An electric train leaves a station starting from rest and attains a speed of 72km/hr in 10 second. Then it undergoes uniform retardation for 20 second to come to halt at the next station then calculate
i) the distance between the train stations.
ii) the average velocity of train.
Well, well, well, let's put on our conductor's hat and solve this riddle, shall we?
i) To find the distance between the train stations, we need to determine the distance traveled during the acceleration phase and the distance traveled during the deceleration phase.
During the acceleration phase, the train starts from rest and reaches a speed of 72 km/hr in 10 seconds. To find this distance, we can use the formula:
Distance = (Initial velocity * Time) + (0.5 * Acceleration * Time^2)
Since the train starts from rest, the initial velocity is 0 km/hr. The acceleration can be found using the formula:
Acceleration = (Final velocity - Initial velocity) / Time
Plugging in the values, we get:
Acceleration = (72 km/hr - 0 km/hr) / 10 s = 7.2 m/s^2
Now, let's find the distance traveled during the acceleration phase:
Distance_acc = (0 * 10) + (0.5 * 7.2 * 10^2)
= 0 + (0.5 * 7.2 * 100)
= 360 meters
During the deceleration phase, the train comes to a halt, meaning its final velocity is 0 km/hr. We have a uniform retardation (basically negative acceleration), so the equation remains the same. The only change is that the initial velocity is 72 km/hr and time is 20 seconds.
Plugging in the values, we get:
Acceleration_ret = (0 km/hr - 72 km/hr) / 20 s
= -3.6 m/s^2
Distance_dec = (72 * 20) + (0.5 * -3.6 * 20^2)
= 1440 - (0.5 * 3.6 * 400)
= 1440 - (0.5 * 3.6 * 400)
= 1440 - 720
= 720 meters
Now, to find the total distance between the train stations, we simply add the distances:
Total distance = Distance_acc + Distance_dec
= 360 meters + 720 meters
= 1080 meters
ii) Now, let's calculate the average velocity of the train. We know that average velocity is equal to the total displacement divided by the total time taken.
The total displacement here is 0 meters since the train starts and ends at the same position (the two stations).
The total time taken is the sum of the acceleration and deceleration times, which is 10 + 20 = 30 seconds.
Average velocity = Total displacement / Total time
= 0 meters / 30 seconds
= 0 m/s
So, the average velocity of the train is 0 m/s, indicating that the train ends up at the same spot it started. Must have been a teleporting train or some quantum physics stuff happening!
To find the distance between the train stations, we first need to find the distance covered by the train during the acceleration phase and the deceleration phase separately.
i) Distance covered during acceleration phase:
Initial speed (u) = 0
Final speed (v) = 72 km/hr = 20 m/s
Time taken (t) = 10 seconds
Using the formula: distance = (initial velocity + final velocity) / 2 * time
distance = (0 + 20) / 2 * 10
distance = 10 * 10
distance = 100 meters
ii) Distance covered during deceleration phase:
Initial speed (u) = 72 km/hr = 20 m/s
Final speed (v) = 0
Time taken (t) = 20 seconds
Using the formula: distance = (initial velocity + final velocity) / 2 * time
distance = (20 + 0) / 2 * 20
distance = 10 * 20
distance = 200 meters
Now, to find the total distance between the train stations, we add the distances covered during the acceleration and deceleration phases:
Total distance = 100 + 200
Total distance = 300 meters
iii) To find the average velocity of the train, we need to calculate the total displacement and divide it by the total time taken.
Total displacement = 0 (since the train starts and stops at the same station)
Total time taken = time taken for acceleration + time taken for deceleration
Total time taken = 10 + 20
Total time taken = 30 seconds
Average velocity = Total displacement / Total time taken
Average velocity = 0 / 30
Average velocity = 0 m/s
Therefore, the answers are:
i) The distance between the train stations is 300 meters.
ii) The average velocity of the train is 0 m/s.
To solve this problem, we can use the equations of motion. Let's break down the problem step by step:
Step 1: Find the acceleration of the train during the first phase.
Given:
Initial velocity, u = 0 (since the train starts from rest)
Final velocity, v = 72 km/hr = (72 * 1000) / 3600 m/s = 20 m/s
Time, t = 10 seconds
We can use the equation: v = u + at, where a is the acceleration.
Rearranging the equation, we have: a = (v - u) / t
Substituting the given values:
a = (20 - 0) / 10
a = 2 m/s²
Step 2: Find the distance traveled during the first phase.
We can use the equation: s = ut + (1/2)at²
Since the initial velocity is zero, the equation simplifies to: s = (1/2)at²
Substituting the values:
s = (1/2) * 2 * (10^2)
s = 100 m
Step 3: Find the retardation (negative acceleration) during the second phase.
Given:
Time, t = 20 seconds
Final velocity, v = 0 (since the train comes to a halt)
Using the equation: v = u + at
Rearranging the equation, we have: a = (v - u) / t
Substituting the values:
a = (0 - 20) / 20
a = -1 m/s²
Step 4: Find the distance traveled during the second phase.
Using the same equation as step 2 (s = ut + (1/2)at²), but with the negative value of acceleration:
s = (1/2) * (-1) * (20^2)
s = -200 m
The negative sign indicates that the displacement is in the opposite direction to the initial motion.
Now, to answer your specific questions:
i) The total distance between the train stations is the sum of the distances traveled in both phases:
Total distance = Distance in Phase 1 + Distance in Phase 2
Total distance = 100 m + (-200 m) = -100 m
ii) The average velocity of the train is the total displacement divided by the total time taken:
Average velocity = Total distance / Total time
Average velocity = (-100 m) / (10 s + 20 s) = -2 m/s
Note: The negative sign in the average velocity indicates that the train is moving in the opposite direction compared to the initial direction.
Therefore, the distance between the train stations is 100 meters and the average velocity of the train is -2 m/s.
a = 7.2km/hr-s
s = 1/2 * 7.2 * (10/3600)^2 + (72*20/3600 - 1/2 * 3.6 * (20/3600)^2) = .4km = 400m