A train at station P accelerate uniformly from rest until it attain a speed of 100 k/h. It then continues at the speed for some time and decelerate uniformly until it comes to a stop at a station, Q 60 km from P. The total time taken for the journey is an hour. If the rate of deceleration is twice that of the acceleration. Calculate the: (i) Time taken during when the constant is maintained. (ii) Acceleration of the train
To solve this problem, we can use the equations of motion for uniformly accelerated motion:
1. The first equation is v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time.
2. The second equation is s = ut + (1/2)at^2, where s is the displacement.
3. The third equation is v^2 = u^2 + 2as.
Let's solve the problem step by step:
(i) Time taken during constant speed:
We know that the total time taken for the journey is one hour. Let's assume that the time taken to accelerate and decelerate is t1, and the time taken during constant speed is t2.
The total distance traveled during acceleration and deceleration is 60 km (from P to Q). So we have:
Distance during acceleration = Distance during deceleration = 60 km / 2 = 30 km.
Using the second equation of motion, for the distance during acceleration:
30 = 0*t1 + (1/2)a*t1^2.
Simplifying this equation, we get:
15 = (1/2)a*t1^2.
Similarly, for the distance during deceleration:
30 = 0*t1 + (1/2)(-2a)t1^2.
Simplifying this equation, we get:
15 = (-1/2)a*t1^2.
Now, let's consider the time taken for constant speed, t2. Since the train is traveling at a constant speed of 100 km/h, the distance covered during this time is:
Distance during constant speed = Speed * Time
= 100 * t2 km.
So, the remaining distance, after acceleration and deceleration, is:
Remaining distance = Total distance - Distance during acceleration - Distance during deceleration
= 60 km - 30 km - 30 km
= 0 km.
We can conclude that the train comes to a stop at Station Q. Therefore, t2 = 0.
(ii) Acceleration of the train:
Given that the rate of deceleration is twice that of the acceleration, we can write:
Deceleration (a2) = 2 * Acceleration (a1).
Using the first equation of motion for the deceleration part:
0 = 100 km/h + a2 * t1.
Since t1 is the same for both acceleration and deceleration, we can substitute the value of a2:
0 = 100 km/h + (2a1) * t1.
Simplifying this equation, we get:
-100 km/h = 2a1 * t1.
Now, using the first equation of motion for the acceleration part:
100 km/h = 0 + a1 * t1.
Simplifying this equation, we get:
100 km/h = a1 * t1.
Now, we can substitute this value in the previous equation:
-100 km/h = 2 * (100 km/h) * t1.
Simplifying this equation, we get:
-1 = 2t1.
Therefore, t1 = -1/2.
Since time cannot be negative, we discard the negative value. Hence, t1 = 1/2 hour.
Now, using the second equation of motion for the distance during acceleration:
30 km = 0 * (1/2) + (1/2)a1 * (1/2)^2.
Simplifying this equation, we get:
30 = (1/8)a1.
Therefore, a1 = 240 km/h^2.
So, the answers to the questions are:
(i) Time taken during constant speed is t2 = 0.
(ii) Acceleration of the train is a1 = 240 km/h^2.