An electric train leaves a station starting from rest and attains a speed of 72km/hr in 10 second. It travels at that speed for 120 seconds. Then it undergoes uniform retardation for 20 seconds to come to halt at the next station calculate :
a) distance between the train stations.
b) the average velocity of train.
Ana is yes
Com
To calculate the distance between the train stations, we can break down the motion of the train into three stages:
1) Acceleration
2) Uniform motion
3) Deceleration
Step 1: Acceleration
Given:
Initial velocity (u) = 0 km/hr
Final velocity (v) = 72 km/hr
Time (t) = 10 seconds
Using the formula:
v = u + at
where a is the acceleration, we can rearrange the formula to solve for acceleration:
a = (v - u) / t
a = (72 - 0) / 10
a = 7.2 m/s^2 (converting km/hr to m/s)
Using the formula for displacement during constant acceleration:
s1 = ut + (1/2) a t^2
s1 = 0 + (1/2) * 7.2 * 10^2
s1 = 360 meters
Step 2: Uniform motion
Given:
Velocity (v) = 72 km/hr
Time (t) = 120 seconds
Using the formula for displacement during uniform motion:
s2 = v * t
s2 = (72 * 1000) / 3600 * 120
s2 = 2400 meters
Step 3: Deceleration
Given:
Initial velocity (u) = 72 km/hr
Final velocity (v) = 0 km/hr (since the train comes to a halt)
Time (t) = 20 seconds
Using the formula:
v = u + at
0 = 72 + (-a) * 20
-72 = -20a
a = 3.6 m/s^2
Using the formula for displacement during constant acceleration:
s3 = ut + (1/2) a t^2
s3 = 72 * 20 + (1/2) * 3.6 * 20^2
s3 = 1440 + 720
s3 = 2160 meters
Now, to calculate the total distance between the train stations, we add the distances from each stage:
Distance = s1 + s2 + s3
Distance = 360 + 2400 + 2160
Distance = 4920 meters
Therefore, the distance between the train stations is 4920 meters.
To calculate the average velocity of the train, we use the formula:
Average velocity = Total distance / Total time
Total distance = 4920 meters (calculated above)
Total time = 10 + 120 + 20 = 150 seconds
Average velocity = 4920 / 150
Average velocity ≈ 32.8 m/s
Therefore, the average velocity of the train is approximately 32.8 m/s.
To calculate the distance between the train stations and the average velocity of the train, we need to apply some physics concepts and equations. Let's break down the problem step by step.
Step 1: Calculate the acceleration of the train.
The train starts from rest and attains a speed of 72 km/hr in 10 seconds. We can calculate the acceleration using the equation:
acceleration = (final velocity - initial velocity) / time
Converting the final velocity to m/s:
final velocity = 72 km/hr = (72 * 1000) / 3600 m/s = 20 m/s
Converting the initial velocity to m/s (since the train starts from rest):
initial velocity = 0 m/s
Plugging in the values into the formula:
acceleration = (20 m/s - 0 m/s) / 10 s = 2 m/s^2
Step 2: Calculate the distance traveled while the train is at a constant speed.
The train travels at a constant speed of 72 km/hr for 120 seconds. We can calculate the distance using the equation:
distance = speed * time
Converting the speed to m/s:
speed = 72 km/hr = (72 * 1000) / 3600 m/s = 20 m/s
Plugging in the values into the formula:
distance = 20 m/s * 120 s = 2400 meters
Step 3: Calculate the deceleration (retardation) of the train.
The train undergoes uniform retardation for 20 seconds to come to a halt. We can calculate the retardation using the equation:
retardation = (final velocity - initial velocity) / time
Since the train comes to a halt, the final velocity is 0 m/s. The initial velocity is the same as the speed when the train was traveling at a constant speed, which is 20 m/s.
Plugging in the values into the formula:
retardation = (0 m/s - 20 m/s) / 20 s = -1 m/s^2
Note: The negative sign indicates deceleration or retardation.
Step 4: Calculate the time it takes for the train to decelerate.
The retardation is given as -1 m/s^2 and we need to find the time it takes for the train to come to a halt. We can use the equation:
retardation = (final velocity - initial velocity) / time
Plugging in the values:
-1 m/s^2 = (0 m/s - 20 m/s) / time
Simplifying the equation:
-1 m/s^2 = -20 m/s / time
Cross-multiplying and solving for time:
time = -20 m/s / -1 m/s^2 = 20 seconds
Step 5: Calculate the total distance between the train stations.
The total distance is the sum of the distance traveled while the train was at a constant speed and the distance traveled while decelerating.
Distance traveled while at constant speed = 2400 meters
Distance traveled while decelerating = 0.5 * acceleration * time^2 (using the equation of motion)
Plugging in the values:
Distance traveled while decelerating = 0.5 * (-1 m/s^2) * (20 s)^2 = 200 meters
Total distance = Distance traveled while at constant speed + Distance traveled while decelerating
Total distance = 2400 meters + 200 meters = 2600 meters
So, the distance between the train stations is 2600 meters.
Step 6: Calculate the average velocity of the train.
The average velocity can be calculated using the formula:
Average velocity = total distance / total time
Total time = time taken to accelerate + time taken while at constant speed + time taken to decelerate
Total time = 10 s + 120 s + 20 s = 150 s
Plugging in the values:
Average velocity = 2600 meters / 150 s = 17.33 m/s (rounded to two decimal places)
So, the average velocity of the train is 17.33 m/s.
hint - it decelerates at half the rate it accelerated.
Now recall your basic equation of motion:
s(t) = vt + 1/2 at^2
and apply it to the three intervals.