Can someone please explain to me on how to solve this Algebra 1 Motion Problem?

I am confused on these Motion Problems. Can someone explain to me how to do this one and explain each step! Thanks.



A speedboat traveled 2 hours and with a 4 km/h current before turning around. The return trip against the same current took 3 hours. Find the speed of the boat in still waters.Can someone please explain to me on how to solve this Algebra 1 Motion Problem?The total distance traveled is x kilometers. The boat travels the same distance but because of the current, it takes a longer time for the boat to get back to the starting position. Since the boat moves at the same speed, we can define two equations:



Forward: x km = (2 h)(4 km/h + y) = 8 + 2y

Reverse: x km = (3 h)(y - 4 km/h) = 3y - 12



Using simple algebra, we calculate:



8 km + 2y = 3y - 12 km

y = 20 km



Plug this back into the first two formulas:



8 + 2(20) = 8 + 40 = 48

3(20) - 12 = 60 - 12 = 48.



Therefore, if there was no current (or moving reference frame), then the boat would have a 20 km/h speed in still waters.Can someone please explain to me on how to solve this Algebra 1 Motion Problem?Let



Vs = speed of the speedboat in still water

Vc = speed of water current = 4 kph (given)



Working formula is



Distance = Velocity * Time



Outbound trip:



D = (Vs + Vc)2



D = (Vs + 4)2



Return trip:



D = (Vs - 4)3



Since



Outbound distance = Return trip distance



then



(V2 + 4)*2 = (Vs - 4)*3



2(Vs) + 8 = 3(Vs) - 12



3(Vs) - 2(Vs) = 12 + 8 = 20



Vs = 20 km/hr.



Hope this helps.Can someone please explain to me on how to solve this Algebra 1 Motion Problem?let speed boat speed in still water is x km/h



So it will travel 2 ( x+4) km in the direction of current in 2 hours.

which is equal to 3 ( x-4) km on return trip in 3 hours.

So 2x +8 = 3x -12

x=20 km /h Answer