Chapter MPS, Problem 9.9MPS

To determine

To find: The time taken by two cyclists to meet at a point

Answer to Problem 9.9MPS

  $1\frac{1}{2}hrs.$

Explanation of Solution

Given:

Distance of the opposite end of course = 45 miles

Speed of one cyclist = 14mph

Speed of second cyclist = 16mph

Formula Used:

  $\text{Distance}=speed\times time$

Calculation:

Let AB the course of length is 45miles

  

One cyclist starts from A with speed 14 mph and other start from B with speed of 16mph and both meet at point C.

Let AC be $x$ miles then BC $=\left(45-x\right)$

Time taken by first cyclist to reach at C from A $=\frac{x}{14}h$

Time taken by second cyclist to reach at C from B $=\frac{\left(45-x\right)}{16}h$

Both cyclists will take same time to reach at C

So, $\begin{array}{l}\frac{x}{14}=\frac{45-x}{16}\\ \Rightarrow 16x=14\times 45-14x\\ \Rightarrow 30x=14\times 45\\ \Rightarrow x=\frac{14\times 45}{30}\\ \Rightarrow x=7\times 3\\ \Rightarrow x=21miles\end{array}$

Therefore, time taken by both cyclists to meet at point C $$

  $\begin{array}{l}=\frac{x}{14}h\\ \Rightarrow \frac{21}{14}=1\frac{1}{2}hr.\end{array}$

Conclusion:

Thus they will meet after $1\frac{1}{2}hrs.$