Chapter 4.8, Problem 2OE

To determine

Tocalculate:Determine the time taken by gracias to overtake smiths in the given problem.

Answer to Problem 2OE

  $\text{9:40 A}\text{.M}$

Explanation of Solution

Given:

Speed of Smiths is $48\text{mi/h}$ and speed of Gracias is $60\text{mi/h}$ .

Smiths left the campground at 8:00A.M and Gracias left the campground at 8:20 A.M.

Calculation:

Since both left the camp and following the same route therefore it is a case of motion in same direction.

Using the property of quotient.

Let $t$ be the time at which Gracias left the camp then $t+\frac{1}{3}$ is the time at which Smiths left the camp.

Distance covered by both will be equal at the time of overtaking therefore calculate time equating distance as given below.

  $\begin{array}{cccc}& Speed\times & Time& =Dis\tan ce\\ Gracias& 60& t& 60t\\ Smiths& 48& t+\frac{1}{3}& 48\left(t+\frac{1}{3}\right)\end{array}$

  $\begin{array}{l}48\left(t+\frac{1}{3}\right)=60t\\ 48t+48\times \frac{1}{3}=60t\\ 16=12t\\ t=\frac{16}{12}\\ t=\frac{4}{3}\text{Hours}\text{or 80 min}\text{.}\end{array}$

Time at Gracias overtake Smiths is.

  $\begin{array}{l}=8:20+1\text{ Hr 20 min}\\ \text{=9:40 A}\text{.M}\end{array}$

Therefore, the simplified answer is $\text{9:40 A}\text{.M}$ .