# Trains a and b are traveling in the same direction on parallel tracks. Train a is traveling...

## Question:

Trains {eq}a {/eq} and {eq}b {/eq} are traveling in the same direction on parallel tracks. Train {eq}a {/eq} is traveling at 60 miles per hour and train {eq}b {/eq} is traveling at 80 miles per hour. Train {eq}a {/eq} passes a station at 9:10 am. If train {eq}b {/eq} passes the same station at 9:25 am, at what time will train {eq}b {/eq} catch up to train {eq}a {/eq}?

## Word Problems Travel and Distance:

A wide variety of math problems are often based on real-life examples, such as travel and distance traveled. Among the most common are those that concern speed, distance and time, since they are very important for the understanding of the properties of moving bodies. If we have two vehicles moving in the same direction with different speeds we can calculate the time it takes them to pass through a reference position. Time is associated with velocity and distance using the following equation: {eq}time = \frac{{distance}}{{speed}}{\text{.}} {/eq}

## Answer and Explanation: 1

{eq}\eqalign{ & {\text{In this case}}{\text{, we have the following::}} \cr & \,\,\,\,\,{\text{Speed of the train }}A:\,\,\,{v_{TA}} = 60\,mph\,\,\, \to {v_{TA}} = \frac{{60\,m}}{{60\,\min }} = 1\,\,{\text{miles}}/\min \cr & \,\,\,\,\,{\text{Speed of the train }}B:\,\,\,{v_{TB}} = 80\,mph \cr & {\text{So}}{\text{, the catch up speed:}} \cr & \,\,\,\,\,{v_{CU}}{\text{ = }}{v_{TB}} - {v_{TA}} = 80 - 60 = 20\,mph{\text{ }} \cr & {\text{In addition:}} \cr & {\text{The train }}A{\text{ passes through the station at 09}}:10\,am\,\,\, \to {t_A} = 9:10\,am \cr & {\text{The train }}B{\text{ passes through the station at }}09:25\,am\,\,\, \to {t_B} = 9:25\,am \cr & {\text{Then}}{\text{, the time difference is determined by:}} \cr & \,\,\,\,\,\Delta t = {t_B} - {t_A} = 9:25 - 9:10 = 0:15\,\,\,\, \Rightarrow \Delta t = 15\,\min \cr & {\text{Thus}}{\text{, the catch up distance:}} \cr & \,\,\,\,\,{d_{CU}} = {v_{TA}} \cdot \Delta t = 1 \times 15 = 15\,miles \cr & {\text{Since}}{\text{, the catch up time is given by:}} \cr & \,\,\,\,\,{t_{CU}} = \frac{{{d_{CU}}}}{{{v_{CU}}}} = \frac{{15\,miles}}{{20\,mph{\text{ }}}} = \frac{3}{4}\,hours = 45\,\min \cr & {\text{It will take }}\frac{3}{4}\,hours\left( {45\,\min } \right){\text{ to catch the train }}A.{\text{ Thus:}} \cr & \,\,\,\,\,9{\text{:25}}\,pm + 40\,\min = 10:05\,am \cr & {\text{Therefore}}{\text{, the train }}B{\text{ will catch up with the train }}A{\text{ at }}\boxed{{\text{10:05 am}}}. \cr} {/eq}