A cyclist gets left behind by $500$ meters every $minute$ by motorcyclist

Question

A cyclist gets left behind by $500$ meters every $minute$ by motorcyclist, because of that he takes $2$ $hour$ and $42$ $minute$ more than motorcyclist to cover $52$ $km$. Find both of their speed.

My approach: $v_2-v_1=30km/h$ (converted 500 meter per minute to km/h)

$v_2=52/t$
$v_1=52/(t+2.42)$

then I plug them in the first formula, seems wrong. How can I solve this?

Answer 1

$2$ hours and $42$ minutes is not $2.42$ hours. It is $2.7$ hours. The denominator in your last equation should be $t+2.7.$ Otherwise you are doing fine.

Answer 2

$500 \frac {meters}{minute} * \frac {km}{1000 meters}*\frac {60 minutes}{hr}= 30 \frac {km}{hr}$

$2 hr 42 minutes = 2\frac {42}{60} hr = 2\frac {7}{10} = 2.7 hr$.

$v_2 - v_1 = 30\frac {km}{hr}$

$v_2 = \frac {52}{t}$

$v_1 = \frac {52}{t + 2.7}$ so

$\frac {52}{t}-\frac {52}{t + 2.7} = 30$

$52(t+2.7) - 52t = 30t(t+2.7)$

$30t^2 + 81t - 140.4 = 0$.

$t = \frac {-81 \pm \sqrt{81^2 +4*30*140.4}}{2*30}$ (but we know it is positive)

$= \frac {-81 +\sqrt {6561 + 16848}}{60}=\frac {-81 + \sqrt{23409}}{60} =\frac {-81 + 153}{60} = {72}{60} =\frac {12}{10} = 1.2$ (which is an $1 hr 12 minutes$ BTW-- which is how long the car took.)

So $v_2 = \frac {52 km}{1.2 hr} = 43 \frac 13 \frac {km}{hr}$ and

An $v_1 = \frac {52 km}{1.2 + 2.7 hr} = \frac {52}{3.9}\frac {km}{hr} = 13\frac 13 \frac {km}{hr}$

(And that confirms that $v_2 - v_1 = 43\frac 13\frac {km}{hr} - 13\frac 13\frac {km}{hr} = 30\frac {km}{hr}$)