A train starts from rest and attains a speed of 15 m/s in 10 seconds and produce acceleration of 2.5 m/s (square) it stops in 6 seconds after the breaks apply find the distance covered by train
Huh?
v = a t
so
a = 15/10 = 1.5 m/s^2 not 2.5 m/s^2
Something is wrong with the problem statement.
To find the distance covered by a train, you need to calculate the total distance traveled during acceleration and the total distance traveled during deceleration.
First, let's calculate the distance traveled during acceleration.
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 15 m/s
Time taken (t) = 10 seconds
Acceleration (a) = 2.5 m/s^2
We can use the equation of motion to calculate the distance traveled during acceleration:
v = u + at
Rearranging the equation, we have:
s = ut + (1/2)at^2
Substituting the given values, we get:
s1 = 0(10) + (1/2)(2.5)(10^2)
s1 = 125 m
So, the distance traveled during acceleration is 125 meters.
Next, let's calculate the distance traveled during deceleration.
Given:
Initial velocity (u) = 15 m/s
Final velocity (v) = 0 m/s
Time taken (t) = 6 seconds
Acceleration (a) = -2.5 m/s^2 (negative because it is deceleration)
Using the equation of motion:
s = ut + (1/2)at^2
Substituting the given values:
s2 = 15(6) + (1/2)(-2.5)(6^2)
s2 = 90 - 45
s2 = 45 m
So, the distance traveled during deceleration is 45 meters.
To find the total distance covered by the train, we need to add the distances traveled during acceleration and deceleration:
Total distance = s1 + s2
Total distance = 125 + 45
Total distance = 170 meters
Therefore, the train covered a distance of 170 meters.