The distance between two stations is 40 km. A train takes 1 hour to travel this distance. The train after starting from the first station moves with constant acceleration for 5 km; then it moves with constant velocity for 20 km and finally its velocity keeps on decreasing continuously for 15 km and it stops at the other station. Find the maximum velocity of the train.
Answers
Answer:
60 km/Hr
Explanation:
Total Distance = 40 km
Time taken = 1 Hr
Let say train take T₁ time for 5 km
Then Distance Covered = 0T₁ + (1/2)aT₁² = 5
=> aT₁² = 10
=> a = 10/T₁²
Velocity after 5 km = 0 + aT₁ = T₁10/T₁² = 10/T₁
then time taken for 20 km = 20/(10/T₁) = 2T₁
Let say train take T₂ time for next 15 km
deceleration = (10/T₁T₂)
Distance Covered = 15 = (10/T₁ )T₂ - (1/2)(10/T₁T₂)T₂²
=> 15 = 5T₂ /T₁
=> 3 = T₂ /T₁
=> T₂ = 3T₁
total time = T₁ + 2T₁ + 3T₁ = 6T₁ = 1 hr = 60 min
=> T₁ = 10 Min = 1/6 hr
Velocity = 10/T₁ = 10/(1/6) = 60 km/Hr
Answer:
60km/h
Explanation:
time taken to travel distance 5km
5=0t1+1/2at1^2
10=at1^2
a=10/t1^2
velocity after 5km is v=10/t1
time taken for 20km =2t1
now t2 for next 15km
deceleration
15=5t2/t1
t2=3t1
total time = t1+2t1+3t1 =6t1=60 min
t1=10min
=1/6 hr
v=10/t1 =10/(1/6)
=60km/h