A boat travels 2.4 miles downstream and in the same amount of time, it travels 1.8 miles upstream. if the boat travels at an average speed of 21 mph in the still water, find the current of the river.

Answer 1

let c=rate of the current

distance=rate*time

d=r*t

2.4=(21+c)*t

1.8=(21-c)*t

2.4/(21+c)=1.8/(21-c)

cross multiply

2.4(21-c)=1.8(21+c)

50.4-2.4c=37.8+1.8c

50.4-37.8=2.4c+1.8c

12.6=4.2c

c=12.6/4.2

c=3 mph

Answer 2

Hi Pa C.,

The distance of the boat traveling down stream (lets call this D_dn) is equal to the average speed of the boat in still water (lets call this V_avg) multiplied by a time (t) plus the speed of the current of the river (lets call this V_cur) multiplied by a time (t), so D_dn = V_avg*(t) + V_cur*(t). The same can be said for the distance traveling up stream except the current is in the opposite (-) direction. D_up = V_avg*(t) - V_cur*(t). V_cur is what we are looking for.

D_dn = V_avg*(t) + V_cur*(t), becomes 2.4 mi = 21 mi/hr*(t) + V_cur*(t),

D_up = V_avg*(t) - V_cur*(t), becomes 1.8 mi = 21 mi/hr*(t) - V_cur*(t),

If we add those two equations we get:

4.2 mi = 42 mi/hr*(t)

t = .1 hr (6 minutes)

We can solve for V_cur by plugging (t) back into one of the equations:

2.4 mi = 21 mi/hr*(.1 hr) + V_cur*(.1 hr)

.3 mi = V_cur*(.1 hr)

3 mi/hr = V_cur, current of the river.

I hope this helps, Joe.