The MRT train takes 30 minutes to travel from EDSA-Pasay Station to Shaw Boulevard Station. The train accelerates from rest at a rate of 0.5 meters per second squared for 60 seconds. It then travels at a constant speed before it is brought to rest in 6 minutes with a constant deceleration. The motion is modeled as a linear motion on a straight railway track. Sketch the velocity-time graph for the journey
To sketch the velocity-time graph for the journey, we need to break down the motion into three phases: acceleration, constant speed, and deceleration.
1. Acceleration Phase:
Since the train starts from rest and accelerates at a rate of 0.5 meters per second squared for 60 seconds, we can calculate the final velocity using the equation v = u + at, where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration, and t is the time.
v = 0 + (0.5 m/s^2) * 60 s
v = 0 + 30 m/s
v = 30 m/s
So during the acceleration phase, the velocity increases linearly from 0 m/s to 30 m/s.
2. Constant Speed Phase:
During this phase, the train travels at a constant speed. Since we know the time it takes to travel from EDSA-Pasay Station to Shaw Boulevard Station is 30 minutes, we can convert it to seconds:
30 minutes * 60 seconds/minute = 1800 seconds
So during this phase, the velocity remains constant at 30 m/s.
3. Deceleration Phase:
The train is brought to rest in 6 minutes, which is 360 seconds.
Since the train decelerates at a constant rate, we need to find the acceleration during the deceleration phase. We can use the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (30 m/s), a is the deceleration, and s is the distance covered.
0 = (30 m/s)^2 + 2 * a * s
900 = 2 * a * s
Given that s is the distance between the two stations, we can use the formula s = ut + 0.5at^2 to calculate the distance.
s = (30 m/s * 360 s) + 0.5 * a * (360 s)^2
s = 10800 m + 64800a
Substituting this value of s back into the previous equation:
900 = 2 * a * (10800 m + 64800a)
900 = 21600a + 129600a^2
0 = 129600a^2 + 21600a - 900
This can be solved using the quadratic formula, giving two possible values for a: a = -0.15 m/s^2 and a = 0.005 m/s^2. Since we are decelerating, we choose the negative value of a.
So during the deceleration phase, the velocity decreases linearly from 30 m/s to 0 m/s.
Based on this information, the velocity-time graph will look as follows:
^ acceleration constant speed deceleration
30 m/s | ___/________________\___ 0 m/s
| / \
| / \
| / \
|/ \
+-------------------------------> time (s)
Please note that the graph is not to scale, and only represents the general trend of the velocity during each phase of the journey.