Question

Two cars leave simultaneously from points A and B, the distance between which is $280km$.

If the cars move to meet each other, they’ll meet in $2$ hours.

But if they move in the same direction, then the car going from point $A$ will catch up with the car going from point $B$ in $14$ hours.

What is the speed of each of the cars?

Answer 1

Explanation:

Given:

Two cars are left simultaneously from points A and B and the distance between them is $280km$

At certain point the cars move to meet each other in $2$ hours.

Also, When they move in the same direction, then the car going from point $A$ will catch up with the car going from point $B$ in $14$ hours.

To find:

The speed of each of the cars.

Explanation:

Speed is defined as, “the ratio of the distance traveled by the time taken”.

That is, $Speed=\frac{dis\tan cetraveled}{timetaken}$

Let $x$ be the speed of the car at point $A$, and $y$ be the speed of the car at point $B$.

Also $d_1$ be the distance of the car from point $A$ and $d_2$ be the distance of the car from point $B$.

From the car at point $A$ and the given condition, if the cars move to meet each other in $2$ hours, $t=2$.

The speed of the car from point A is,

$\begin{array}{rcl}x& =& \frac{d_1}{2}\\ d_1& =& 2x\end{array}$

Similarly, the speed of the car from point $A$ is,

$\begin{array}{rcl}y& =& \frac{d_2}{2}\\ d_2& =& 2y\end{array}$

If the distance between both cars is $280km$, then we get

$\begin{array}{rcl}{d}_1+d_2& =& 280\\ 2x+2y& =& 280\\ x+y& =& 140\end{array}$

Also, if they move in the same direction, then the car going from point $A$ will catch up with the car going from point $B$ in $14$ hours. we get

$\begin{array}{rcl}x& =& \frac{d_1}{14}\\ d_1& =& 14x\end{array}$

And,

$\begin{array}{rcl}y& =& \frac{d_2}{14}\\ d_2& =& 14y\end{array}$

So,

$\begin{array}{rcl}{d}_1& =& d_2+280\\ 14x& =& 14y+280\\ 14x-14y& =& 280\\ x-y& =& 20................\left(2\right)\end{array}$

Now add both sides of the equation (1) and equation (2).

$\begin{array}{rcl}x+y+x-y& =& 140+20\\ 2x-0y& =& 160\\ 2x& =& 160\\ x& =& \frac{160}{2}\\ x& =& 80\\ & & \end{array}$

Substitute the obtained value of $x$ in equation (2) to find the value of $y$.

$\begin{array}{rcl}80-y& =& 20\\ y& =& 80-20\\ y& =& 60\end{array}$

Therefore, the value of $x$ is $80$, and the value of $y$ is $60$.

Hence, the speed of the car from point A is $80\frac{km}{h}$ and the speed of the car from point B is $60\frac{km}{h}$