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Two trains ‘A’ and ‘B’ leave the same station on parallel lines. Train ‘A’ starts with a uniform acceleration of 1/6 m/s2 and attains the speed of 24 km/hr, then stem is reduced to keep the speed constant. Train ‘B’ leaves 40 seconds after, with uniform acceleration of 1/3 m/s2 to attain the maximum speed of 48 km/hr. When will it overtake the train ‘A’?

User RealHowTo
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Answer:

Step-by-step explanation:

Let's first convert the speeds of the trains from km/hr to m/s:

Train A: 24 km/hr = (24 x 1000) / (60 x 60) = 6.67 m/s

Train B: 48 km/hr = (48 x 1000) / (60 x 60) = 13.33 m/s

Now, let's find the distance covered by train A until it reaches its constant speed:

v = u + at

where

u = initial velocity = 0

a = acceleration = 1/6 m/s^2

t = time taken to reach constant speed

At constant speed, v = 6.67 m/s

So, 6.67 = (1/6)t + 0

t = 40 seconds

Using the formula for distance covered during uniform acceleration:

s = ut + (1/2)at^2

The distance covered by train A during the acceleration phase is:

s = (1/2)(1/6)(40^2) = 133.33 m

Now, let's find the equation of motion for train B:

s = ut + (1/2)at^2

where

u = initial velocity = 0

a = acceleration = 1/3 m/s^2

t = time taken to overtake train A

At the time of overtaking, both trains will cover the same distance. Let's call this distance "d". So we have:

d = 133.33 + 6.67t (distance covered by train A + distance covered by train B)

Setting the equations for both trains equal to each other, we get:

133.33 + 6.67t = (1/2)(1/3)t^2 + (1/3)t^2

Simplifying and solving for t, we get:

t = 180 seconds

Therefore, train B will overtake train A after 180 seconds or 3 minutes.

User Mozzbozz
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