Answer

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Hint: First of all, consider speed of car A and car B as \[{{S}_{A}}\] and \[{{S}_{B}}\] respectively. When they are moving in same direction, use \[\left( {{S}_{A}}-{{S}_{B}} \right)=\dfrac{\text{Distance between them}}{\text{Time taken by them to meet}}\]. Solve these two equations to find \[{{S}_{A}}\] and \[{{S}_{B}}\].

Complete step-by-step answer:

We are given that place A and B are 80 km apart. Also A and B starts at the same time from positions A and B respectively. When they move in the same direction, they meet in 8 hours and when they move in the opposite direction that is towards each other they meet in 1 hour 20 minutes. We have to find the speeds of car A and B.

Let us diagrammatically consider the situation,

Case 1: When both cars are moving in the same direction.

Let us consider that these cars meet at point C.

Let \[BC={{d}_{1}}\]

Let the speed of car A be \[{{S}_{A}}\] and speed of car B be \[{{S}_{B}}\].

In this case, we are given that car A and car B meet after 8 hours. Since they meet at C, therefore this means that car A takes 8 hours to reach C and car B also takes 8 hours to reach C. Hence, we get,

Time taken by car A to reach car C = Time taken by car B to reach C = 8 hours.

As we know that time taken \[=\dfrac{\text{Distance travelled}}{\text{Speed}}\]

Therefore, we get time taken by car A to reach car C \[=\dfrac{\text{Distance travelled by car A}}{\text{Speed of car A}}\]

By putting the value of distance travelled = AB + BC = \[80+{{d}_{1}}\]

Speed of car A \[={{S}_{A}}\] and time taken = 8 hours, we get

\[\dfrac{80+{{d}_{1}}}{{{S}_{A}}}=8.....\left( i \right)\]

Also, we get time taken by car B to reach C \[=\dfrac{\text{Distance travelled by car B}}{\text{Speed of car B}}\]

By putting the value of distance travelled \[={{d}_{1}}\]

Speed of car \[={{S}_{B}}\] and time taken = 8 hours, we get,

\[\dfrac{{{d}_{1}}}{{{S}_{B}}}=8\]

By cross multiplying above equation, we get

\[{{d}_{1}}=8{{S}_{B}}\]

By putting the value of \[{{d}_{1}}\] in equation (i), we get,

\[\dfrac{80+8{{S}_{B}}}{{{S}_{A}}}=8\]

By cross multiplying the above equation, we get,

\[80+8{{S}_{B}}=8{{S}_{A}}\]

\[8{{S}_{A}}-8{{S}_{B}}=80\]

By taking out 8 common in above equation, we get,

\[{{S}_{A}}-{{S}_{B}}=10....\left( ii \right)\]

Case 2: When cars are moving towards each other or in opposite direction

Let us consider that these cars meet at C.

Let \[AC={{d}_{1}}\]

Since, AC + CB = 80 km

Therefore, \[CB=80-AC=\left( 80-{{d}_{1}} \right)km\]

In this case, we are given that car A and car B meet after 1 hour 20 minutes \[=\left( 1+\dfrac{20}{60} \right)\text{hours}=1+\dfrac{1}{3}=\dfrac{4}{3}\text{hours}\]

Since, they meet at C, therefore this means that car A takes \[\dfrac{4}{3}\] hours to reach C and car B also takes \[\dfrac{4}{3}\] hours to reach C. Hence, we get

Time taken by car A to reach C = Time taken by car B to reach C = \[\dfrac{4}{3}\] hours.

As, we know that time taken \[=\dfrac{\text{distance travelled}}{\text{speed}}\]

Therefore, we get time taken by car A to reach car C \[=\dfrac{\text{Distance travelled by car A}}{\text{Speed of car A}}\]

By putting the value of distance travelled \[=AC={{d}_{1}}\]

Speed of car A \[={{S}_{A}}\] and time taken \[=\dfrac{4}{3}\text{hours}\], we get

\[\dfrac{{{d}_{1}}}{{{S}_{A}}}=\dfrac{4}{3}\]

By cross multiplying above equation, we get

\[{{d}_{1}}=\dfrac{4{{S}_{A}}}{3}....\left( iii \right)\]

Also, we get the time taken by car B to reach C \[=\dfrac{\text{Distance travelled by car B}}{\text{Speed of car B}}\]

By putting the value of distance travelled \[=BC=80-{{d}_{1}}\]

