Question

From the same place at $7a.m.$ $A$ started walking in the north at the speed of $4km/hr$. After $1hr$, $B$ started cycling in the east at a speed of $8km/hr$. At what time they will be at a distance of $20km$ apart from each other?

Answer 1

Let $t=$ travel time of bicycle
then $(t+1)=$ Walking time
Distance $=$ Speed $\times$ Time
This is a Pythagoras problem
$a^2+b^2=c^2$
Where $a=8t$, bicycle distance
$b=4(t+1)$, Walking distance
$c=20$, distance apart in $t$ hrs from $8a.m$.
$\therefore (8t{)}^2+[4(t+1){]}^2=(20{)}^2$
$64t^2+16(t^2+1+2t)=400$
$64t^2+16t^2+32t+16=400$
$80t^2+32t-384=0$
$16(5t^2+2t-24)=0$
$5t^2+2t-24=0$
$5t^2+(12-10)t-24=0$
$5t^2+12t-10t-24=0$
$5t^2-10t+12t-24=0$
$5t(t-2)+12(t-2)=0$
$(t-2)(5t+12)=0$
$t=2$
and $t=-\frac{12}{15}$ which is not possible.
Positive solution $t=2hrs$.
At $10a.m.$ they will be $20km$ apart.