Physics, asked by Stera, 11 months ago

Hlw Everyone

Question : A Bus starts at 6 :00 its journey from P and end it at Q at 11 : 00 another bus starts its journey at 9 :00 from Q and end its journey at 11 : 30 at P. Find the time when both the buses cross each other

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Answers

Answered by shadowsabers03

Assume the buses are moving with uniform velocity, i.e., no acceleration.

The time duration taken by the first bus to move from P to Q is,

$\longrightarrow\sf{t_1=5\ hours}$

The time duration taken by the second bus to move from Q to P is,

$\longrightarrow\sf{t_2=2.5\ hours}$

We know distance is the product of time and speed.

$\longrightarrow\sf{s=vt}$

For a constant displacement (the two buses move through same distance),

$\longrightarrow\sf{v\propto\dfrac{1}{t}}$

Thus,

$\longrightarrow\sf{\dfrac{v_1}{v_2}=\dfrac{t_2}{t_1}}$

$\longrightarrow\sf{\dfrac{v_1}{v_2}=\dfrac{2.5}{5}}$

$\longrightarrow\sf{\dfrac{v_1}{v_2}=\dfrac{1}{2}}$

$\longrightarrow\sf{v_2=2v_1\quad\quad\dots(1)}$

where $\sf{v_1}$ and $\sf{v_2}$ are speed of first and second buses respectively. But if $\overrightarrow{\sf{v_1}}$ is velocity of first bus, the velocity of second bus will be $-\overrightarrow{\sf{v_2}}$ because both are moving opposite to each other.

Since the first bus travels PQ in 5 hours with the speed $\sf{v_1,}$ we have,

$\longrightarrow\sf{PQ=5v_1}$

$\longrightarrow\sf{\dfrac{PQ}{v_1}=5\quad\quad\dots(2)}$

The diagram for the motion of first and second buses is given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\line(1,0){60}}\put(0,-4){$\sf{P}$}\put(60,-4){$\sf{Q}$}\put(36,-4){$\sf{R}$}\put(44,-4){$\sf{S}$}\put(0,3){\vector(1,0){10}}\put(60,3){\vector(-1,0){10}}\put(3,5){$\overrightarrow{\sf{v_1}}$}\put(52,5){$-\overrightarrow{\sf{v_2}}$}\end{picture}$

Let R be the point on the track PQ, where the first bus reaches at 9:00. We have to find how much the length of RQ is to that of PQ.

Here S is the point where both the buses cross each other. We are asked to find the time at which both buses reach S, or after which the first bus travels PS, actually.

[Note:- S can be calculated only if we consider the path RQ, because the two buses are not starting their journey at the same time, but they appear to move at the same time since the first bus reaches R.]

Since the first bus reaches R at 9:00, the time duration taken by the bus to reach R is 3 hours. So the bus can travel the distance PR in 3 hours, hence it can travel the rest RQ in 2 hours, because it can travel PQ in 5 hours.

For 5 hours the first bus can travel the distance PQ.

For 1 hour the first bus can travel one-fifth the distance PQ.

For 2 hours the first bus can travel two - fifth the distance PQ.

For 2 hours the first bus can travel RQ too. Hence,

$\longrightarrow\sf{RQ=\dfrac{2\,PQ}{5}\quad\quad\dots(3)}$

Now here is the path RQ.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\line(1,0){60}}\put(0,-4){$\sf{R}$}\put(60,-4){$\sf{Q}$}\put(20,-4){$\sf{S}$}\put(0,5){\vector(1,0){10}}\put(60,5){\vector(-1,0){10}}\put(3,7){$\overrightarrow{\sf{v_1}}$}\put(52,7){$-\overrightarrow{\sf{v_2}}$}\end{picture}$

Here the first bus moves with $\sf{v_1}$ and the second bus moves with $\sf{-v_2,}$ and let $\sf{t}$ be the time taken by first bus (in hours) to meet the second, or to travel RS.

Here we use relative motion concept, where motion of first bus with respect to the second is considered, thereby the second bus being at rest, whose diagram is given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\line(1,0){60}}\put(0,-4){$\sf{R}$}\put(60,-4){$\sf{Q}$}\put(20,-4){$\sf{S}$}\put(0,5){\vector(1,0){20}}\put(4,7){$\overrightarrow{\sf{v_1}}+\overrightarrow{\sf{v_2}}$}\end{picture}$

Here the first bus moves with $\sf{v_1+v_2}$ and the second bus is at rest, but $\sf{t}$ is the time taken by the first bus to meet the second, or to travel RQ.