Question

6. Two trains each of length 100 m, moving in opposite directions along parallel lines, meet each other
with speeds of 50 kmph & 40 kmph. If their accelerations are 30 cm/s" and 20 cm/s, respectively,
find the time they will take to pass each other.​

Answer 1

Given data :-

Two trains each of length 100 m, moving in opposite directions along parallel lines, meet each other with speeds of 50 kmph & 40 kmph. Their accelerations are 30cm/s² and 20 cm/s² respectively.

Solution :-

Let, first train be A & second train be B

→ length of A = 100 m

→ length of B = 100 m

→ speed of A = 50 km/hr

→ speed of A = {[50×1000]/[60×60]}km/hr

→ speed of A = {50000/3600}km/hr

→ speed of A = {125/9} km/hr

→ speed of B = 40 km/hr

→ speed of B = {[40×1000]/[60×60]}km/hr

→ speed of B = {40000/3600}km/hr

→ speed of B = {100/9} km/hr

→ Acceleration of A = 30 cm/s²

→ Acceleration of A = 0.3 m/s²

→ Acceleration of B = 20 cm/s²

→ Acceleration of B = 0.2 m/s²

Now,

→ Relative displacement of trains, s{relative} = length of A + length of B

→ Relative displacement of trains, s{relative} = [100 + 100] m

→ Relative displacement of trains, s{relative} = 200 m .....( 1 )

→ Realative speed of train, u{relative}

= speed of A + speed of B

→ Realative speed of train, u{relative}

= [125/9 + 100/9] m/s

→ Realative speed of train, u{relative}

= [225/9] m/s

→ Realative speed of train, u{relative}

= 25 m/s ......( 2 )

→ Relative acceleration of trains, a{relative} = Acceleration of A + Acceleration of B

→ Relative acceleration of trains, a{relative}

= [0.3 + 0.2] m/s²

→ Relative acceleration of trains, a{relative}

= 0.5 m/s² or 1/2 m/s² ......( 3 )

Now to find time taken by trains to pass each other.

→ s{relative} = u{relative }× t + 1/2 × a{relative} × t²

{from ( 1 ) ( 2 ) & ( 3 )}

→ 200 = 25 × t + 1/2 × 1/2 × t²

→ 200 = 25 × t + 1/4 × t²

Multiply both side by 4

→ 800 = 100t + t² i.e.

→ - t² - 100t + 800 = 0 i.e.

→ t² + 100t - 800 = 0

Compair above equation with

at² + bt + c = 0

Hence, a = 1, b = 100 & c = -800

Now,

→ t = {-b ± √b²-4ac}/2a

→ t = {-100 ± √(100)²- 4×1×(-800)}/2×1

→ t = {-100 ± √10000 + 3200}/2

→ t = {-100 ± √13200}/2

→ t = {-100 ± 20√33}/2

→ t = - 50 ± 10√33

→ t = - 50 + 10√33 or t = - 50 - 10√33 i.e

→ t = 10√33 - 50 or t = - 10√33 - 50

We know that time is never in negative hence, t = 10√33 - 50 sec

Hence, time taken by trains to pass each other is 10√33 - 50 sec.

{Note:- 10√33 - 50 sec = 7.446 sec}