I cant solve a simple Acceleration/Deceleration rate of time problem.

[smashbrohamme]

A train with a maximum speed of 105km/h has an acceleration rate of 0.25m/s² and a deceleration rate of 0.7 m/s². Determine the minimum running time between stations 7km apart, if it stops at all stations.


We are using the basic kinematics equations.
S= VoT + 1/2aT²
V=Vo +aT
V² = Vo² +2as


My main problem is what would acceleration be considered, I am giving two variables.
.25m/s² and -0.7 m/s².

So far I have S= 7km which I converted to 7000m
V= 105 km/h which I converted to 29.16m/s
But what about acceleration?

 

[cupid.callin]

first find the distance train travels while it accelerates to max velocity and also distance is travels while decelerating from max velocity to 0 .. rest distance left is traveled at max velocity.

Find corresponding times and add them

 

[smashbrohamme]

Ok first I took the kinematics equation V=Vo+at

29.16m/s= 0+(.25m/s²)t

T= 116.6s for the train to accelerate to a full velocity of 29.16m/s
S = 1700M for the train to travel to full acceleration.
So there is a distance of 5300M left to go.

It takes the train 607.3M to go from full velocity to 0.
T= 41.6s

4692.7M to go.

so at full velocity of 29.16m/s
I divide the 4692.7 M by 29.16m/s to get a time of 160s...

Man is there a easier way to do this, like more kinematics formulas that I might need to know?

 

[cupid.callin]

I'm afraid No

 

[rwisz]

I would suggest breaking down the problem into smaller parts to better understand and solve it. First, let's define the variables that we have been given:

- Maximum speed (Vmax) = 105 km/h = 29.16 m/s
- Acceleration rate (a) = 0.25 m/s²
- Deceleration rate (a') = 0.7 m/s²
- Distance between stations (d) = 7 km = 7000 m

Now, let's consider the train's motion between the two stations:

1. Acceleration phase: The train starts from rest (Vo = 0) and accelerates to its maximum speed (Vmax) with an acceleration rate of 0.25 m/s². Using the equation V = Vo +aT, we can find the time (T) taken for the train to reach Vmax:

Vmax = Vo + aT
29.16 = 0 + (0.25)xT
T = 116.64 seconds

2. Constant speed phase: The train travels at a constant speed of Vmax for a certain distance (d/2) between the two stations. Using the equation S = VoT + 1/2aT², we can find the distance (d/2) traveled during this phase:

d/2 = (29.16)x(116.64) + 1/2(0.25)(116.64)²
d/2 = 1698.36 + 1700.26
d/2 = 3398.62 meters

3. Deceleration phase: The train now needs to decelerate from its maximum speed (Vmax) to a complete stop at the next station, with a deceleration rate of 0.7 m/s². Using the equation V² = Vo² + 2a'S, we can find the distance (S) required for the train to come to a complete stop:

0 = (29.16)² + 2(-0.7)S
S = 120.64 meters

4. Total time taken: Now, we can find the total time taken for the train to travel between the two stations by adding the times taken for each phase:

Total time (Ttotal) = Acceleration time + Constant speed time + Deceleration time
=