Between two stations a train acclerates uniformly at first, then moves with constant velocity and finally s uniformly. If the ratio of the time taken be 1:8:1 and the max speed attained be 60km/h then what is the average speed over the whole journey?
12km/hr
To find the average speed of the entire journey, we need to calculate the total distance traveled and divide it by the total time taken.
Let's assume the total distance traveled is D.
The ratio of the time taken for acceleration, constant velocity, and retardation is given as 1:8:1. So, let's calculate the time taken for each phase.
Let T1 be the time taken for acceleration, T2 be the time taken for constant velocity, and T3 be the time taken for retardation.
Given that the ratio of the time taken is 1:8:1, we can write:
T1 : T2 : T3 = 1 : 8 : 1
Let's assume the common ratio between the time taken is x. Therefore, we have:
T1 = x,
T2 = 8x,
T3 = x.
Now, let's calculate the distance traveled during each phase.
During acceleration, the speed increases uniformly from 0 to the maximum speed of 60 km/h. Since we know that speed = (acceleration) × (time), and the acceleration is uniform, we can calculate the distance traveled during acceleration using the formula: distance = (initial speed) × (time) + (1/2) × (acceleration) × (time)^2.
The initial speed during acceleration is 0 km/h, acceleration = (change in speed) / (time). As the maximum speed is reached, the change in speed is 60 km/h, and time taken is T1 = x.
distance1 = 0 × x + (1/2) × [(60 - 0) / x] × x^2.
Simplifying, we have:
distance1 = (1/2) × (60) × x.
During constant velocity, the speed remains constant at 60 km/h. Therefore, the distance traveled during this phase is: distance2 = (60 km/h) × (T2) = 60 × 8x.
During retardation, the speed decreases uniformly from 60 km/h to 0 km/h. Similar to acceleration, we can calculate the distance traveled using the formula: distance = (initial speed) × (time) - (1/2) × (acceleration) × (time)^2.
The initial speed during retardation is 60 km/h, acceleration = (change in speed) / (time). As the speed decreases to 0 km/h, the change in speed is 60 km/h, and time taken is T3 = x.
distance3 = (60 km/h) × (x) - (1/2) × [(60 - 0) / x] × x^2.
Simplifying, we have:
distance3 = 60x - (1/2) × (60) × x.
Now, let's calculate the total distance traveled (D):
D = distance1 + distance2 + distance3
D = (1/2) × 60 × x + 60 × 8x + 60x - (1/2) × 60 × x
D = 30x + 480x + 60x - 30x
D = 540x
Now, let's calculate the total time taken:
Total time = T1 + T2 + T3
Total time = x + 8x + x
Total time = 10x
Average speed = Total distance / Total time
Average speed = D / (Total time)
Average speed = (540x) / (10x)
Average speed = 54 km/h
Therefore, the average speed over the whole journey is 54 km/h.
avg velocity when accelerating or deaccelerating = 30km/hr
So assume t=10 hrs
first hour, distance=30km/hr*1hr
last hour, distnace=30km/hr*1hr
middle time, distance=60km/hr*8=480km
avg velocity=total distance/time=you do it.