Question

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Answer 1

Hey..... ☺️

Given,

The initial velocity of both train A and train B, u = 72 km/h

$\frac{72 \: \times 1000}{60 \times 60} = 20ms$

Length of each train, l = 400 m

Acceleration of train A = 0 ms-2

Acceleration of train B = 1 ms-2

Time to overtake, t = 50 s

Distance to be covered to overtake,

Let, Distance traveled by train B in "t" time= SB

Distance traveled by train A in "t" time = SA

We have,

sB + sA= initial distance between the trains (d)

From 1st equation of motion,

v = u + at

where,

v = Final velocity

u = Initial velocity

a = Acceleration/Deceleration

t = Time

Final velocity of train A, vA = 20 + 0 = 20 m/s

Final velocity of train B, vB = 20 + (1×50) = 70 m/s

From 2nd equation of motion,

s = u × t + 0.5 × a × t2

where,

u = Initial velocity

a = Acceleration or Deceleration

s = Distance covered

t = Time

∴

sA = (20×50) + (0.5×0×502) = 1000 m

sB = (20×50) + (0.5×1×502) = 2250 m

∴ Initial distance between trains, d = 2250 -1000

= 1250 m

Hope this helps ☺️

Answer 2

☆$\red {HELLO\:DEAR}$☆

Given,

The initial velocity of both train A and train B, u = 72 km/h

\frac{72 \: \times 1000}{60 \times 60} = 20ms

60×60

72×1000

=20ms

Length of each train, l = 400 m

Acceleration of train A = 0 ms-2

Acceleration of train B = 1 ms-2

Time to overtake, t = 50 s

Distance to be covered to overtake,

Let, Distance traveled by train B in "t" time= SB

Distance traveled by train A in "t" time = SA

We have,

sB + sA= initial distance between the trains (d)

From 1st equation of motion,

v = u + at

where,

v = Final velocity

u = Initial velocity

a = Acceleration/Deceleration

t = Time

Final velocity of train A, vA = 20 + 0 = 20 m/s

Final velocity of train B, vB = 20 + (1×50) = 70 m/s

From 2nd equation of motion,

s = u × t + 0.5 × a × t2

where,

u = Initial velocity

a = Acceleration or Deceleration

s = Distance covered

t = Time

∴

sA = (20×50) + (0.5×0×502) = 1000 m

sB = (20×50) + (0.5×1×502) = 2250 m

∴ Initial distance between trains, d = 2250 -1000

= 1250 m

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