Speed of car A \[={{S}_{B}}\] and time taken \[=\dfrac{4}{3}\text{hours}\], we get

\[\dfrac{80-{{d}_{1}}}{{{S}_{B}}}=\dfrac{4}{3}\]

By cross multiplying above equation, we get,

\[3\left( 80-{{d}_{1}} \right)=4{{S}_{B}}\]

By putting the value of \[{{d}_{1}}\] from equation (iii), we get

\[3\left( 80-\dfrac{4{{S}_{A}}}{3} \right)=4{{S}_{B}}\]

\[=3\times 80-4{{S}_{A}}-4{{S}_{B}}\]

Or, \[4\left( {{S}_{A}}+{{S}_{B}} \right)=240\]

By dividing by 4 on both sides, we get

\[\left( {{S}_{A}}+{{S}_{B}} \right)=60....\left( iv \right)\]

By adding equation (ii) and (iv), we get

\[\left( {{S}_{A}}+{{S}_{B}} \right)+\left( {{S}_{A}}-{{S}_{B}} \right)=60+10\]

\[\Rightarrow 2{{S}_{A}}=70\]

Therefore, we get \[{{S}_{A}}=\dfrac{70}{2}=35km/hr\]

By putting the value of \[{{S}_{A}}\] in equation (iv), we get

\[\begin{align}

& 35+{{S}_{B}}=60 \\

& \Rightarrow {{S}_{B}}=60-35=25km/hr \\

\end{align}\]

Therefore, we get the speed of car A as 35 km/hr and car B as 25 km/hr.

Note: Students must note that whenever two cars are moving in the same direction, they will never meet at a point between them. In the same way, if they are moving towards each other, they will always meet a point between them. Also, students can cross check their answer by putting \[{{S}_{A}}\] and \[{{S}_{B}}\] back into the equation and checking if they are satisfying the equations or not.

Here, students can also calculate \[{{S}_{A}}\] and \[{{S}_{B}}\] by using direct formula, that is \[{{S}_{A}}=\dfrac{\dfrac{D}{{{T}_{1}}}+\dfrac{D}{{{T}_{2}}}}{2}\] and \[{{S}_{B}}=\dfrac{\dfrac{D}{{{T}_{1}}}+\dfrac{D}{{{T}_{2}}}}{2}\] where D is the distance between A and B, \[{{T}_{1}}\] is the time taken by them to meet when they are moving towards each other and \[{{T}_{2}}\] is the time taken by them to meet when they are moving in the same direction.

Complete step-by-step answer:

We are given that place A and B are 80 km apart. Also A and B starts at the same time from positions A and B respectively. When they move in the same direction, they meet in 8 hours and when they move in the opposite direction that is towards each other they meet in 1 hour 20 minutes. We have to find the speeds of car A and B.

Let us diagrammatically consider the situation,

Case 1: When both cars are moving in the same direction.

Let us consider that these cars meet at point C.

Let \[BC={{d}_{1}}\]

Let the speed of car A be \[{{S}_{A}}\] and speed of car B be \[{{S}_{B}}\].

In this case, we are given that car A and car B meet after 8 hours. Since they meet at C, therefore this means that car A takes 8 hours to reach C and car B also takes 8 hours to reach C. Hence, we get,

Time taken by car A to reach car C = Time taken by car B to reach C = 8 hours.

As we know that time taken \[=\dfrac{\text{Distance travelled}}{\text{Speed}}\]

Therefore, we get time taken by car A to reach car C \[=\dfrac{\text{Distance travelled by car A}}{\text{Speed of car A}}\]

By putting the value of distance travelled = AB + BC = \[80+{{d}_{1}}\]

Speed of car A \[={{S}_{A}}\] and time taken = 8 hours, we get

\[\dfrac{80+{{d}_{1}}}{{{S}_{A}}}=8.....\left( i \right)\]

Also, we get time taken by car B to reach C \[=\dfrac{\text{Distance travelled by car B}}{\text{Speed of car B}}\]

By putting the value of distance travelled \[={{d}_{1}}\]

Speed of car \[={{S}_{B}}\] and time taken = 8 hours, we get,

\[\dfrac{{{d}_{1}}}{{{S}_{B}}}=8\]

By cross multiplying above equation, we get

\[{{d}_{1}}=8{{S}_{B}}\]

By putting the value of \[{{d}_{1}}\] in equation (i), we get,

\[\dfrac{80+8{{S}_{B}}}{{{S}_{A}}}=8\]

By cross multiplying the above equation, we get,

\[80+8{{S}_{B}}=8{{S}_{A}}\]

\[8{{S}_{A}}-8{{S}_{B}}=80\]

By taking out 8 common in above equation, we get,

\[{{S}_{A}}-{{S}_{B}}=10....\left( ii \right)\]

Case 2: When cars are moving towards each other or in opposite direction

Let us consider that these cars meet at C.

