A train starts from rest with uniform acceleration of 0.25 m/s^2 and attains a speed of 45 kmph which subsequently remains constant. One minute after the start of train A, another train B steams off on a parallel track with uniform acceleration of 0.45 m/s^2. If the maximum speed gained by train B is 60 kmph, how much time it will take for train B to overtake train A?
Answers
A train starts from rest with uniform acceleration of 0.25 m/s^2 and attains a speed of 45 kmph which subsequently remains constant. One minute after the start of train A, another train B steams off on a parallel track with uniform acceleration of 0.45 m/s^2. If the maximum speed gained by train B is 60 kmph
we have to find the time it when train B overtakes train A.
for train A :
u₁ = 0, v₁ = 45 km/h = 45 × 5/18 = 12.5 m/s
a₁ = 0.25 m/s²
using formula, v = u + at
⇒12.5 = 0 + 0.25 × t
⇒t = 50 seconds
hence, after 50 sec, velocity of train A becomes 12.5 m/s which subsequently remains constant.
means acceleration after 50 seconds of train A is zero.
distance travelled in 50 sec , s = 1/2 × 0.25 × (50)²
= 1/2 × 0.25 × 2500
= 1/2 × 625
= 312.5 m
distance travelled in 50 to 60 , s' = 12.5 m/s × 10s
= 125 m
now total distance travelled by train A in 60 second , S = s + s' = 312.5 + 125 = 437.5 m
for train B :
u₂ = 0 , a₂ = 0.45 m/s² , v₂ = 60 km/h = 16.67 m/s
using formula, v = u + at
⇒ 16.67 = 0 + 0.45 × t
t = 37 seconds
so, distance travelled in 37 seconds , S" = 0 + 1/2 × 0.45 × (37)²
≈ 308 m
hence after 37 seconds train B moves constantly.
let after time t, train B overtakes train A
distance travelled by train A in 1 min + distance travelled by train A in t sec = distance travelled by train B in t sec
⇒437.5 + 12.5t = 308 + 16.67 t
⇒129.5 = 4.17t
⇒t = 129.5/4.17 = 31 seconds
therefore, 31 seconds would take for train B to overtake train A