A train traveling between two stations. The train starts from rest and accelerates uniformly for 150 seconds. It then travels at a constant speed for 300 seconds and finally decelerates uniformly for 200 seconds.
Given that the distance between the two stations is 10450 m, calculate the:
(a) Maximum speed, in km/h, the train attained;
(b) Acceleration,
(c) Distance the train traveled during the last 100 seconds;
(d) Time the train takes to travel the first half of the journey.
apply your basic equations of motion
v = at
s = v0t + 1/2 at^2
To find the answers to these questions, we can use the equations of motion for uniformly accelerated motion.
(a) To find the maximum speed attained by the train, we need to calculate the velocity at the end of the acceleration phase, which occurs after 150 seconds. The equation relating initial velocity (u), final velocity (v), acceleration (a), and time (t) is:
v = u + at
Since the train starts from rest (u = 0), the equation simplifies to:
v = at
Given that the acceleration is uniform and lasts for 150 seconds, we can substitute the values into the equation:
v = a * 150
Now we need the value of acceleration, which can be found using the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the final velocity is the maximum velocity attained by the train, which is what we're trying to find. The initial velocity (u) is 0 because the train starts from rest, and the time (t) is 150 seconds because that's the duration of the acceleration phase. Plugging in these values, the equation becomes:
v = 0 + a * 150
Simplifying this equation, we find that:
v = 150a
Plugging this value back into the equation v = a * 150:
150a = 150a
This indicates that the maximum speed attained by the train is 150a m/s.
To convert this to km/h, we can multiply by a conversion factor:
1 m/s = 3.6 km/h
So the maximum speed attained by the train is 150a * 3.6 km/h.
(b) To find the acceleration, we can use the formula:
v = u + at
The initial velocity (u) is 0 because the train starts from rest, and the final velocity (v) is found in part (a) as 150a. The time (t) is 150 seconds, which is the duration of the acceleration phase. Plugging in these values, the equation becomes:
150a = 0 + a * 150
Simplifying, we find that:
150a = 150a
This indicates that the acceleration of the train is a.
(c) To find the distance the train traveled during the last 100 seconds, we need to calculate the distance traveled during the constant speed phase and subtract it from the total distance between the two stations (10450 m). The formula for distance (s), initial velocity (u), time (t), and acceleration (a) is:
s = ut + 0.5at^2
During the constant speed phase, the acceleration is 0, so the equation becomes:
s = ut
The initial velocity (u) is found in part (b) as a, and the time (t) is 300 seconds, which is the duration of the constant speed phase. Plugging in the values, we get:
s = a * 300
Now we subtract this distance from the total distance between the stations:
10450 - a * 300
(d) To find the time the train takes to travel the first half of the journey, we need to know the distance traveled during the first half. This can be found by dividing the total distance (10450 m) by 2:
10450 / 2 = 5225 m
We can then use the equation for distance:
s = ut + 0.5at^2
Let's assume the time to travel the first half is t_half. The initial velocity (u) is 0 since the train starts from rest, the acceleration (a) is equal to the previously found value, and the distance (s) is 5225 m. Plugging in these values, the equation becomes:
5225 = 0.5 * a * t_half^2
Simplifying, we find:
t_half = √(10450 / a)