Question

Two trains take 3 seconds to pass one another when going in opposite directions but only 2.5 second if the speed of one is increased by 50%. the time one would take to pass the other when going in the same direction at the original speed is

(a) 10 seconds

(b) 12 seconds

(c) 15 seconds

(d) 18 seconds

Answer 1

ofcourse ans will be (c)

see, t1=3 & t2=2.5

1/T=1/t2-1/t1

1/T=1/2.5-1/3

1/T=1/15

T=15sec

Answer 2

Answer:

One train would take 15 sec to pass the other train.

c is correct option.

Explanation:

Given that,

Time t = 3 sec

Let us consider,

Speed of first train = u

Speed of second train = v

The total distance traveled = l

When going in opposite direction with original speed,

The speed is defined as,

$v = \dfrac{d}{t}$

Where, d  = distance

t = time

Put the value into the formula

$v-(-u)=\dfrac{l}{3}$

$l = 3v+3u$.....(I)

Now, if the speed of first is increased by 50% and time taken 2.5 sec.

New speed of first train is,

$v'=v+0.5v=1.5v$

So, $1.5v+u=\dfrac{l}{2.5}$

$l=3.75v+2.5u$.....(II)

On subtracting equation (II) from equation (I)

$0.75v-1.5u=0$

$1.5v=u$....(III)

Now, when going in the same direction at the original speed is

The relative velocity = u-v

The time taken is defined as,

$t = \dfrac{D}{v}$

Put the value into the formula

$t=\dfrac{l}{u-v}$.....(IV)

Now, Put the value of l from equation (I) and u from equation (III) in the equation (IV)

$t =\dfrac{3\times1.5v+3v}{1.5v-v}$

$t = \dfrac{7.5}{0.5}$

$t=15\ sec$

Hence, One train would take 15 sec to pass the other train.