Question 178081
For the bike, {{{d[b] = r[b]*t[b]}}}
For the car, {{{d[c] = r[c]*t[c]}}}
When they meet, {{{d[b] = d[c]}}}, so I'll just call it {{{d}}}
given:
{{{r[b] = 18}}} mi/hr
{{{r[c] = 45}}} mi/hr
So far I have
For the bike, {{{d = 18*t[b]}}}
For the car, {{{d = 45*t[c]}}}
Since they're both equal to {{{d}}}, I can say
{{{18*t[b] = 45*t[c]}}}
Suppose I have a stopwatch and I start it when the car leaves
and I know that the bike left 1 hour ago
I can say {{{t[c] = t[b] - 1}}}
{{{18*t[b] = 45*(t[b] - 1)}}}
{{{18*t[b] = 45t[b] - 45}}}
{{{27t[b] = 45}}}
{{{t[b] = 5/3}}}hr , or in minutes
{{{t[b] = (5/3)*60}}}
{{{(5/3)*60 = 100}}} min
That's 1 hr and 40 min
Since the bike had a 1 hr headstart, the car took 40 min
to catch the bike
check answer:
For the bike, {{{d = 18*t[b]}}}
For the car, {{{d = 45*t[c]}}}
{{{d = 18*(5/3)}}}
{{{d = 30}}}
{{{d = 45*(2/3)}}}
{{{d = 15*2}}}
{{{d = 30}}}
The distance should be the same for both and it is