# Kinematics Problem on river

1. Jan 22, 2008

### dougr81

A boat takes 2.6 hours to travel 28.43 km down a river, then 5.5 hours to return. How fast is the river flowing?

So we want to find the velocity the river has.

I'm not sure I even started this right.
First I converted 2.6 hours to seconds = 9360 s
Then 5.5 hours to seconds = 19800 s
And 28.43 km to meters = 28430 m

I then used the linear uniform motion equation (xf = xi + vi*t). Am I right in doing this, there is no gravity, correct?

28430 = vi*9360s ---> vi = 3.037 m/s
28430 = vi*19800 ---> vi = 1.436 m/s

This doesn't seem right at all, I don't think I'm understanding the question completely.

2. Jan 22, 2008

### Leong

Is the answer 0.80 m/s @2.88 km/h?

3. Jan 22, 2008

### dougr81

4. Jan 22, 2008

### Leong

5. Jan 23, 2008

### physixguru

let x be the speed of boat and let y be the speed of river...

28.43/x+y=2.6 >>x+y=10.93 [downstream]

28.43/x-y=5.5>>x-y=5.18 [upstream]

>>2x=16.11

>>x=8.05 km/hr=2.236 m/s

>>y=2.88 km/hr=.8 m/s

verify and tell if the answer is correct or not..

explanation shall be done later....

6. Jan 23, 2008

### dougr81

physixguru, I've tried 0.8 m/s as the answer to the velocity of the river and it is incorrect.

7. Jan 23, 2008

### Karrava

Hi, im new here but i think ive got the answer to this, this is assuming the boat is travelling at a constant velocity.

So far the two values youve worked out are the same as mine(im not familiar with the equation (xf = xi + vi*t) i just used v=s/t )

What weve worked out is the velocity of the boat up and downstream, not the river itself.

v1 = 3.03 m/s
v2 = 1.43 m/s

So would the Resultant velocity of the River not just be this

vr = v1 - v2

vr = 3.03 - 1.46

vr= 1.6 m/s

Try looking at it like this perhaps

3.03 m/s
------------------------->
<---------
1.43m/s

(Could someone else verify this though? :P)

8. Jan 24, 2008

### physixguru

if the answer above is incorrect then...u shall have to apply vector algebra equations to solve the problem...directions... i mean....