A train starts from rest and accelerates uniformly at a rate of 4m/s^2 attains to maximum velocity of 40m/s in certain time t seconds. It then maintains a constant speed for 100s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50s then, Calculate: (i) Time ‘t’ (ii) Distance covered during the acceleration (iii) Total Distance covered (iv) Retardation
Plz answer with proper steps
Answers
Given:
Initial velocity, u = 0
Acceleration, a = 4 m/s²
Final velocity, v = 40 m/s
Time, t = t sec
(i) To calculate time t
We know,
v = u + at
40 = 0 + (a × t)
40 = 4t
t = 40/4 = 10 sec
(ii) To calculate distance (s1) covered during the acceleration
We know,
s = ut + 1/2at²
s1 = (0 × 10) + (1/2 × 4 × 10²)
s1 = 2 × 100
s1 = 200 m
(iii) To calculate total distance covered
In the initial 10 seconds, distance travelled (s1) = 200 m
For the next 100 seconds, distance travelled (s2)
= Velocity × time
= 40 × 100
= 4000 m
For the next 50 seconds while retardation,
u’ = 40 m/s
v’ = 0
t = 50 seconds
Retardation = r
=> v' = u' - rt
=> 0 = 40 - r(50)
=> r50 = 40
=> r = 40/50
=> r = 0.8 m/s²
Therefore, distance (s3) travelled while retardation
s3 = u't + 1/2rt²
s3 = (40 × 50) - 1/2 × 0.8 × 50²
s3 = 2000 - 1000
s3 = 1000
Total distance, T = s1 + s2 + s3
T = 200 + 4000 + 1000
T = 5200 m
(iv) To calculate retardation
As calculated above,
Retardation, r = 0.8 m/s²
__________________________
(i) Time ‘t’ = 10 seconds
(ii) Distance covered during the acceleration = 200 m
(iii) Total Distance covered = 5200 m
(iv) Retardation = 0.8 m/s²