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 = 0.278 m/s
If it accelerated at a m/s^2, then it took t=0.278/a seconds to achieve that speed.
The distance covered while accelerating/decelerating is thus
1/2 at^2 + 0.278(3t) - 1/2 (a/3)(3t)^2
= 1/2 a(0.278/a)^2 + 0.278(3(0.278/a))-1/2 (a/3)(3(0.278/a))^2
= 0.155/a
Now, if it travels at constant speed for x seconds, we have
0.155/a + 0.278x = 3000
4(0.278/a) + x = 300
Solve for a. As always, double-check my math.
Big OOOPS! I used a speed of 1 km/hr, not 40. See Damon's "physics" solution below.
To find the time taken by the train to accelerate, let's break down the problem step by step:
Step 1: Find the time taken to deaccelerate:
The problem states that it takes three times as long to decelerate as it does to accelerate. Let's denote the time taken to accelerate as 't' minutes. Therefore, the time taken to decelerate will be 3t minutes.
Step 2: Convert minutes to hours:
Since the given speed is in km/h, let's convert the time from minutes to hours for uniformity. There are 60 minutes in an hour, so we divide the time in minutes by 60 to get the time in hours.
Step 3: Calculate the acceleration and deceleration rates:
To find the acceleration and deceleration rates, we need to use the formula:
Speed = (Initial Velocity) + (Acceleration × Time)
Since the train starts from rest, the initial velocity is 0 km/h. At the end of the acceleration phase, the speed reaches 40 km/h. We already know that the time for acceleration is 't' hours. Therefore, we can rearrange the formula to find the acceleration rate:
40 km/h = 0 km/h + (Acceleration × t)
Acceleration = 40 km/h ÷ t
Similarly, at the end of the deceleration phase, the speed becomes 0 km/h. The time for deceleration is 3t hours. Using the same formula, we can find the deceleration rate:
0 km/h = 40 km/h + (Deceleration × 3t)
Deceleration = -40 km/h ÷ (3t)
Note: The negative sign indicates that the train is slowing down.
Step 4: Calculate the distance traveled during acceleration and deceleration:
To calculate the distance traveled during acceleration and deceleration, we use the formula:
Distance = (Initial Velocity × Time) + (0.5 × Acceleration × Time²)
During acceleration:
The initial velocity is 0 km/h, and the acceleration is given by 40 km/h ÷ t. The time is 't' hours.
Distance during acceleration = (0 km/h × t hours) + (0.5 × (40 km/h ÷ t) × t²)
During deceleration:
The final velocity is 0 km/h, and the deceleration is given by -40 km/h ÷ (3t). The time is 3t hours.
Distance during deceleration = (40 km/h × 3t hours) + (0.5 × (-40 km/h ÷ (3t)) × (3t)²)
Step 5: Calculate the remaining distance:
The total distance between the two stations is given as 3 km. The distance traveled during acceleration and deceleration should sum up to 3 km.
Remaining distance = Total distance - (Distance during acceleration + Distance during deceleration)
3 km = [0 km/h × t hours + 0.5 × (40 km/h ÷ t) × t²] + [40 km/h × 3t hours + 0.5 × (-40 km/h ÷ (3t)) × (3t)²]
Step 6: Solve the equation to find 't':
Now, we have an equation with one unknown 't'. Simplify the equation and solve for 't'.
After solving the equation, you will find the value of 't', which represents the time taken by the train to accelerate.