a boat travel at a speed of 20 km/h in still water. the current in a river flows at 5 km/h so that downstream the boat can travel at 25 km/h and upstream it travels at only 15 km/h.
The boat has only enough fuel for 3 hours. The boat leaves it’s base and travels downstream. Draw a distance time graph and draw lines to indicate the outward and return journeys. After what time does the boat turn around so that it has enough fuel to return to base?
Just a hint: denote the time that the boat travels downstream with $t$. Distance travelled downstream is velocity 25 multiplied by time $t$. The boat has to travel the same distance upstream with speed 15 over time $(3-t)$. Now go ahead and create an equation.
EDIT: Ok, I see from your comment that you cannot proceed.
Both distances must be equal:
$$25t=15(3-t)$$
$$40t=45$$
$$t=\frac98\text{h}=1\text{h}\space 7\text{m}\space 30\text{s}$$
The distance traveled downstream is $25*9/8=28.125$km and the total distance is twice as big: 56.25km.
Now draw a distance time graph. You have three important points:
$O(t=0, s=0)$, this is the starting point.
$A(t=\frac98\text{h}, s=28.125\text{km})$, this is the turning point.
$B(t=3\text{h}, s=56.25\text{km})$, this is the final point.
Connect dots $O,A,B$ with two straight lines and you're done.