Question 157346
Remember:
{{{Speed=distance/time}}} -------------> working eqn
{{{Alan=A=55m/hr}}}, speed of Alan
{{{Dave=D=65m/hr}}}, speed of Dave
We have to remember Alan has a head start of 1 hour & travelled a certain distance and lets' see how far he is:
{{{A[1]=(55m/hr)(1cross(hr))=55miles}}}
IN our working eqn we get: {{{time=distance/speed}}}
The {{{distance}}} will be the max range = 250 miles but we need to deduct {{{A[1]}}} because we'll take the {{{speed}}} for both.
In just make sense to deduct because that 55 miles run, Dave was not part of it.
{{{t=d/s}}}
{{{t=(250-55)cross(m)/((55+65)(cross(m)/hr))}}}
{{{t=(195/120)hr}}}
{{{t=1.625hr}}}= 1 hour & 37.5 minutes ------------------> This long they can talk on their car phones. FINAL ANSWER
In doubt? Go back working eqn we get, {{{d=(speed)(time)}}}
this time we have to add {{{A[1]}}} because the {{{time}}} we'll use we deduct it. We take speed for both and {{{d=max}}} range.
{{{d=(55+65)(m/cross(hr))(1.625cross(hr))+55m}}}
{{{250m=195m+55m}}}
{{{250m=250m}}}
Thank you.
Jojo