at a rate of 2.2 [m/s^2]

so... This yields...

(1) = a = -2.2 [m/s^2](negative because train is slowing down, remember coord. system setup)
(2) = v(t) = a*t + v_0 = -2.2*(t) + 21 = 0 [m/s]; v_0 = 21 due to the fact that the train was slowing down from the initial speed of 21 [m/s]. also, we set v(t) = 0 because train has to stop at the end. from this we can find the time it takes for the train to slow to a stop. Solve the above equation for 't' and it is equal to 9.55 [s]=t=time for train to stop.
(3) = p(t) = a/2*t^2 + v_0*t + p_0 = -1.1*(9.55)^2 + 21*(9.55) + (157.5+1100) = 1447.55 [m] ; p_0 = (157.5+1100) due to the fact that the train traveled 157.5 [m] in section 1 and 1100 [m] in section 2 so the initial position of the train in section 3 was (157.5+1100) [m].

From this we can see...

-it took the train 9.55 seconds to slow from its initial speed of 21 meters per second to 0 [m/s]
-the total distance the train traveled was 1447.55 meters. this was found by solving for the distance the train traveled while accelertation (section 1), knowing the train traveled 1100 meters (give in problem, section 2), and by using the sum of those to as the initial position (p_0) for section 3 we found the total distanced traveled.

END SECTION 3

-----------------------------------ANSWERS--------------------------------

already know the total distance traveled between stations: 1447.55 [m]

to find total, sum the three times of each section (1,2, and 3):

1 + 2 + 3 = (15) + (52.38) + (9.55) = 76.93 [s]

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I hope all of my work is correct but other verifications would be appreciated. Enjoy