Physics, asked by wisdomhopempozi5, 10 months ago

Two electric trains A and B leave the same station on parallel lines. The train A starts from rest with a uniform acceleration of 0.2 m/s2 and attains a speed of 45 km.p.h., which is maintained constant afterwards. The train B leaves 1 minute after with a uniform acceleration of 0.4 m/s2 to attain a maximum speed of 72 km.p.h., which is maintained constant afterwards. When will the train B overtake the train A ?

Answers

Answered by aristocles
28

Answer:

Train B will overtake Train A after 2.74 min of start

Explanation:

As we know that train A is moving with uniform acceleration a = 0.2 m/s/s

It starts from rest and reached to its maximum speed of 45 kmph

so we have

v = 45 \times \frac{5}{18}

v = 12.5 m/s

time to reach this speed is given as

12.5 = 0 + 0.2 t

t = 62.5 s

distance covered by the train in such time

d = 0 + \frac{1}{2}(0.2)62.5^2

d = 390.625 m

Now train B start 1 min after train A with acceleration a = 0.4 m/s/s

Its maximum speed is 72 kmph

so we will have

v_b = 72 \times \frac{5}{18}

v_b = 20 m/s

now time to reach this speed

20 = 0 + 0.4 t

t = 50 s

distance covered by train B in this time

d = \frac{1}{2}(0.4)(50^2)

d = 500 m

now let say t second after this train B overtake train A

so we will have

total time of motion for train A is (t + 60) s

d_a = d_b

390.625 + 12.5(60 + t - 62.5) = 500 + 20(t - 50)

390.625 + 12.5 t - 31.25 = 500 - 1000 + 20 t

859.375 = 7.5 t

t = 114.6 s

So train B will overtake train A after

t = 114.6 + 50 = 164.6 s = 2.74 min

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Topic : kinematics

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