Let \[AC={{d}_{1}}\]

Since, AC + CB = 80 km

Therefore, \[CB=80-AC=\left( 80-{{d}_{1}} \right)km\]

In this case, we are given that car A and car B meet after 1 hour 20 minutes \[=\left( 1+\dfrac{20}{60} \right)\text{hours}=1+\dfrac{1}{3}=\dfrac{4}{3}\text{hours}\]

Since, they meet at C, therefore this means that car A takes \[\dfrac{4}{3}\] hours to reach C and car B also takes \[\dfrac{4}{3}\] hours to reach C. Hence, we get

Time taken by car A to reach C = Time taken by car B to reach C = \[\dfrac{4}{3}\] hours.

As, we know that time taken \[=\dfrac{\text{distance travelled}}{\text{speed}}\]

Therefore, we get time taken by car A to reach car C \[=\dfrac{\text{Distance travelled by car A}}{\text{Speed of car A}}\]

By putting the value of distance travelled \[=AC={{d}_{1}}\]

Speed of car A \[={{S}_{A}}\] and time taken \[=\dfrac{4}{3}\text{hours}\], we get

\[\dfrac{{{d}_{1}}}{{{S}_{A}}}=\dfrac{4}{3}\]

By cross multiplying above equation, we get

\[{{d}_{1}}=\dfrac{4{{S}_{A}}}{3}....\left( iii \right)\]

Also, we get the time taken by car B to reach C \[=\dfrac{\text{Distance travelled by car B}}{\text{Speed of car B}}\]

By putting the value of distance travelled \[=BC=80-{{d}_{1}}\]

Speed of car A \[={{S}_{B}}\] and time taken \[=\dfrac{4}{3}\text{hours}\], we get

\[\dfrac{80-{{d}_{1}}}{{{S}_{B}}}=\dfrac{4}{3}\]

By cross multiplying above equation, we get,

\[3\left( 80-{{d}_{1}} \right)=4{{S}_{B}}\]

By putting the value of \[{{d}_{1}}\] from equation (iii), we get

\[3\left( 80-\dfrac{4{{S}_{A}}}{3} \right)=4{{S}_{B}}\]

\[=3\times 80-4{{S}_{A}}-4{{S}_{B}}\]

Or, \[4\left( {{S}_{A}}+{{S}_{B}} \right)=240\]

By dividing by 4 on both sides, we get

\[\left( {{S}_{A}}+{{S}_{B}} \right)=60....\left( iv \right)\]

By adding equation (ii) and (iv), we get

\[\left( {{S}_{A}}+{{S}_{B}} \right)+\left( {{S}_{A}}-{{S}_{B}} \right)=60+10\]

\[\Rightarrow 2{{S}_{A}}=70\]

Therefore, we get \[{{S}_{A}}=\dfrac{70}{2}=35km/hr\]

By putting the value of \[{{S}_{A}}\] in equation (iv), we get

\[\begin{align}

& 35+{{S}_{B}}=60 \\

& \Rightarrow {{S}_{B}}=60-35=25km/hr \\

\end{align}\]

Therefore, we get the speed of car A as 35 km/hr and car B as 25 km/hr.

Note: Students must note that whenever two cars are moving in the same direction, they will never meet at a point between them. In the same way, if they are moving towards each other, they will always meet a point between them. Also, students can cross check their answer by putting \[{{S}_{A}}\] and \[{{S}_{B}}\] back into the equation and checking if they are satisfying the equations or not.

Here, students can also calculate \[{{S}_{A}}\] and \[{{S}_{B}}\] by using direct formula, that is \[{{S}_{A}}=\dfrac{\dfrac{D}{{{T}_{1}}}+\dfrac{D}{{{T}_{2}}}}{2}\] and \[{{S}_{B}}=\dfrac{\dfrac{D}{{{T}_{1}}}+\dfrac{D}{{{T}_{2}}}}{2}\] where D is the distance between A and B, \[{{T}_{1}}\] is the time taken by them to meet when they are moving towards each other and \[{{T}_{2}}\] is the time taken by them to meet when they are moving in the same direction.

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