Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

There is a roadway between city A and B . A car P starts at 5:00 am from A and reaches B at 10:00 am. Another car Q starts from B at 7:00 am and reaches A at 9:00 am. Find the time when car P meets car Q ?

I did as follows

for car P , travelling from 5:00 am to 7:00 am cannot definitely meet car Q as Q has not even started . Lets say at 7:00 am car P reached a point R, and lets assume total road way distance be D

so $D_{AR} = \frac{2\times D}{5} $

so lets see from 7:00 am to 9:00 am when they will meet . Let the cars P and Q are said to meet at a distance $D_{1}$ from R and $D_{1}$ from B. The velocity ratio is inversely proportional to time ratio so it is $2/5$

$$ \frac{D_{1}}{D_{2}} = \frac{2}{5}$$ As distance travelled by car P between 7:00 to 9:00 (that is 2 hrs ) is $\frac{2\times D}{5} $

so the cars meet at $D_{AR} + \frac{2}{7}\times \frac{2\times D}{5} $ = $ \frac{18 \times D }{35} $ from city A

SO Time is calculated as like

To travel D distance if it takes 5 hours , then to travel $ \frac{18 \times D }{35} $ it should take

$\displaystyle\frac{5}{D} \times \frac{18D}{35}$

So $\frac{18}{7}$ hours . Am I right ? If wrong please correct .

Even if my answer is correct please suggest if there are easier and logical ways to solve this sort of problems without equations . In the above also I have tried avoiding equations by using simple logic Distance proportional to velocity when speed is constant.

share|improve this question

3 Answers 3

up vote 3 down vote accepted

Let the meeting time be $t$, a real number with the whole portion being hours. At time $t$, $P$ has covered $\frac {t-5}5$ of the distance. $Q$ has covered $\frac {t-7}2$ of the distance. These have to add to the whole distance, so $$\frac {t-5}5+\frac {t-7}2=1\\2t-10+5t-35=10\\7t=55\\t=7\frac67\approx 7:51:25.714$$

share|improve this answer
    
Thanks for the reply.Please, Can you also state whats the error in my working ? Because I am getting 18/7 hours which is 7:34 am. Please correct me. –  Harish Kayarohanam Aug 15 '13 at 21:25
    
in your line "so the cars meet at", the part $\frac {2D}5$ should be $\frac {3D}5$ because that is how much is left. –  Ross Millikan Aug 15 '13 at 22:06
    
thanks for pointing out the problem . –  Harish Kayarohanam Aug 15 '13 at 22:57

Let's say that cars velocity is constant.

Then car B travells $\frac{1}{2}$ of the path in one hour and car A travells $\frac{1}{5}$ of the path. We know that they move each to each other so when they meet the sum of the distances they've passed will be equals as the distance between place A and place B.

Then we'll have:

$$\frac{x}{2} + \frac{x+2}{5} = 1$$

$$\frac{5x + 4 + 2x}{10} = 1$$

$$7x + 4 = 10$$

$$7x = 6$$

$$x = \frac{6}{7}$$

It means that they'll meat $\frac{6}{7}$ hours after the second car have started travelling.

share|improve this answer
    
Ya I have edited it , It is waiting his approval –  Harish Kayarohanam Aug 15 '13 at 21:34
1  
Thanks for editing, it's just a typo and i continued later using $3x$, instead of $2x$ –  Stefan4024 Aug 15 '13 at 21:38

Normalize the distance to $1$. Let $x$ be the location they meet at time $t$. The distance traveled by P is $x=v_Pt, $ similarly for Q the distance traveled is $1-x=v_Q(t-2).$ $v_P$ and $v_Q$ will be $\frac15,$ and $\frac12$ respectively based on the total travel time. Solve the two equations for $t$ and $x$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.