a train takes 5 minutes to cover a distance of 3 km between two stations p and q . Starting from rest at p , it accelerate at a constant rate to a speed 40 km/h and maintain this speed untill it is brought to rest at q . If train takes 3 times as long to deaccelerate as it does to accelerate , find time taken by the train to accelerate .

40 km/h = 40,000 m/ 3600 s = 11.1 m/s

5 min = 300 s

time to accelerate = t

d1 = (1/2) a t^2
11.1 = a t
so
d1 = (1/2)(11.1) t
d1 = 5.56 t

runs at 11.1 m/s for t2 seconds
distance = d2 = 11.1 t2

deaccelerate time = 3 t
0 = 11.1 + 3 a2 t
a2 t = -11.1/3 = - 3.7 t
a2 = -3.7/t
d3 = 11.1 (3t) + (1/2) (-3.7 /t )(3t)^2
d3 = 33.3 t - 16.7 t
d3 = 16.7 t

so

5.56 t + 11.1 t2 + 16.7 t = 3000 meters
t + t2 + 3 t = 300 seconds

so t2 = (300 -4t)
solve for t

T = 148

Let's assume the time taken to accelerate is 't' minutes.

To find the time taken to accelerate, we need to consider two parts of the journey: the acceleration phase and the deceleration phase.

During the acceleration phase:
- The train starts from rest and accelerates at a constant rate, reaching a speed of 40 km/h.
- The distance covered during the acceleration phase can be found using the formula: distance = (initial velocity * time) + (0.5 * acceleration * time^2).
- The initial velocity in this case is 0 km/h, and the acceleration can be calculated using the formula: acceleration = (final velocity - initial velocity) / time.

Given that the train covers a distance of 3 km during this phase, we can set up the following equation:

3 = (0.5 * ((40 - 0) / t) * t^2)

Simplifying the equation, we get:

3 = 20t

Dividing both sides by 20:

t = 3/20

So, the time taken to accelerate is 3/20 minutes.

Now, according to the problem, the time taken to decelerate is 3 times the time taken to accelerate. Therefore, the time taken to decelerate is:

3 * (3/20) = 9/20 minutes.

Therefore, the time taken by the train to accelerate is 3/20 minutes.

To find the time taken by the train to accelerate, let's break down the problem into different stages.

1. Acceleration Stage:
We are given that the train accelerates at a constant rate until it reaches a speed of 40 km/h. Let's assume the time taken for acceleration is 't' minutes.

We know that acceleration is the rate of change of velocity. Therefore, we can use the equation:
Acceleration (a) = Change in velocity (v) / Time taken (t)

The initial velocity is 0 km/h, and the final velocity is 40 km/h.

Change in velocity = Final velocity - Initial velocity
Change in velocity = 40 km/h - 0 km/h = 40 km/h

Now, we need to convert the time taken ('t') to hours to match the units of velocity.

1 hour = 60 minutes

Therefore, time taken (t) = t/60 hours.

Using the formula for acceleration, we can rewrite the equation as:
a = v / t
a = 40 km/h / (t/60) hours

2. Deceleration Stage:
We are told that the train takes three times as long to decelerate as it does to accelerate. So, the time taken for deceleration is 3t minutes.

3. Total Time:
The total time taken by the train is the sum of the acceleration time, the constant speed time, and the deceleration time.
Since the distance covered is constant, we can use the formula:
Distance = Speed x Time

Given that the distance between the stations P and Q is 3 km,

Distance = 3 km
Total Time = Time for Acceleration + Time at Constant Speed + Time for Deceleration

Time for Acceleration + Time at Constant Speed + Time for Deceleration = 5 minutes

Now, let's substitute the values and solve the equations to find the time taken for acceleration.

Using the distance formula,
Distance = Speed x Time

During acceleration: Distance = 0.5 * a * (t/60)^2 -----(1) [Using v = u + at = 0 + a(t/60)]

During constant speed: Distance = 40 * (5 - t - 3t)/60 -----(2) [Using v = u + at -> 0 = 40 - a(t/60)]

During deceleration: Distance = 0.5 * a * (3t/60)^2 -----(3) [Using v = u + at -> 0 = 40 - 3a(t/60)]

Total Distance = Distance during acceleration + Distance during constant speed + Distance during deceleration
3 km = 0.5 * a * (t/60)^2 + 40 * (5 - t - 3t)/60 + 0.5 * a * (3t/60)^2

Simplifying this equation will give us the value of 't', representing the time taken for acceleration.