Problems on Trains

Question

A train Leaves station $x$ at $5 AM$ and reaches station $y$ at $9AM$. Another train leaves station $y$ at $7AM$ and reaches station $x$ at $10.30 AM$ . At what time do the two trains cross each other?

I have tried:

Let the trains cover the (same) distance $X$. The time taken by the first train to cover the distance $X$ is $4$ Hours & the time taken by the second train to cover the distance $X$ is $3.5$ Hours.

Equating the distance I got,

$$Speed *4=speed*3.5$$

Speed of First Train be $4$, which will have time $3.5$ hours, again second Train speed be $3$, time $4$ hours

Total Distance be $13.2$

Relative speed $= 7.5$

Time taken to cross each other $= 13.2/7.5=1.76$

so $8.76AM$

But the Answer is $7.56 AM$

What have I done wrong? Please anyone guide me.

Answer 1

Let's start with where you have gone wrong.

Firstly, the total distance is $4\times3.5=14$ rather than $13.2$.

Secondly, as user69468 has rightly pointed out, the second train starts $2$ hours later, by which time the first train has covered half the distance, which is $7$ - this is when the relative speed becomes $7.5$. As a result, the time taken will be $7/7.5$, which is $56$ minutes, hence the time will be $7.56$.

Rather than taking the relative velocity, a less confusing approach is to consider the following.

Let the distance between stations x and y be $X$. The earlier train has speed $X/4$, the later train has speed $X/3.5=2X/7$. Let $t$ be the time the trains cross each other after the later train has started, which happens when the total distance cross covered by both trains is $X$.

The earlier train starts $2$ hours ahead, so it has covered distance $(2+t)\times(\frac{X}{4})$. As for the later train, it has covered distance $t\times\frac{2X}{7}$, so we end up with the following equation:- $$(2+t)\left(\frac{X}{4}\right)+\frac{2tX}{7}=X$$ solving for which leads to $t=\frac{14}{15}\text{ hours}$.

The trains will meet at $7+ t=7.56$

Answer 2

There is a time difference one starts at 5. Another at 7. So only after two hours you take relative velocity.

So after 2 hours at 7 they have already crossed half the distance. Then add the time you got by dividing half the distance by relative velocity

Answer 3

Let $d$ denote the distance in km between the stations.

The $1$st train drives $d$ km within $9-5=4$ hours, hence at a speed of $\frac14d$ km per hour.

The $2$nd train drives $d$ km within $10.5-7=3.5$ hours, hence at a speed of $\frac27d$ km per hour.

Let $t$ denote the time at which the trains meet.

Hence $(t-5)\cdot\frac14d+(t-7)\cdot\frac27d=d$.

Hence $7d(t-5)+8d(t-7)=28d$.

Hence $7(t-5)+8(t-7)=28$.

Hence $7t-35+8t-56=28$.

Hence $15t=119$.

Hence $t=7+\frac{14}{15}$.

Hence $t=7:56$.

Answer 4

Let $v_1$ , $v_2$ be the velocities of the two trains and $d$ the distance. You know that:

$$ v_1\cdot \Delta t_1=v_2 \Delta t_2=d $$

where $\Delta t_1=4$ and $\Delta t_2=3.5$, so that: $4v_1=3.5 v_2$. From this you cannot deduce the values of $v_1$ and $v_2$ or the distance $d$, but only that: $$ v_2=\frac{8}{7} v_2 \quad \mbox{and}\quad d=4v_1 $$

(this is your mistake, I suppose).

Now, at $7$ am, the position of the first train is: $$ s_{1,0}=v_1 \cdot (7-5)=2v_1 $$ and the position of the second train is $$s_{2,0}=d$$

so the equation of motion of the two trains, starting at $7$ am are:

$$ s_1=v_1t+s_{1,0}=v_1(t+2) $$ $$ s_2=s_{2,0}-v_2t $$

and they meet at the time $t$ ( from $7$ am) when $s_1=s_2$. Substituting this gives the equation $$ t+2=4-\frac{8}{7}t $$ that, solved, gives you the time $t_m$ such that $7+t_m$ is the time of meeting.